Patent Publication Number: US-2010131215-A1

Title: Insulation monitoring system &amp; insulation detecting method for electric power supply system

Description:
TECHNICAL FIELD 
     The present invention relates to an insulation detecting apparatus and an insulation detecting method for a power line. More specifically, the present invention relates to an insulation detecting apparatus and method, in which an insulation state can be correctly detected by calculating insulation resistance and electrostatic capacitance of a power line including a load even when electrostatic capacitance between a three-phase power line including the load and the ground is in an unbalanced state, as well as in a balanced state. 
     According to the present invention, active component leakage current (or insulation resistance) between a power line including a load and the ground that is directly related to an insulation state or reactive component leakage current (or electrostatic capacitance) that is not directly related to the insulation state, but generated by ever-existing electrostatic capacitance, is calculated, displayed, and outputted together with an alarm. The insulation detecting apparatus can be remotely controlled through a communication unit. 
     BACKGROUND ART 
     In a grounding system where the secondary side of a transformer is grounded, a conventional method of monitoring the insulation state of a power line including a load uses a method of detecting a zero-phase leakage current to component flowing through a ground line  5  or detecting zero-phase leakage current Io flowing between the power line  3  with a load and the ground as shown in  FIG. 1 . In the present invention, the power line means a power distribution line including a load, a high voltage distribution line, and a low voltage distribution line, and a zero-phase current transformer means a current transformer that can detect a zero-phase leakage current component. A voltage detection line means a component that can detect a voltage component by directly connecting to a wire or detect a voltage component of the power line using a non-contact method. 
     Hereinafter, the convention technique will be described in more detail. 
     As shown in  FIG. 1 , commercial alternating voltage is supplied to a load  4  through a transformer  1  for transforming voltage, a switch  2 , and a power line  3 . 
     In  FIG. 1 , the secondary side of the transformer  1  is a Y-connection, and the neutral point of the Y-connection is connected to the ground  6  through a ground line  5 . Active component leakage current Ir and reactive component leakage current Ic flow between the power line  3  and the ground. The active component leakage current Ir is generated by an insulation resistance  9  component directly related to an insulation state, and the reactive component leakage current Ic is generated by an electrostatic capacitance  8  component that is not directly related to the insulation state, but generated if the power line  3  is long or a noise filter is installed at the input terminal of the load  4 . 
     Zero-phase leakage current Io=Ir+Ic, which is a vector sum of the two components, flows through the ground line  5  of the transformer. Generally, mainly used is a method of detecting an insulation state only with the zero-phase leakage current Io, which is leakage current detected at the middle of the ground line  5  connected to the secondary side of the transformer  1  or at the secondary side of the zero-phase current transformer  10  that flows three phases of the power line  3  all together. Electrical devices using the method of detecting zero-phase leakage current to described above include an electric leakage breaker, electric leakage alarm, electric leakage detector, ground fault detector, and the like. 
     In a conventional method of detecting an insulation state only with zero-phase leakage current Io shown in  FIG. 1 , if unbalance of electrostatic capacitance  8  between the power line  3 , or the load  4 , and the ground is large among three phases, reactive leakage current Ic generated by the electrostatic capacitance  8  component that is not directly related to the insulation state is increased. Therefore, the zero-phase leakage current to is detected high even at a power line where insulation resistance  9  is small, i.e., where the insulation state is favorable, and thus the insulation state is displayed as unfavorable, or although constant active component leakage current Ir generated by the insulation resistance  9  component flows, since the active component leakage current Ir is small due to a magnetic field characteristic of the zero-phase current transformer  10 , the detected zero-phase leakage current Io varies depending on the magnitude of the reactive component leakage current Ic generated by the electrostatic capacitance  8  component, and therefore, there is a problem in that the insulation state cannot be correctly detected. 
     Accordingly, required are an insulation detecting apparatus and an insulation detecting method that can correctly monitor an insulation state between the power line  3  including the load  4  and the ground even when electrostatic capacitance existing between the three-phase power line and the ground is in an unbalanced state, as well as in a balanced state. 
     DISCLOSURE OF INVENTION 
     Technical Problem 
     Accordingly, the present invention has been made in order to solve the above problems, and it is an object of the invention to provide an insulation detecting apparatus, in which an insulation state of a power line including a load can be correctly detected even when electrostatic capacitance between the three-phase power line including a load and the ground is in an unbalanced state, as well as in a balanced state. 
     Another object of the invention is to provide an insulation detecting apparatus, in which active component leakage current (or insulation resistance) between a power line and the ground that is directly related to an insulation state or reactive component leakage current (or electrostatic capacitance) that is not directly related to the insulation state, but generated by ever-existing electrostatic capacitance, can be calculated, displayed, and outputted together with an alarm, and which can be remotely controlled through a communication unit. 
     A further object of the invention is to provide an insulation detecting method, in which an insulation state of a power line including a load can be correctly detected even when electrostatic capacitance between the three-phase power line including a load and the ground is in an unbalanced state, as well as in a balanced state. 
     Technical Solution 
     In order to accomplish the above objects of the invention, according to one aspect of the invention, there is provided an insulation detecting apparatus for detecting an insulation state, comprising: a voltage detecting means for transforming a voltage component of each of three phases of the power line including a load into a voltage component of a certain magnitude and collectively extracting voltage of each of the three phases; a zero-phase current transformer for detecting zero-phase leakage current flowing between the power line and ground; a leakage current detecting means for converting a leakage current component detected by the zero-phase current transformer into a voltage component and extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band; a phase comparing means for detecting a phase difference between an output value of each of the three phases of the voltage detecting means and an output value of the leakage current detecting means; an analog-to-digital conversion unit for converting an analog component of the output value of the leakage current detecting means into a digital component; an operation controller for reading, outputting, operating, and controlling a variety of data; and an input-output unit for inputting and displaying a variety of data. 
     The leakage current detecting means comprises: the zero-phase current transformer for detecting leakage current between the power line and the ground; a current-to-voltage conversion unit for converting a leakage current component detected by the zero-phase current transformer into a voltage component; an amplification unit for amplifying the leakage current component converted by the current-to-voltage conversion unit; and a current filter unit for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the leakage current component amplified by the amplification unit  42 . 
     The voltage detecting means comprises: a voltage detecting unit for transforming a voltage component of each of three phases of the power line including a load into a voltage component of a certain magnitude and collectively detecting three phase voltage; and a voltage filter unit for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component transformed by the voltage detecting unit. 
     The phase comparing means comprises: a voltage component waveform shaping unit for shaping a waveform of the voltage component outputted from the voltage detecting means; a current component waveform shaping unit for shaping a waveform of the leakage current component outputted from the leakage current detecting means; and a phase difference detecting unit for detecting a phase difference of an output of the current component waveform shaping unit from an output of the voltage component waveform shaping unit. 
     The voltage detecting unit is configured with any one of a resistor, a condenser, and a transformer having the same impedance between each phase of the three-phase power line and the ground. 
     The input-output unit comprises an input unit for inputting a variety of data, a display unit for displaying and outputting a variety of data, and a memory unit for storing a variety of data. 
     According to another aspect of the invention, there is provided an insulation detecting apparatus for detecting an insulation state, comprising: a voltage detecting means for transforming a voltage component of the power line including a load into a voltage component of a certain magnitude and sequentially extracting a voltage component of each of the three phases, one phase at a time; a zero-phase current transformer for detecting zero-phase leakage current flowing between the power line and ground; a leakage current detecting means for converting a leakage current component detected by the zero-phase current transformer into a voltage component and extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band; a phase comparing means for detecting a phase difference between an output value of each of the three phases of the voltage detecting means and an output value of the leakage current detecting means; an analog-to-digital conversion unit for converting an analog component of the output value of the leakage current detecting means into a digital component; an operation controller for reading, outputting, operating, and controlling a variety of data; and an input-output unit for inputting and displaying a variety of data. 
     The voltage detecting means comprises: a voltage detecting unit for detecting a voltage component of each of three phases of the power line including a load and transforming the voltage component into a voltage component of a certain magnitude; a phase selection unit for selecting a voltage component of only one phase of three phases from the voltage component transformed by the voltage detecting unit; and a voltage filter unit for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component of the phase selected by the phase selection unit  32 . 
     The leakage current detecting means comprises: the zero-phase current transformer for detecting leakage current between the power line and the ground; a current-to-voltage conversion unit for converting a leakage current component detected by the zero-phase current transformer  10  into a voltage component; a current filter unit for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the leakage current component converted by the current-to-voltage conversion unit; and an amplification unit for amplifying the leakage current component extracted by the current filter unit. 
     Preferably, the insulation detecting apparatus of the present invention further comprises a communication unit for remotely monitoring the insulation detecting apparatus from outside. 
     According to another aspect of the invention, there is provided insulation detecting method for a power line, which can detect an insulation state of the power line even when electrostatic capacitance between the power line and ground is in an unbalanced state, as well as in a balanced state, comprising the steps of: allowing the leakage current detecting means to detect a leakage current component Io 1  from a zero-phase leakage current component detected at a secondary side of a zero-phase current transformer, allowing a voltage detecting means to detect a voltage component Vf by extracting only a frequency component, and detecting a phase difference of the leakage current component Io 1  from the voltage component Vf of each of the three phases outputted from the voltage detecting means; calculating an in-phase component and a 90° phase-shifted component of the leakage current component Io 1  of each phase; calculating a 90° phase-shifted zero value of each phase and/or an in-phase component zero value of each phase; verifying data on active component leakage current or reactive component leakage current of each phase calculated and stored in the memory unit in the step of calculating an in-phase component zero value of each phase and/or in the step of calculating a 90° phase-shifted component zero value of each phase; and externally displaying and/or outputting a combination of data recalculated in the step of verifying calculated data and the data used in the step of detecting Io 1 , Vf, and of each phase. 
     In a first insulation detecting method of the present invention, an active component leakage current generated by insulation resistors or reactive component leakage current generated by electrostatic capacitance is detected by calculating a reactive component zero leakage current, with which the reactive component leakage current generated by the electrostatic capacitance between the power line and the ground becomes zero in each of three phases. 
     In a second insulation detecting method of the present invention, an active component leakage current generated by insulation resistors or reactive component leakage current generated by electrostatic capacitance is detected by calculating an active component zero leakage current, with which the active component leakage current generated by the insulation resistors between the power line and the ground becomes zero in each of three phases. 
     In a third insulation detecting method of the present invention, an active component leakage current generated by insulation resistors or reactive component leakage current generated by electrostatic capacitance is detected by calculating an active component zero leakage current and a reactive component zero leakage current, with which the active component leakage current generated by the insulation resistors and the reactive component leakage current generated by the electrostatic capacitance between the power line and the ground become zero in each of three phases. 
     ADVANTAGEOUS EFFECTS 
     According to the present invention, active leakage current generated by insulation resistors or insulation resistance, which is an insulation state of a power line including a load, can be correctly detected even when electrostatic capacitance between the three-phase power line including a load and the ground is in an unbalanced state, as well as in a balanced state, and in addition, information on the most unfavorable phase can be informed. 
     In addition, an alarm state is displayed on a display unit by comparing current leakage with an alarm level additionally inputted through an input unit or stored in a memory unit, or a variety of data or alarm states detected at a remote site can be remotely monitored through a remote control apparatus installed in the remote site and a communication unit having a communication and control function. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         FIG. 1  shows a connection diagram describing a conventional method of monitoring leakage current. 
         FIG. 2  shows a connection diagram of an insulation detecting apparatus according to a first embodiment of the present invention. 
         FIG. 3  shows a connection diagram of an insulation detecting apparatus according to a second embodiment of the present invention. 
         FIG. 4  shows a connection diagram of an insulation detecting apparatus according to a third embodiment of the present invention. 
         FIG. 5  shows a connection diagram of an insulation detecting apparatus according to a fourth embodiment of the present invention. 
         FIG. 6  shows a connection diagram of an insulation detecting apparatus according to a fifth embodiment of the present invention. 
         FIG. 7  shows a connection diagram of an insulation detecting apparatus according to a sixth embodiment of the present invention. 
         FIG. 8  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a first embodiment of the insulation detecting apparatus. 
         FIG. 9  shows a detailed circuit diagram of the insulation detecting apparatus used in  FIG. 8 . 
         FIG. 10  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a second embodiment of the insulation detecting apparatus. 
         FIG. 11  shows a detailed circuit diagram of the insulation detecting apparatus used in  FIG. 10 . 
         FIG. 12  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 8 and 9  according to a first embodiment of the voltage detecting means. 
         FIG. 13  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 8 and 9  according to a second embodiment of the voltage detecting means. 
         FIG. 14  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a first embodiment of the voltage detecting means. 
         FIG. 15  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a second embodiment of the voltage detecting means. 
         FIG. 16  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a third embodiment of the voltage detecting means. 
         FIG. 17  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a fourth embodiment of the voltage detecting means. 
         FIG. 18  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a fifth embodiment of the voltage detecting means. 
         FIG. 19  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a sixth embodiment of the voltage detecting means. 
         FIG. 20  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a seventh embodiment of the voltage detecting means. 
         FIG. 21  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to an eighth embodiment of the voltage detecting means. 
         FIG. 22  shows a detailed circuit diagram of the leakage current detecting means used in  FIGS. 8 to 11  according to another embodiment of the leakage current detecting means. 
         FIG. 23  is a flowchart illustrating an insulation detecting method according to a first embodiment of the insulation detecting apparatus used in  FIGS. 8 to 11 . 
         FIG. 24  is a flowchart illustrating an insulation detecting method according to a second embodiment of the insulation detecting apparatus used in  FIGS. 8 to 11 . 
         FIG. 25  is a flowchart illustrating an insulation detecting method according to a third embodiment of the insulation detecting apparatus used in  FIGS. 8 to 11 . 
     
    
    
     MODE FOR THE INVENTION 
     Hereinafter, the preferred embodiments of the present invention will be described in detail with reference to the accompanying drawings. 
       FIG. 2  shows a connection diagram of an insulation detecting apparatus according to a first embodiment of the present invention,  FIG. 3  shows a connection diagram of an insulation detecting apparatus according to a second embodiment of the present invention,  FIG. 4  shows a connection diagram of an insulation detecting apparatus according to a third embodiment of the present invention,  FIG. 5  shows a connection diagram of an insulation detecting apparatus according to a fourth embodiment of the present invention,  FIG. 6  shows a connection diagram of an insulation detecting apparatus according to a fifth embodiment of the present invention, and  FIG. 7  shows a connection diagram of an insulation detecting apparatus according to a sixth embodiment of the present invention. 
       FIG. 8  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a first embodiment of the insulation detecting apparatus,  FIG. 9  shows a detailed circuit diagram of the insulation detecting apparatus used in  FIG. 8 ,  FIG. 10  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a second embodiment of the insulation detecting apparatus, and  FIG. 11  shows a detailed circuit diagram of the insulation detecting apparatus used in  FIG. 10 . 
       FIG. 12  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 8 and 9  according to a first embodiment of the voltage detecting means,  FIG. 13  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 8 and 9  according to a second embodiment of the voltage detecting means,  FIG. 14  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a first embodiment of the voltage detecting means,  FIG. 15  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a second embodiment of the voltage detecting means,  FIG. 16  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a third embodiment of the voltage detecting means,  FIG. 17  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a fourth embodiment of the voltage detecting means,  FIG. 18  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a fifth embodiment of the voltage detecting means,  FIG. 19  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a sixth embodiment of the voltage detecting means,  FIG. 20  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to a seventh embodiment of the voltage detecting means,  FIG. 21  shows a detailed circuit diagram of the voltage detecting means used in  FIGS. 10 and 11  according to an eighth embodiment of the voltage detecting means,  FIG. 22  shows a detailed circuit diagram of the leakage current detecting means used in  FIGS. 8 to 11  according to another embodiment of the leakage current detecting means, and  FIGS. 23 to 25  show flowcharts according to embodiments of the present invention used in  FIGS. 8 to 11 . 
       FIG. 2  shows an embodiment, in which the secondary side connection of the transformer  1  is a Y-connection, the neutral point is grounded, phase voltage to the ground is detected using voltage detection lines  12 ,  13 , and  14  for detecting a voltage component of the power line  3 , and a zero-phase current transformer  10  for detecting a zero-phase leakage current component of the power line  3  flowing to the ground is installed in the middle of the power line  3 .  FIG. 3  shows an embodiment almost similar to the embodiment shown in  FIG. 2 , in which the zero-phase current transformer  10  for detecting a zero-phase leakage current component of the power line  3  flowing to the ground is installed in the middle of the ground line  5  connected to the neutral point of the transformer  1 . 
       FIG. 4  shows an embodiment almost similar to the embodiment shown in  FIG. 2 , which describes that the leakage detecting apparatus can be used in a three-phase four-wire system where the neutral phase (N-phase) of the transformer is laid together.  FIG. 5  is an embodiment for describing that connection points of the voltage detection lines  12 ,  13 , and  14  also can be implemented at a position further closer to the load than the zero-phase current transformer  10 , in which although the secondary side connection of the transformer  1  is a delta connection, T-phase is grounded.  FIG. 6  is an embodiment of a non-grounding method, in which the secondary side connection of the transformer  1  is a delta connection, and  FIG. 7  is an embodiment almost similar to the embodiment shown in  FIG. 2 . Although  FIG. 2  shows a method of detecting phase voltage, line voltage is detected using the voltage detection lines  12  and  14  in the embodiment of  FIG. 7 . 
     Although only six embodiments are described in the present invention as shown in  FIGS. 2 to 7 , like the embodiments of the voltage detecting means  30  that will be described below, there can be a variety of embodiments, such as an embodiment where the voltage component of one phase among three phases is detected and the other two voltage components are obtained by shifting the phase of the detected voltage component by 120 degrees, an embodiment where the voltage component of two phases among three phases is detected and the other one voltage component is obtained by shifting the phase of either of the detected voltage components by 120 degrees (or −120 degrees), an embodiment where line voltage is detected instead of phase voltage, and an embodiment of a resistance grounded neutral system where resistors are installed between the neutral point of the transformer  1  and the ground  6  to limit the magnitude of ground fault current. 
       FIGS. 8 to 11  show detailed circuit diagrams of the embodiments of the insulation detecting apparatus  20  shown in  FIGS. 2 to 7 .  FIGS. 8 and 9  show an embodiment where a three phase voltage component between the three-phase power line  3  and the ground is inputted into a phase comparing means  50 .  FIGS. 10 and 11  show an embodiment where three phase voltage components between the three-phase power line  3  and the ground are sequentially inputted into the phase comparing means  50  one phase after another in response to an RST voltage control signal of the operation controller  70 . 
       FIGS. 12 and 13  show detailed circuit diagrams of the embodiment of the voltage detecting means  30  shown in  FIGS. 8 and 9 . If the voltage component of each of three phases of the power line  3  is inputted through the voltage detection lines  12 ,  13 , and  14 , voltage of each phase is split by resistors Rv 1  and Rv 2  as shown in  FIG. 12 .  FIG. 13  shows an embodiment, where the voltage component of the power line  3  is detected after the voltage is lowered to a certain level by a transformer TR, and a voltage filter unit  33  is configured to extract a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the detected voltage component. Although only two embodiments are shown above, there also can be an embodiment where condensers are used instead of the resistors as shown in  FIG. 16  that will be described below or an embodiment where a line voltage component is detected instead of the phase voltage component of the power line  3  described above. 
       FIGS. 14 to 21  show detailed circuit diagrams of the embodiments of the voltage detecting means  30  shown in  FIGS. 10 and 11 , where a variety of embodiments for detecting a three phase or single-phase voltage component is shown. However, a variety of embodiments can be considered, such as an embodiment where a line voltage component is detected, an embodiment where a phase is shifted by −120 degrees instead of using a 240° phase shift unit  312 , or the like. Describing  FIGS. 14 to 21  further more,  FIG. 14  shows an embodiment where if the voltage component of each of three phases of the power line  3  is inputted through the voltage detection lines  12 ,  13 , and  14 , voltage of each phase is split by resistors Rv 1  and Rv 2 , and the voltage component of the power line  3  is detected by a phase selection unit  32  configured like a switch sw 1  for selecting one phase at a time among three RST phases, in response to an RST voltage control signal outputted from the operation controller  70 . In this embodiment, a voltage filter unit  33  is configured to extract a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component of one of three phases detected by the phase selection unit  32 .  FIG. 15  shows an embodiment where voltage is split by a transformer TR instead of the resistors used in the embodiment of  FIG. 14 .  FIG. 16  shows an embodiment where voltage is split by condensers Cv 1  and Cv 2 , not by the resistors used in the embodiment of  FIG. 14 .  FIG. 17  shows an embodiment where resistors Rv installed between the secondary side of the transformer TR and the ground are added to the embodiment of  FIG. 15 , and there can be an embodiment where condensers are used instead of the resistors Rv in  FIG. 17 . 
       FIG. 18  shows an embodiment, where if the voltage component of each of three phases of the power line  3  is inputted through the voltage detection lines  12 ,  13 , and  14 , the voltage is lowered to a certain level by a transformer TR, and a voltage filter unit  33  is coupled to extract a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component that is split again by an upper resistor Rv 1  and a resistor Rv 2  connected to the ground. The upper resistor Rv 1  is selected by the phase selection unit  32  for selecting one phase at a time among three RST phases in response to an RST voltage control signal outputted from the operation controller  70 . 
       FIG. 19  shows an embodiment, where the transformer TR of the embodiment of  FIG. 18  is not used, and the resistors Rv 1  used in this embodiment preferably have resistance higher than those of the resistors Rv 1  shown in  FIGS. 12 to 18 .  FIGS. 20 and 21  show embodiments, where in order to detect the voltage component of the power line  3 , only voltage component of one phase is detected, and voltage components of the other two phases are detected by shifting the phase of the voltage component detected by the voltage detecting unit  31  by 120 degrees and 240 degrees. In  FIG. 20 , in order to detect the voltage component of the power line  3 , the voltage component of R-phase, for example, inputted through one voltage detection line  12  is split by two resistors Rv 1  and Rv 2  connected between the voltage detection line  12  and the ground. The split voltage component of R-phase is connected to a of the phase selection unit  32 , and voltage components of the other S-phase and T-phase respectively use a 120° phase shift unit  311  for shifting the voltage component of R-phase by 120 degrees and a 240° phase shift unit  312  for shifting the voltage component of R-phase by 240 degrees. The voltage component having a phase difference of 120 degrees from the voltage component of R-phase, which is outputted from the 120° phase shift unit  311 , is connected to b of the phase selection unit  32 . The voltage component having a phase difference of 240 degrees from the voltage component of R-phase, which is outputted from the 240° phase shift unit  312 , is connected to c of the phase selection unit  32 . The voltage component of the power line  3  is detected by the phase selection unit  32  configured like a switch sw 1  for selecting one phase at a time among three RST phases in response to an RST voltage control signal outputted from the operation controller  70 , and a voltage filter unit  33  is configured to extract a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component of one of three phases detected by the phase selection unit  32 . The resistors Rv 1  and Rv 2  preferably have high resistance. 
       FIG. 21  shows an embodiment that is different from the embodiment of  FIG. 20  only in that condensers Cv 1  and Cv 2  are used instead of the resistors Rv 1  and Rv 2 , where the condensers preferably have small capacitance. Although it is not shown in the figure of the embodiment, and the 120° phase shift unit  311  and the 240° phase shift unit  312  are used in  FIGS. 20 and 21 , there can be an embodiment where a −120° (minus)120° phase shift unit for shifting a phase by −120° is used instead of the 240° phase shift unit  312 . Those skilled in the art will readily understood that there also can be an embodiment, where the voltage component of the power line  3  is detected by detecting two phase voltages or a line voltage using two voltage detection lines. 
       FIG. 22  shows a detailed circuit diagram according to another embodiment of the leakage current detecting means  40  of  FIGS. 8 to 11 , in which the leakage current detecting means  40  is configured in a sequence of a current-to-voltage conversion unit  41 , an amplification unit  42 , and a current filter unit  43  as shown in  FIGS. 9 and 11 . Contrarily, although  FIG. 22  shows an embodiment where the leakage current detecting means  40  is configured in a different sequence of a current-to-voltage conversion unit  41 , a current filter unit  43 , and an amplification unit  42 , there can be an embodiment where functions of the current-to-voltage conversion unit  41  and the amplification unit  42  are implemented as a single function. A variety of embodiments can be considered, such as an embodiment where the current-to-voltage conversion unit  41  is not implemented within the leakage current detecting means  40 , but is embedded in the secondary winding of the zero-phase current transformer  10 . 
       FIGS. 23 to 25  are flowcharts illustrating the operation of the insulation detecting apparatus  20  of the present invention shown in  FIGS. 8 to 11  or illustrating the insulation detecting method of the present invention. 
       FIGS. 2 ,  8 ,  9 , and  23  of the present invention will be described first. Referring to  FIG. 2 , the transformer  1  for transforming voltage provides power to the power line  3  through the switch  2 . Reference symbol  5  denotes a ground line for connecting the neutral point of the transformer to the ground  6  for safety. On the other hand, while the power is provided to the load  4  through the switch  2  and the power line  3 , active component leakage current Ir of each of three phases flows between the power line  3  including a load  4  and the ground through an insulation resistor  9  that is directly related to insulation aging. 
     The active component leakage current Ir is Ir=Irr+Irs+Irt that is a vector sum of Irr flowing through the insulation resistor Rr between R-phase and the ground, Irs flowing through the insulation resistor Rs between S-phase and the ground, and Irt flowing through the insulation resistor Rt between T-phase and the ground. In addition, between the power line  3  including the load  4  and the ground, reactive component leakage current Ic of each of three phases flows to the ground through electrostatic capacitance  8  that is not directly related to the insulation state, but generated if the power line  3  is long or an equipment such as a noise filter for reducing noises is installed at the input terminal of the load  4 . The reactive component leakage current Ic is Ic=Icr+Ics+Ict that is a vector sum of Icr flowing through the electrostatic capacitor Cr between R-phase and the ground, Ics flowing through the electrostatic capacitor Cs between S-phase and the ground, and Ict flowing through the electrostatic capacitor Ct between T-phase and the ground. Accordingly, the zero-phase leakage current Io, which is leakage current flowing between the power line  3  and the ground, is represented as a vector sum of the active component leakage current Ir=Irr+Irs+Irt and the reactive component leakage current Ic=Icr+Ics+Ict. 
     If the zero-phase leakage current to component and the voltage component of each phase between the power line  3  and the ground are known, the active component leakage current Ir related to an insulation state of the power line  3  and the reactive component leakage current Ic that is not directly related to an insulation state, but flows between the power line  3  and the ground can be calculated. 
     The zero-phase current transformer (ZCT)  10  is used to detect the zero-phase leakage current Io component, and the voltage detection lines  12 ,  13 , and  14  are used to detect the voltage component of each of three phases between the power line  3  and the ground. The voltage detection lines  12 ,  13 , and  14  and the secondary side of the zero-phase current transformer  10  are connected to the insulation detecting apparatus  20  shown in  FIG. 8 . 
     First Embodiment 
       FIG. 8  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a first embodiment of the insulation detecting apparatus. 
     The insulation detecting apparatus  20  of the present invention comprises a voltage detecting means  30  for detecting a voltage component between a power line  3  and the ground, transforming the detected voltage component into voltage of a certain magnitude, and extracting a frequency component lower than a certain frequency or a frequency component of a certain band; a leakage current detecting means  40  for converting zero-phase leakage current Io component, which is detected at the secondary side of the zero-phase current transformer  10  for detecting zero-phase leakage current Io between the power line  3  including the load  4  and the ground, into a voltage component, amplifying the converted voltage component, and extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band; a phase comparing means  50  for comparing each of three phases of an output value of the voltage detecting means  30  with a phase of an output value of the leakage current detecting means  40 ; an analog-to-digital conversion unit  60  for converting an analog component of the output value of the leakage current detecting means  40  into a digital component; an operation controller  70  having an operation and control function for reading and outputting a variety of data; an input-output means  80  for inputting and displaying a variety of data; and a communication unit  90  for remotely controlling the insulation detecting apparatus from outside. The input-output means  80  includes an input unit  82 , a display unit  84 , and a memory unit  86 . 
     Referring to  FIG. 9 , the voltage detecting means  30  for detecting a three phase voltage component of the power line  3  including a load comprises a voltage detecting unit  31  for transforming the three phase voltage component detected by the voltage detection lines  12 ,  13 , and  14  into voltage of a certain magnitude, and a voltage filter unit  33  for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the three phase voltage component transformed by the voltage detecting unit  31 . The leakage current detecting means  40  comprises a current-to-voltage conversion unit  41  for converting leakage current component, which is detected at the secondary side of the zero-phase current transformer  10  for detecting the zero-phase leakage current Io between the power line  3  including the load  4  and the ground, into a voltage component, an amplification unit  42  for amplifying the leakage current component Ia converted by the current-to-voltage conversion unit  41 , and a current filter unit  43  for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from a leakage current component corresponding to the zero-phase leakage current Io component amplified by the amplification unit  42 . The phase comparing means  50  includes a voltage component waveform shaping unit  51  for shaping the waveform of the voltage component of each of three phases outputted from the voltage detecting means  30 , a current component waveform shaping unit  52  for shaping the waveform of the leakage current component Io 1  corresponding to the zero-phase leakage current Io component outputted from the leakage current detecting means  40 , and a phase difference detecting unit  53  for detecting a phase difference between an output of the voltage component waveform shaping unit  51  and an output of the current component waveform shaping unit  52 . Only one output value of the leakage current detecting means  40  is inputted into the analog-to-digital conversion unit  60  in  FIG. 8 . However, two analog components, i.e., the output value of the leakage current detecting means  40  and the output value of the voltage detecting means  30  for additionally detecting the magnitude of the voltage component, are inputted into the analog-to-digital conversion unit  60  in  FIG. 9 . 
     Accordingly, if the number of components inputted into the analog-to-digital conversion unit  60  is two as described above, a voltage component value of the power line  3  is read to calculate both leakage current and insulation resistance, and if one component is inputted, only the leakage current is calculated and the insulation resistance is not calculated, choice of which will vary depending on embodiments. In the present invention, an embodiment where even the insulation resistance is calculated will be described to express a variety values needed for monitoring an insulation state. 
     Hereinafter, the present invention is described in more detail referring to  FIG. 9  showing a first embodiment of the insulation detecting apparatus  20  and  FIG. 23  showing a flowchart of the insulation detecting apparatus  20 . 
     A first insulation detecting method of the present invention shown in  FIG. 23 , capable of detecting an insulation state of a power line even when electrostatic capacitance between the power line and the ground is in an unbalanced state, as well as in a balanced state, comprises the steps of: allowing the leakage current detecting means  40  to detect a leakage current component Io 1  from a zero-phase current component detected at the secondary side of the zero-phase current transformer, allowing the voltage detecting means to detect a voltage component Vf of each of three phases by extracting only a frequency component, and detecting a phase difference of the leakage current component Io 1  from the voltage component Vf of each of three phases outputted from the voltage detecting means  30 ; calculating an in-phase component and a 90° phase-shifted component of the leakage current component Io 1  of each phase; calculating a 90° phase-shifted component zero value of each phase; verifying data on active component leakage current or reactive component leakage current of each phase calculated and stored in the memory unit in the step of calculating a 90° phase-shifted component zero value of each phase; and externally displaying and/or outputting a combination of data recalculated in the step of verifying calculated data and the data used in the step of detecting Io 1 , Vf, and of each phase. 
     As shown in  FIGS. 9 and 23 , on the main flow stored in the memory unit  86  of the insulation detecting apparatus  20 , the input unit  82  sets a variety of data used in the insulation detecting apparatus  20  using a constitutional component such as a keypad or a switch  100 , in which the input unit has a function for setting a variety of data, e.g., a number address, an alarm setting value, and the like of each insulation detecting apparatus  20  if a plurality of insulation detecting apparatuses  20  is installed. Next, a read operation is performed on a variety of data set by the input unit  82 , previously stored in the memory unit  86 , or inputted from a remote external site through the communication unit  90   110 . 
     If the step of detecting Io 1 , Vf, and of each phase  120  is performed, the zero-phase leakage current component Io detected at the secondary side of the zero-phase current transformer  10  shown in  FIGS. 8 and 9  is converted into a voltage component by the current-to-voltage conversion unit  41  that converts current to voltage and amplified by the amplification unit  42 . Then, a current component Io 1  corresponding to the zero-phase leakage current, from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the current filter unit  43 , is outputted to the analog-to-digital conversion unit  60  and the phase comparing means  50 . The component Io 1  value corresponding to the zero-phase leakage current inputted into the analog-to-digital conversion unit  32  ( 60 ) is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . 
     Then, as described above, the voltage detecting unit  31  of  FIG. 16  used in the embodiments shown in  FIGS. 12 to 16  or another embodiment splits the voltage component of each of three phases between the power line  3  and the ground inputted through the voltage detection lines  12 ,  13 , and  14  into voltages that can be used in the insulation detecting apparatus  20 , using resistors, condensers, or a transformer. The voltage component Vf of each of three phases (Vf_r of R-phase, Vf_s of S-phase, or Vf_t of T-phase), i.e., a split voltage from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the voltage filter unit  33 , is outputted to the phase comparing means  50  and the analog-to-digital conversion unit  60 . The voltage component Vf of each of three phases to the ground inputted into the analog-to-digital conversion unit  60  is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . 
     Using the values of three voltage components Vf_r, Vf_s, and Vf_t to the ground of three phases inputted into the phase comparing means  50 , whose waveforms are shaped by the voltage component waveform shaping unit  51 , and the value of one leakage current component Io 1  corresponding to the zero-phase leakage current component outputted from the leakage current detecting means  40 , whose waveform is shaped by the current component waveform shaping unit  52 , the phase difference detecting unit  53  detects three phase differences between the three voltage components of three phases outputted from the voltage component waveform shaping unit  51  and the one leakage current component outputted from the current component waveform shaping unit  52 , i.e., a phase difference r of the leakage current component Io 1  for the voltage component of R-phase Vf_r, a phase difference s of the leakage current component Io 1  for the voltage component of S-phase Vf_s, and a phase difference t of the leakage current component Io 1  for the voltage component of R-phase Vf_t, and the operation controller  70  reads and stores the three phase differences into the memory unit  86 . 
     An example of calculating the aforementioned Vf, Io 1 , and is described. For the convenience of explanation, it is assumed that an amplification-related coefficient of the leakage current detecting means  40  including the zero-phase current transformer  10  is one, and an amplification-related coefficient of the voltage detecting means  30  is 0.001 (i.e. 1/1000). 
     The voltage between the three-phase power line  3  and the ground is 220V, and the frequency is 60 Hz. The leakage currents flowing through the insulation resistors between the three-phase power line and the ground are respectively Irr=1 mA for R-phase, Irs=40 mA for S-phase, and Irt=1 mA for T-phase. The leakage currents flowing through the electrostatic capacitors between the three-phase power line and the ground are respectively Icr=60 mA for R-phase, Ics=20 mA for S-phase, and Ict=20 mA for T-phase. 
     The values detected and stored in the memory unit  86  in the step of detecting the voltage component Vf, leakage current component Io 1 , and phase difference  120  are such that Io 1  is 76.3 mA, Vf_r, Vf_s, and Vf_t are 220 mV respectively, r is 104.8, s is −15.2, and t is −135.2. 
     Next, if the step of calculating an in-phase and a 90° phase-shifted components of the leakage current component Io 1  of each phase  130  is performed, the leakage current component Io 1  and the phase differences r, s, and t detected and stored in the memory unit  86  in the step of detecting the voltage component Vf, leakage current component Io 1 , and phase difference  120  are read. For each of three phases, an in-phase component cos value and a 90° phase-shifted component sin value for the voltage of the leakage current Io 1  corresponding to the zero-phase leakage current are calculated and stored in the memory unit  86 . Describing it more specifically, the in-phase component leakage current of R-phase Io 1   rr  is Io 1 ×cos r, the 90° phase-shifted component leakage current of R-phase Io 1   cr  is Io 1 ×sin r, the in-phase component leakage current of S-phase Io 1   rs  is Io 1 ×cos s, the 90° phase-shifted component leakage current of S-phase Io 1   cs  is Io 1 ×sin s, the in-phase component leakage current of T-phase Io 1   rt  is Io 1 ×cos t, and the 90° phase-shifted component leakage current of T-phase Io 1   ct  is Io 1 ×sin t. 
     Substituting the values described in the above example and approximately calculating zero-phase leakage currents, Io 1   rr =−19.5 mA, Io 1   cr= 73.8 mA, Io 1   rs= 73.6 mA, Io 1   cs =−20.0 mA, Io 1   rt =−54.1 mA, and Io 1   ct =−53.8 mA, and therefore, the zero-phase leakage current Io for the R-phase voltage is −19.5+j73.8 mA, the zero-phase leakage current to for the S-phase voltage is 73.6−j20 mA, and the zero-phase leakage current to for the T-phase voltage is −54.1−j53.8 mA. 
     Next, before describing the step of calculating a 90° phase-shifted component zero value of each phase  140 , for the values described above, a functional relation between an in-phase component leakage current value and a 90° phase-shifted component leakage current value for voltage of each of three phases will be described first. The zero-phase leakage current Io component flowing between the power line  3  including the load  4  and the ground can be expressed as shown in Equation 1. Next, the zero-phase leakage current for a three phase voltage component shown in Equation 1 is converted into a voltage component value of R-phase, and a value corresponding to the zero-phase leakage current component of a voltage component that is in-phase with the R-phase voltage, i.e., Io 1   rr , is expressed as shown in Equation 2, and a value corresponding to the zero-phase leakage current component of the voltage component that is 90° phase-shifted from the R-phase voltage, i.e., Io 1   cr , is expressed as shown in Equation 3. 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             Io 
                             ° 
                           
                           = 
                             
                            
                           
                             
                               
                                 V 
                                 
                                   R 
                                   ° 
                                 
                               
                               Rr 
                             
                             + 
                             
                               
                                 V 
                                 
                                   S 
                                   ° 
                                 
                               
                               Rs 
                             
                             + 
                             
                               
                                 V 
                                 
                                   T 
                                   ° 
                                 
                               
                               Rt 
                             
                             + 
                           
                         
                       
                     
                     
                       
                         
                             
                            
                           
                             j 
                              
                             
                               ( 
                               
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Cr 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       R 
                                       ° 
                                     
                                   
                                 
                                 + 
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Cs 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       S 
                                       ° 
                                     
                                   
                                 
                                 + 
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Ct 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       T 
                                       ° 
                                     
                                   
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     
                       
                         
                           = 
                             
                            
                           
                             
                               Irr 
                               ° 
                             
                             + 
                             
                               Ics 
                               ° 
                             
                             + 
                             
                               Ict 
                               ° 
                             
                             + 
                             
                               j 
                                
                               
                                 ( 
                                 
                                   
                                     Icr 
                                     ° 
                                   
                                   + 
                                   
                                     Ics 
                                     ° 
                                   
                                   + 
                                   
                                     Ict 
                                     ° 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Here 
                     , 
                     
                       V 
                        
                       
                           
                       
                        
                       and 
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       I 
                        
                       
                           
                       
                        
                       are 
                        
                       
                           
                       
                        
                       vector 
                        
                       
                           
                       
                        
                       
                         functions 
                         . 
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     1 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     Irr 
                     - 
                     
                       Irs 
                       2 
                     
                     - 
                     
                       Irt 
                       2 
                     
                     - 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Ics 
                     
                     + 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Ict 
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Here 
                     , 
                     
                       I 
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       a 
                        
                       
                         
                             
                         
                          
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       real 
                        
                       
                           
                       
                        
                       
                         value 
                         . 
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     2 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     Icr 
                     - 
                     
                       Ics 
                       2 
                     
                     - 
                     
                       Ict 
                       2 
                     
                     + 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Irs 
                     
                     - 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Irt 
                     
                   
                    
                   
                     
 
                   
                    
                   Here 
                   , 
                   
                     I 
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     a 
                      
                     
                         
                     
                      
                     real 
                      
                     
                         
                     
                      
                     
                       value 
                       . 
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     3 
                   
                   &gt; 
                 
               
             
           
         
       
     
     The relational expression between the in-phase component leakage current and the 90° phase-shifted component leakage currents of S-phase and T-phase is that they respectively have 120° and −120° phase differences from the in-phase component leakage current and the 90° phase-shifted component leakage current of R-phase. 
     As shown in Equations 2 and 3, it is understood that the in-phase component leakage current of R-phase includes not only active component leakage current Irr flowing due to the insulation resistance of R-phase, but also active component leakage currents Irs and Irt flowing due to the insulation resistances of S-phase and T-phase and reactive component leakage currents Ics and Ict flowing due to the electrostatic capacitances of S-phase and T-phase. The 90° phase-shifted component leakage current of R-phase includes not only reactive component leakage current Irr flowing due to the electrostatic capacitance of R-phase, but also reactive component leakage currents Ics and Ict flowing due to the electrostatic capacitances of S-phase and T-phase and active component leakage currents Irs and Irt flowing due to the insulation resistances of S-phase and T-phase. Then, it can be conjectured from Equations 2 and 3 that an insulation state cannot be correctly obtained by a conventional method of detecting zero-phase leakage current Io generated by a zero-phase current transformer. 
     Three methods of calculating an in-phase component zero value of each phase will be described. 
     That is, a first method is for calculating a reactive component zero leakage current value with which the reactive component leakage current Ic becomes zero, a second method is for calculating an active component zero leakage current value with which the active component leakage current Ir becomes zero, and a third method is for calculating an active component zero leakage current value and a reactive component zero leakage current value with which the reactive component leakage current Ic and the active component leakage current Ir become zero. 
     1) A Case of Calculating a Reactive Component Zero Leakage Current Value with which the Reactive Component Leakage Current Ic Becomes Zero 
     First, the step of calculating a 90° phase-shifted component zero value of each phase  140  is described (refer to  FIG. 23 ). 
     A reactive component leakage current value generated by electrostatic capacitance with which a reactive component leakage current becomes zero is calculated for each of three phases. This value is calculated to understand that active component leakage current of which component flows at the secondary side of the zero-phase current converter  10  if the reactive component leakage current Ic generated by the electrostatic capacitance becomes zero. Describing it easily, it is to make reactive component leakage currents generated by electrostatic capacitances be balanced in all three phases. However, a predetermined value that is slightly larger than the zero value can be selected. A reactive component leakage current value of R-phase is calculated first to find out the phase and magnitude of reactive component zero leakage current Ic′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   cr  zero, i.e., the value of Equation 3 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Icr′, Ics′, and Ict′ values are calculated so that the 90° phase-shifted reactive component leakage current of R-phase becomes zero, i.e., the value of Equation 3 becomes zero. If the Icr′, Ics′, and Ict′ values are put into Equation 2 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 4. If the Icr′, Ics′, and Ict′ values are put into Equation 3 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 5. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Icr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Irr 
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       - 
                       19.5 
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ics 
                             ′ 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ict 
                             ′ 
                           
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     4 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lcr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Icr 
                           + 
                           
                             Icr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     
                       73.8 
                       + 
                       
                         Icr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ics 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ict 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     5 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 5 becomes zero when Icr′=−73.8, Ics′=147.6, and Ict′=147.6, if the values are put into Equation 4 respectively, the zero-phase leakage currents become −19.5+j0, −147.3+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The reactive component leakage current value of S-phase becomes zero when Icr′=−40.0, Ics′=20.0, and Ic=−40.0, and if the values are put as described above, the zero-phase leakage currents become 39.9+j0, 73.6+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of S-phase are stored in the memory unit  86 . 
     The reactive component leakage current value of T-phase becomes zero when Icr′=−107.5, Ics′=−107.5, and Ict′=53.8, and if the values are put as described above, the zero-phase leakage currents become 39.0+j0, −147.3+j0, and −54.1+j0, and the Icr′, Ics′, and Ict′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of verifying the calculated data  160  is described. 
     Reactive component zero leakage current Ic′ values calculated and stored in the memory unit  86  in the step of calculating a 90° phase-shifted component zero value of each phase  140 , with which the reactive component leakage current of each phase becomes zero, are read. Then, using the zero-phase leakage current for R-phase voltage Io=19.5+j73.8 mA, the zero-phase leakage current for S-phase voltage Io=73.6−j20.0 mA, and the zero-phase leakage current for T-phase voltage Io=−54.1−j53.8 mA, which are calculated in the step of calculating an in-phase and a phase-shifted components of the leakage current component Io 1  of each phase  130 , it is examined that an in-phase component and a 90° phase-shifted component of which phase is large or small. Examining a large value first, since the largest in-phase component is +73.6 of S-phase, if the in-phase component of S-phase is added + (plus), the active component leakage current of S-phase Irs or the reactive component leakage current of R-phase Icr is larger than the active component leakage current of T-phase Irt, the active component leakage current of R-phase Irr, or the reactive component leakage current of T-phase Ict. Since the largest 90° phase-shifted component is +73.8 of R-phase, if the 90° phase-shifted component of R-phase is added + (plus), the reactive component leakage current of R-phase Icr or the active component leakage current of S-phase Irs is larger than the reactive component leakage current of S-phase Ics, the reactive component leakage current of T-phase Ict, or the active component leakage current of T-phase lit. Therefore, this case satisfies two conditions, and thus, although a small value is examined, among the zero-phase leakage current components flowing between the power line  3  and the ground, active component leakage current generated by insulation resistors is largest in S-phase, and reactive component leakage current generated by electrostatic capacitance is largest in R-phase. Accordingly, in the cases where reactive component leakage current becomes zero, a condition corresponding to R-phase where in-phase component leakage current of S-phase is plus + is Icr′=40 with which the reactive component leakage current of S-phase becomes zero. At this point, the in-phase active component leakage current of S-phase is +39. Accordingly, it is understood that actual reactive component leakage current of R-phase flowing between the power line  3  and the ground due to the electrostatic capacitance is larger than those of the other phases by +40 mA, and actual active component leakage current of S-phase flowing between the power line  3  and the ground due to the insulation resistors is larger than those of the other phases by +39 mA. 
     2) A Case of Calculating an Active Component Zero Leakage Current Value with which the Active Component Leakage Current Ir Becomes Zero 
     A second insulation detecting method of the present invention shown in  FIG. 24 , capable of detecting an insulation state of a power line even when electrostatic capacitance of the power line to the ground is in an unbalanced state, as well as in a balanced state, comprises the steps of: allowing the leakage current detecting means  40  to detect a leakage current component Io 1  from a zero-phase current component detected at the secondary side of the zero-phase current transformer, allowing the voltage detecting means to detect a voltage component Vf of each of three phases by extracting only a frequency component, and detecting a phase difference of the leakage current component Io 1  from the voltage component Vf of each of three phases outputted from the voltage detecting means  30 ; calculating an in-phase component and a 90° phase-shifted component of the leakage current component Io 1  of each phase; calculating an in-phase component zero value of each phase; verifying data on active component leakage current or reactive component leakage current of each phase calculated and stored in the memory unit in the step of calculating an in-phase component zero value of each phase; and externally displaying and/or outputting a combination of data recalculated in the step of verifying calculated data and the data used in the step of detecting Io 1 , Vf, and of each phase. 
     The steps common to those of  FIG. 23  are omitted, and the step of calculating an in-phase component zero value of each phase  150  will be described. Active component leakage current generated by insulation resistors Ir, with which the active component leakage current value Ir becomes zero, is calculated for each of three phases. This value is calculated to understand that how much reactive component leakage current Ic flows at the secondary side of the zero-phase current converter  10  if the active component leakage current Ir generated by the insulation resistors becomes zero. Describing it easily, it is to make active component leakage currents generated by the insulation resistors Ir be balanced in all three phases. However, a predetermined value that is slightly larger than the zero value can be selected. 
     An active component leakage current value of R-phase is calculated first to find out the phase of active component zero leakage current that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   rr  zero, i.e., the value of Equation 2 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Irs′, and Irt′ values are calculated so that the in-phase active component leakage current of R-phase becomes zero, i.e., the value of Equation 2 becomes zero. If the Irr′, Irs′, and Irt′ values are put into Equation 3 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 6. If the Irr′, Irs′, and Irt′ values are put into Equation 2 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 7. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Irr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Icr 
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     73.8 
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         + 
                         
                           
                             
                               3 
                             
                              
                             
                               Irs 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             
                               3 
                             
                              
                             
                               Irt 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     6 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lrr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Irr 
                           - 
                           
                             Irr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       
                         - 
                         19.5 
                       
                       + 
                       
                         Irr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irs 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irt 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     7 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 7 becomes zero when Irr′=19.5, Irs′=−39, and Irt′=−39, if the values are put into Equation 6 respectively, the zero-phase leakage currents become 0+j73.8, 0+j40, and 0+j107.6, and the Irr′, Irs′, and Irt′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The active component leakage current value of S-phase becomes zero when Irr′=147.3, Irs′=−73.7, and Irt′=147.3, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0−j20, and 0+j107.5, and the Irr′, Irs′, and Irt′ values of S-phase are stored in the memory unit  86 . 
     The active component leakage current value of T-phase becomes zero when Irr′=−108.3, Irs′=−108.3, and Irt′=54.1, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0+j40.0, and 0−j53.8, and the Irr′, Irs′, and Irt′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of verifying the calculated data  160  is described. 
     Active component zero leakage current Ir′ values calculated and stored in the memory unit  86  in the step of calculating an in-phase component zero value of each phase  150 , with which the active component leakage current Ir of each phase becomes zero, are read. Then, using the zero-phase leakage current for R-phase voltage Io=19.5+j73.8 mA, the zero-phase leakage current for S-phase voltage Io=73.6−j20.0 mA, and the zero-phase leakage current for T-phase voltage Io=−54.1−j53.8 mA, which are calculated in the step of calculating an in-phase and a phase-shifted components of the leakage current component Io 1  of each phase  130 , it is examined that an in-phase component and a 90° phase-shifted component of which phase is large or small. 
     Examining a large value first, since the largest in-phase component is +73.6 of S-phase, if the in-phase component of S-phase is added + (plus), the active component leakage current of S-phase Irs or the reactive component leakage current of R-phase Icr is larger than the active component leakage current of T-phase Irt, the active component leakage current of R-phase Irr, or the reactive component leakage current of T-phase Ict. Since the largest 90° phase-shifted component is +73.8 of R-phase, if the 90° phase-shifted component of R-phase is added + (plus), the reactive component leakage current of R-phase Icr or the active component leakage current of S-phase Irs is larger than the reactive component leakage current of S-phase Ics, the reactive component leakage current of T-phase Ict, or the active component leakage current of T-phase Irt. Therefore, this case satisfies two conditions, and thus although a small value are examined, among the zero-phase leakage current components flowing between the power line  3  and the ground, active component leakage current generated by insulation resistors is largest in S-phase, and reactive component leakage current generated by electrostatic capacitance is largest in R-phase. Accordingly, in the cases where active component leakage current becomes zero, a condition corresponding to R-phase where 90° phase-shifted component leakage current of R-phase is plus + is Irs′=39 with which the active component leakage current of R-phase becomes zero. At this point, the 90° phase-shifted reactive component leakage current of R-phase is +40. Accordingly, it is understood that actual reactive component leakage current of R-phase flowing between the power line  3  and the ground due to the electrostatic capacitance is larger than those of the other phases by +40 mA, and actual active component leakage current of S-phase flowing between the power line  3  and the ground due to the insulation resistors is larger than those of the other phases by +39 mA. 
     3) A Case of Calculating an Active Component Zero Leakage Current Value and a Reactive Component Zero Leakage Current Value when the Reactive Component Leakage Current Ic is Zero and the Active Component Leakage Current Ir is Zero 
     A third insulation detecting method of the present invention shown in  FIG. 25  is a combined method of  FIGS. 23 and 24 . The steps already shown in  FIGS. 23 and 24  will not be described, and the step of calculating a 90° phase-shifted component zero value of each phase  140  is described. 
     A reactive leakage current value generated by electrostatic capacitance, with which a reactive component leakage current value becomes zero, is calculated for each of three phases. This value is calculated to understand that active component leakage current of which component flows at the secondary side of the zero-phase current converter  10  if the reactive component leakage current Ic generated by the electrostatic capacitance becomes zero. Describing it easily, it is to make reactive component leakage currents generated by the electrostatic capacitances be balanced in all three phases. A reactive component leakage current value of R-phase is calculated first to find out the phase and magnitude of reactive component zero leakage current Ic′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   cr  zero, i.e., the value of Equation 3 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and lobes of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Icr′, Ics′, and Ict′ values are calculated so that the 90° phase-shifted reactive component leakage current of R-phase becomes zero, i.e., the value of Equation 3 becomes zero. If the Icr′, Ics′, and Ict′ values are put into Equation 2 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 8. If the Icr′, Ics′, and Ict′ values are put into Equation 3 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 9. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Icr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         - 
                         19.5 
                       
                     
                      
                     
                       
 
                     
                      
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ics 
                             ′ 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ict 
                             ′ 
                           
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     8 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lcr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Icr 
                           + 
                           
                             Icr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     
                       73.8 
                       + 
                       
                         Icr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ics 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ict 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     9 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 9 becomes zero when Icr′=−73.8, Ics′=147.6, and Ict′=147.6, if the values are put into Equation 8 respectively, the zero-phase leakage currents become −19.5+j0, −147.3+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The reactive component leakage current value of S-phase becomes zero when Icr′=−40.0, Ics′=20.0, and Ict′=−40.0, and if the values are put as described above, the zero-phase leakage currents become 39.9+j0, 73.6+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of S-phase are stored in the memory unit  86 . 
     The reactive component leakage current value of T-phase becomes zero when Icr′=−107.5, and Ics′=−53.8, and if the values are put as described above, the zero-phase leakage currents become 39.0+j0, −147.3+j0, and −54.1+j0, and the Icr′, Ics′, and Ict′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of calculating an in-phase component zero value of each phase  150  is described. 
     In a method almost similar to the step of calculating a 90° phase-shifted component zero value of each phase  140 , an active component leakage current generated by insulation resistors, with which the active component leakage current value becomes zero, is calculated for each of three phases. This value is calculated to understand that how much reactive component leakage current flows at the secondary side of the zero-phase current converter  10  if the active component leakage current generated by the insulation resistors becomes zero. Describing it easily, it is to make active component leakage currents generated by the insulation resistors Ir be balanced in all three phases. An active component leakage current value of R-phase is calculated first to find out the phase of active component zero leakage current Ir′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   rr  zero, i.e., the value of Equation 2 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and to of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Irr′, Irs′, and Irt′ values are calculated so that the in-phase active component leakage current of R-phase becomes zero, i.e., the value of Equation 2 becomes zero. If the Irr′, Irs′, and Irt′ values are put into Equation 3 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 10. If the Irr′, Irs′, and Irt′ values are put into Equation 2 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 11. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Irr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Icr 
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     73.8 
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         + 
                         
                           
                             
                               3 
                             
                              
                             
                               Irs 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             
                               3 
                             
                              
                             
                               Irt 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     10 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lrr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Irr 
                           - 
                           
                             Irr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       
                         - 
                         19.5 
                       
                       + 
                       
                         Irr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irs 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irt 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     11 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 11 becomes zero when Irr′=19.5, Irs′=−39, and Irt′=−39, if the values are put into Equation 10 respectively, the zero-phase leakage currents become 0+j73.8, 0+j40, and 0+j107.6, and the Irr′, Irs′, and Irt′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The active component leakage current value of S-phase becomes zero when Irr′=147.3, Irs′=−73.7, and Irt′=147.3, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0−j20, and 0+j107.5, and the Irr′, Irs′, and Irt′ values of S-phase are stored in the memory unit  86 . 
     The active component leakage current value of T-phase becomes zero when Irr′=−108.3, Irs′=−108.3, and Irt′=54.1, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0+j40.0, and 0−j53.8, and the Irr′, Irs′, and Irt′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of verifying the calculated data  160  is performed. Reactive component zero leakage current Ic′ values calculated and stored in the memory unit  86  in the step of calculating a 90° phase-shifted component zero value of each phase  140 , with which the reactive component leakage current of each phase becomes zero, are combined by each case with Ir′ values calculated and stored in the memory unit  86  in the step of calculating an in-phase component zero value of each phase  150 , with which the active component leakage current of each phase becomes zero, and a combination where an Io value corresponding to the zero-phase leakage current becomes zero is searched. The Ic′ values and the Ir′ values are combined case by case, and the recalculated Io value becomes zero. In this case, since the leakage current generated by the insulation resistor of S-phase is the largest, a selected combination is Irs′=−39.0 and Icr′=−40.0. Analyzing the meaning of the result, the active component leakage current generated by the insulation resistor of S-phase is larger than those of R-phase and T-phase by about 39 mA, and the reactive component leakage current generated by the electrostatic capacitance of R-phase is larger than those of S-phase and T-phase by about 40 mA. That is, it is understood that the insulation resistance of S-phase is the lowest, which means poor insulation, and the electrostatic capacitance of R-phase to the ground is the highest. 
     The step of displaying and outputting a variety of data shown in  FIGS. 23 to 25   170  is described. 
     This is a step of displaying the combinations recalculated in the step of verifying the calculated data  160  and a result of performing the step of detecting Io 1 , and Viol each phase  120 , in which the detected data, such as the active component leakage current Ior=39 mA, reactive component leakage current Ioc=40 mA, zero-phase leakage current Io=76.3 mA, information on a phase where insulation is the most poor (e.g., S-phase in the example described above), information on a phase where the reactive component leakage current generated by electrostatic capacitance is the highest (e.g., T-phase in the example described above), or the like, is displayed on the display unit  84 . In addition, insulation resistance R for the active component leakage current Ior=39 mA or electrostatic capacitance C for the reactive component leakage current Ioc=40 mA of the zero-phase voltage between the power line  3  and the ground detected in the step of detecting Io 1 , and Vf of each phase  120  can also be displayed. 
     Here, the insulation resistance R can be expressed as shown in Equation 12, and the electrostatic capacitance C can be expressed as shown in Equation 13. In Equations 12 and 13, the voltage amplification coefficient is an amplification-related coefficient of the voltage detecting means  30 , and it is assumed that the amplification-related co-efficient of the leakage current detecting means  40  including the zero-phase current transformer  10  is one. 
     
       
         
           
             
               
                 
                   
                     Insulation 
                      
                     
                       
                           
                       
                        
                       
                           
                       
                     
                      
                     resistance 
                      
                     
                         
                     
                      
                     
                       ( 
                       R 
                       ) 
                     
                   
                   = 
                   
                     
                       Voltage 
                        
                       
                           
                       
                        
                       Amplification 
                        
                       
                           
                       
                        
                       Coefficient 
                       × 
                       Vf 
                     
                     Icr 
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     12 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     Electrostatic 
                      
                     
                         
                     
                      
                     
                       capacitance 
                        
                       
                         ( 
                         C 
                         ) 
                       
                     
                   
                   = 
                   
                     Ic 
                     
                       
                         
                           
                             2 
                              
                             π 
                              
                             
                                 
                             
                              
                             f 
                             × 
                           
                         
                       
                       
                         
                           
                             Voltage 
                              
                             
                                 
                             
                              
                             Amplification 
                              
                             
                                 
                             
                              
                             Coefficient 
                             × 
                             Vf 
                           
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     13 
                   
                   &gt; 
                 
               
             
           
         
       
     
     In addition, using various types of communication methods (RS-232, RS-485, RS-422, code division multiple access (CDMA), or powerline communication), the data described above can be outputted to outside through the communication unit  90 . 
     In addition, a variety of data described above is compared with an alarm setting value that is previously stored in the memory unit  86 , inputted through the input unit  82 , or received through the communication unit  90 . If the alarm setting value is larger than the active leakage current Ior or Ir, or smaller than the insulation resistance R, a warning alarm can be displayed on the display unit  84  or outputted through the communication unit  90 . 
     Second Embodiment 
     A second embodiment of the insulation detecting apparatus of the present invention is described below. 
       FIGS. 23 to 25  are flowcharts illustrating an insulation detecting apparatus and an insulation detecting method that can be used in a second embodiment in the same manner as the first embodiment described referring to  FIGS. 10 and 11 . 
       FIG. 10  shows a block diagram of the insulation detecting apparatus used in  FIGS. 2 to 7  according to a second embodiment of the insulation detecting apparatus. 
     The insulation detecting apparatus  20  of the present invention comprises a voltage detecting means  30  for detecting a voltage component between the power line  3  and the ground, transforming the detected voltage component into voltage of a certain magnitude, and extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from a voltage component of a sequentially selected phase; a leakage current detecting means  40  for converting the zero-phase leakage current Io component, which is detected at the secondary side of the zero-phase current transformer  10  for detecting zero-phase leakage current Io between the power line  3  including a load  4  and the ground, into a voltage component, amplifying the converted voltage component, and extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band; a phase comparing means  50  for comparing each of three phases of an output value of the voltage detecting means  30  with a phase of an output value of the leakage current detecting means  40 ; an analog-to-digital conversion unit  60  for converting an analog component of the output value of the leakage current detecting means  40  to a digital component; an operation controller  70  having an operation and control function for reading and outputting a variety of data; an input-output means  80  for inputting and displaying a variety of data; and a communication unit  90  for remotely controlling the insulation detecting apparatus from outside. The input-output means  80  includes an input unit  82 , a display unit  84 , and a memory unit  86 . 
     Referring to  FIG. 11 , the voltage detecting means  30  for detecting a three phase voltage component of the power line  3  including a load comprises a voltage detecting unit  31  for transforming the three phase voltage component detected by the voltage detection line  12 ,  13 , and  14  into voltage of a certain magnitude, a phase selection unit  32  for sequentially selecting a voltage component of a phase from the three phase voltage component transformed by the voltage detecting unit  31  in response to an RST voltage control signal outputted from the operation controller  70 , and a voltage filter unit  33  for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from the voltage component of the phase selected by the phase selection unit  32 . The leakage current detecting means  40  comprises a current-to-voltage conversion unit  41  for converting leakage current component, which is detected at the secondary side of the zero-phase current transformer  10  for detecting the zero-phase leakage current Io between the power line  3  including the load  4  and the ground, into a voltage component, an amplification unit  42  for amplifying the zero-phase leakage current component Io converted by the current-to-voltage conversion unit  41 , and a current filter unit  43  for extracting a frequency component lower than a certain frequency or a frequency component of a commercial frequency band from a leakage current component corresponding to the zero-phase leakage current Io component amplified by the amplification unit  42 . 
     The phase comparing means  50  includes a voltage component waveform shaping unit  51  for shaping the waveform of the voltage component of each of three phases outputted from the voltage detecting means  30 , a current component waveform shaping unit  52  for shaping the waveform of the leakage current component corresponding to the zero-phase leakage current Io component outputted from the leakage current detecting means  40 , and a phase difference detecting unit  53  for detecting a phase difference of an output of the current component waveform shaping unit  52  from an output of the voltage component waveform shaping unit  51 . Only one output value of the leakage current detecting means  40  is inputted into the analog-to-digital conversion unit  60  in  FIG. 10 . However, two analog components, i.e., the output value of the leakage current detecting means  40  and the output value of the voltage detecting means  30 , are inputted into the analog-to-digital conversion unit  60  in  FIG. 11 . 
     The voltage component value of the power line  3  is read, and whether the read voltage component is used for calculating insulation resistance, as well as calculating a leakage current value, or whether only the leakage current value is calculated and the insulation resistance is not calculated will be different among embodiments. However, in an embodiment of the present invention, the insulation resistance will be calculated in order to express a variety of values for monitoring insulation conditions. 
     Hereinafter, a second embodiment of the insulation detecting apparatus  20  shown  FIG. 11  and a flowchart illustrating an operational flow of the insulation detecting apparatus  20  will be described in detail. 
     As shown in  FIGS. 11 and 23 , on the main flow stored in the memory unit  86  of the insulation detecting apparatus  20 , the input unit  82  sets a variety of data used in the insulation detecting apparatus  20  using a constitutional component such as a keypad or a switch  100 , in which the input unit has a function for setting a variety of data, e.g., a number address, an alarm setting value, and the like of each insulation detecting apparatus  20  if a plurality of insulation detecting apparatuses  20  is installed. Next, a read operation is performed on a variety of data set by the input unit  82 , previously stored in the memory unit  86 , or inputted from a remote external site through the communication unit  90   110 . 
     If the step of detecting Io 1 , Vf, and of each phase  120  is performed, the zero-phase leakage current component to detected at the secondary side of the zero-phase current transformer  10  as shown in  FIGS. 10 and 11  is converted into a voltage component by the current-to-voltage conversion unit  41  that converts current to voltage and amplified by the amplification unit  42 . Then, a current component Io 1  corresponding to the zero-phase leakage current from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the current filter unit  43  is outputted to the analog-to-digital conversion unit  60  and the phase comparing means  50 . The component Io 1  value corresponding to the zero-phase leakage current inputted into the analog-to-digital conversion unit  32  ( 60 ) is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . 
     Then, the voltage detecting unit  31  used in the embodiments shown in  FIGS. 14 to 21  splits the voltage component of each of three phases between the power line  3  and the ground inputted through the voltage detection lines  12 ,  13 , and  14  into voltages that can be used in the insulation detecting apparatus  20 , using resistors, condensers, a transformer, the 120° phase shift unit  311 , or the 240° phase shift unit  312 . 
     In order to detect a voltage component of each of three phases outputted from the voltage detecting unit  31 , if the switch sw 1  in the phase selection unit  32  is connected to terminal a in response to an RST voltage control signal for selecting R-phase received from the operation controller  70 , the voltage component of R-phase is inputted into the voltage filter unit  33 . The voltage component Vf value (Vf_r of R-phase), i.e., the voltage component of R-phase selected by the phase selection unit  32  from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the voltage filter unit  33 , is outputted to the phase comparing means  50  and the analog-to-digital conversion unit  60 . The voltage component Vf_r value to the ground inputted into the analog-to-digital conversion unit  60  is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . Using the value of the voltage component Vf_r to the ground inputted into the phase comparing means  50 , whose waveform is shaped by the voltage component waveform shaping unit  51 , and the value of the leakage current component Io 1  corresponding to the zero-phase leakage current component outputted from the leakage current detecting means  40 , whose waveform is shaped by the current component waveform shaping unit  52 , the phase difference detecting unit  53  detects a phase difference between the voltage component outputted from the voltage component waveform shaping unit  51  and the one leakage current component outputted from the current component waveform shaping unit  52 , i.e., a phase difference r of the leakage current component Io 1  for the voltage component Vf_r of R-phase, and the operation controller  70  reads and stores the phase difference into the memory unit  86 . 
     Next, if the switch sw 1  in the phase selection unit  32  is connected to terminal b in response to an RST voltage control signal for selecting S-phase received from the operation controller  70 , the voltage component of S-phase is inputted into the voltage filter unit  33 . The voltage component Vf value (Vf_s of S-phase), i.e., the voltage component of S-phase selected by the phase selection unit  32  from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the voltage filter unit  33 , is outputted to the phase comparing means  50  and the analog-to-digital conversion unit  60 . The voltage component Vf_s value to the ground inputted into the analog-to-digital conversion unit  60  is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . Using the value of the voltage component Vf_s to the ground inputted into the phase comparing means  50 , whose waveform is shaped by the voltage component waveform shaping unit  51 , and the value of the leakage current component Io 1  corresponding to the zero-phase leakage current component outputted from the leakage current detecting means  40 , whose waveform is shaped by the current component waveform shaping unit  52 , the phase difference detecting unit  53  detects a phase difference between the voltage component outputted from the voltage component waveform shaping unit  51  and the one leakage current component outputted from the current component waveform shaping unit  52 , i.e., a phase difference s of the leakage current component Io 1  for the voltage component Vf_s of S-phase, and the operation controller  70  reads and stores the phase difference into the memory unit  86 . 
     Next, if the switch sw 1  in the phase selection unit  32  is connected to terminal c in response to an RST voltage control signal for selecting T-phase received from the operation controller  70 , the voltage component of T-phase is inputted into the voltage filter unit  33 . The voltage component Vf value (Vf_t of T-phase), i.e., the voltage component of T-phase selected by the phase selection unit  32  from which a frequency component lower than a certain frequency or a frequency component of a commercial frequency band is extracted by the voltage filter unit  33 , is outputted to the phase comparing means  50  and the analog-to-digital conversion unit  60 . The voltage component Vf_t value to the ground inputted into the analog-to-digital conversion unit  60  is converted into a digital value, and the operation controller  70  reads and stores the digital value into the memory unit  86 . Using the value of the voltage component Vf_t to the ground inputted into the phase comparing means  50 , whose waveform is shaped by the voltage component waveform shaping unit  51 , and the value of the leakage current component Io 1  corresponding to the zero-phase leakage current component outputted from the leakage current detecting means  40 , whose waveform is shaped by the current component waveform shaping unit  52 , the phase difference detecting unit  53  detects a phase difference between the voltage component outputted from the voltage component waveform shaping unit  51  and the one leakage current component outputted from the current component waveform shaping unit  52 , i.e., a phase difference t of the leakage current component Io 1  for the voltage component Vf_t of T-phase, and the operation controller  70  reads and stores the phase difference into the memory unit  86 . 
     An example of calculating the voltage component Vf, leakage current component Io 1 , and phase difference described above is described. For the convenience of explanation, it is assumed that the amplification-related coefficient of the leakage current detecting means  40  including the zero-phase current transformer  10  is one, and the amplification-related coefficient of the voltage detecting means  30  is 0.001 (i.e., 1/1000). 
     The voltage between the three-phase power line  3  and the ground is 220V, and the frequency is 60 Hz. The leakage currents flowing through the insulation resistors between the three-phase power line and the ground are respectively Irr=1 mA for R-phase, Irs=40 mA for S-phase, and Irt=1 mA for T-phase. The leakage currents flowing through the electrostatic capacitances between the three-phase power line and the ground are respectively Icr=60 mA for R-phase, Ics=20 mA for S-phase, and Ict=20 mA for T-phase. 
     The values detected and stored in the memory unit  86  in the step of detecting the voltage component Vf, leakage current component Io 1 , and phase difference  120  are such that Io 1  is 76.3 mA, Vf_r, Vf_s, and Vf_t are 220 mV respectively, r is 104.8, s is −15.2, and t is −135.2. 
     Next, if the step of calculating an in-phase and a 90° phase-shifted components of the leakage current component Io 1  of each phase  130  is performed, the leakage current component Io 1  and the phase differences r, s, and t detected and stored in the memory unit  86  in the step of detecting the voltage component Vf, leakage current component Io 1 , and phase difference  120  are read. For each of three phases, an in-phase component cos value and a 90° phase-shifted component sin value for the voltage of the leakage current Io 1  corresponding to the zero-phase leakage current are calculated and stored in the memory unit  86 . Describing more specifically, the in-phase component leakage current of R-phase Io 1   rr  is Io 1 ×cos r, the 90° phase-shifted component leakage current of R-phase Io 1   cr  is Io 1 ×sin r, the in-phase component leakage current of S-phase Io 1   rs  is Io 1 ×cos s, the 90° phase-shifted component leakage current of S-phase Io 1   cs  is Io 1 ×sin s, the in-phase component leakage current of T-phase Io 1   rt  is Io 1 ×cos t, and the 90° phase-shifted component leakage current of T-phase Io 1   ct  is Io 1 ×sin t. 
     Substituting the values described in the above example and approximately calculating the zero-phase leakage currents, Io 1   rr =−19.5 mA, Io 1   cr= 73.8 mA, Io 1   rs =73.6 mA, Io 1   cs =−20.0 mA, Io 1   rt =−54.1 mA, and Io 1   ct =−53.8 mA, and therefore, the zero-phase leakage current Io for the R-phase voltage is −19.5+j73.8 mA, the zero-phase leakage current to for the S-phase voltage is 73.6−j20 mA, and the zero-phase leakage current Io for T-phase voltage is −54.1−j53.8 mA. 
     For the values calculated above, a functional relation between an in-phase component leakage current value for voltage and a 90° phase-shifted component leakage current value for voltage of each of three phases will be described first. The zero-phase leakage current to component flowing between the power line  3  including the load  4  and the ground can be expressed as shown in Equation 14. Next, the zero-phase leakage current for a three phase voltage component shown in Equation 14 is converted to a voltage component value of R-phase, and a value corresponding to the zero-phase leakage current component of a voltage component that is in-phase with the R-phase voltage, i.e., Io 1   rr , is expressed as shown in Equation 15, and a value corresponding to the zero-phase leakage current component of the voltage component that is 90° phase-shifted from the R-phase voltage, i.e., Io 1   cr , is expressed as shown in Equation 16. 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             Io 
                             ° 
                           
                           = 
                             
                            
                           
                             
                               
                                 V 
                                 
                                   R 
                                   ° 
                                 
                               
                               Rr 
                             
                             + 
                             
                               
                                 V 
                                 
                                   S 
                                   ° 
                                 
                               
                               Rs 
                             
                             + 
                             
                               
                                 V 
                                 
                                   T 
                                   ° 
                                 
                               
                               Rt 
                             
                             + 
                           
                         
                       
                     
                     
                       
                         
                             
                            
                           
                             j 
                              
                             
                               ( 
                               
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Cr 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       R 
                                       ° 
                                     
                                   
                                 
                                 + 
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Cs 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       S 
                                       ° 
                                     
                                   
                                 
                                 + 
                                 
                                   ω 
                                    
                                   
                                       
                                   
                                    
                                   Ct 
                                    
                                   
                                       
                                   
                                    
                                   
                                     V 
                                     
                                       T 
                                       ° 
                                     
                                   
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     
                       
                         
                           = 
                             
                            
                           
                             
                               Irr 
                               ° 
                             
                             + 
                             
                               Ics 
                               ° 
                             
                             + 
                             
                               Ict 
                               ° 
                             
                             + 
                             
                               j 
                                
                               
                                 ( 
                                 
                                   
                                     Icr 
                                     ° 
                                   
                                   + 
                                   
                                     Ics 
                                     ° 
                                   
                                   + 
                                   
                                     Ict 
                                     ° 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Here 
                     , 
                     
                       V 
                        
                       
                           
                       
                        
                       and 
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       I 
                        
                       
                           
                       
                        
                       are 
                        
                       
                           
                       
                        
                       vector 
                        
                       
                           
                       
                        
                       
                         functions 
                         . 
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     14 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     Irr 
                     - 
                     
                       Irs 
                       2 
                     
                     - 
                     
                       Irt 
                       2 
                     
                     - 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Ics 
                     
                     + 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Ict 
                     
                   
                    
                   
                     
 
                   
                    
                   
                     Here 
                     , 
                     
                       I 
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       a 
                        
                       
                         
                             
                         
                          
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       real 
                        
                       
                           
                       
                        
                       
                         value 
                         . 
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     15 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     Icr 
                     - 
                     
                       Ics 
                       2 
                     
                     - 
                     
                       Ict 
                       2 
                     
                     + 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Irs 
                     
                     - 
                     
                       
                         
                           3 
                         
                         2 
                       
                        
                       Irt 
                     
                   
                    
                   
                     
 
                   
                    
                   Here 
                   , 
                   
                     I 
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     a 
                      
                     
                         
                     
                      
                     real 
                      
                     
                         
                     
                      
                     
                       value 
                       . 
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     16 
                   
                   &gt; 
                 
               
             
           
         
       
     
     The relational expression between the in-phase component leakage current and the 90° phase-shifted component leakage current of S-phase and T-phase is that they respectively have 120° and −120° phase differences from the in-phase component leakage current and the 90° phase-shifted component leakage current of R-phase. 
     As shown in Equations 15 and 16, it is understood that the in-phase component leakage current of R-phase includes not only active component leakage current Irr flowing due to the insulation resistance of R-phase, but also active component leakage currents Irs and Irt flowing due to the insulation resistances of S-phase and T-phase and reactive component leakage currents Ics and Ict flowing due to the electrostatic capacitances of S-phase and T-phase. The 90° phase-shifted component leakage current of R-phase includes not only reactive component leakage current Icr flowing due to the electrostatic capacitance of R-phase, but also reactive component leakage currents Ics and Ict flowing due to the electrostatic capacitances of S-phase and T-phase and active component leakage currents Irs and Irt flowing due to the insulation resistances of S-phase and T-phase. Then, it can be conjectured from Equations 15 and 16 that an insulation state cannot be correctly obtained by a conventional method of detecting zero-phase leakage current Io generated by a zero-phase current transformer. 
     1) The step of calculating a 90° phase-shifted component zero value of each phase  140  is described referring to  FIG. 23 . 
     A reactive component leakage current value generated by electrostatic capacitance with which a reactive component leakage current becomes zero is calculated for each of three phases. This value is calculated to understand that active component leakage current of which component flows at the secondary winding of the zero-phase current converter  10  if the reactive component leakage current Ic generated by the electrostatic capacitance becomes zero. Describing it easily, it is to make reactive component leakage currents generated by electrostatic capacitances be balanced in all three phases. A reactive component leakage current value of R-phase is calculated first to find out the phase and magnitude of reactive component zero leakage current Ic′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   cr  zero, i.e., the value of Equation 16 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then. Icr′, Ics′, and Ict′ values are calculated so that the 90° phase-shifted reactive component leakage current of R-phase becomes zero, i.e., the value of Equation 3 becomes zero. If the Icr′, Ics′, and Ict′ values are put into Equation 15 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 17. If the Icr′, Ics′, and Ict′ values are put into Equation 16 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 18. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Icr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Irr 
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       - 
                       19.5 
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ics 
                             ′ 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ict 
                             ′ 
                           
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     17 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lcr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Icr 
                           + 
                           
                             Icr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     
                       73.8 
                       + 
                       
                         Icr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ics 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ict 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     18 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 18 becomes zero when Icr′=−73.8, Ics′=147.6, and Ict′=147.6, if the values are put into Equation 17 respectively, the zero-phase leakage currents become −19.5+j0, −147.3+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The reactive component leakage current value of S-phase becomes zero when Icr′=−40.0, Ics′=20.0, and Ict′=−40.0, and if the values are put as described above, the zero-phase leakage currents become 39.9+j0, 73.6+j0, and 108.3+j0, and the Icr′, and Ict′ values of S-phase are stored in the memory unit  86 . 
     The reactive component leakage current value of T-phase becomes zero when Icr′=−107.5, and Ic=53.8, and if the values are put as described above, the zero-phase leakage currents become 39.0+j0, −147.3+j0, and −54.1+j0, and the Icr′, Ics′, and Ict′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of verifying the calculated data  160  is described. 
     Reactive component zero leakage current Ic′ values calculated and stored in the memory unit  86  in the step of calculating a 90° phase-shifted component zero value of each phase  140 , with which the reactive component leakage current of each phase becomes zero, are read. Then, using the zero-phase leakage current for R-phase voltage Io=19.5+j73.8 mA, the zero-phase leakage current for S-phase voltage Io=73.6−j20.0 mA, and the zero-phase leakage current for T-phase voltage Io=−54.1−j53.8 mA, which are calculated in the step of calculating an in-phase and a phase-shifted components of the leakage current component Io 1  of each phase  130 , it is examined that an in-phase component and a 90° phase-shifted component of which phase is large or small. Examining a large value first, since the largest in-phase component is +73.6 of S-phase, if the in-phase component of S-phase is added + (plus), the active component leakage current of S-phase Irs or the reactive component leakage current of R-phase Icr is larger than the active component leakage current of T-phase Irt, the active component leakage current of R-phase Irr, or the reactive component leakage current of T-phase Ict. Since the largest 90° phase-shifted component is +73.8 of R-phase, if the 90° phase-shifted component of R-phase is added + (plus), the reactive component leakage current of R-phase Icr or the active component leakage current of S-phase Irs is larger than the reactive component leakage current of S-phase Ics, the reactive component leakage current of T-phase Ict, or the active component leakage current of T-phase Irt. Therefore, this case satisfies two conditions, and thus, although a small value is examined, among the zero-phase leakage current components flowing between the power line  3  and the ground, active component leakage current generated by insulation resistors is largest in S-phase, and reactive component leakage current generated by electrostatic capacitance is largest in R-phase. Accordingly, in the cases where reactive component leakage current becomes zero, a condition corresponding to R-phase where in-phase component leakage current of S-phase is plus + is Icr′=40 with which the reactive component leakage current of S-phase becomes zero. At this point, the in-phase active component leakage current of S-phase is +39. Accordingly, it is understood that actual reactive component leakage current of R-phase flowing between the power line  3  and the ground due to the electrostatic capacitance is larger than those of the other phases by +40 mA, and actual active component leakage current of S-phase flowing between the power line  3  and the ground due to the insulation resistors is larger than those of the other phases by +39 mA. 
     2) The Step of Calculating an In-Phase Component Zero Value of Each Phase  150  is Described 
     If the step of calculating an in-phase component zero value of each phase  150  is performed, active component leakage current generated by insulation resistors Ir, with which the active component leakage current value Ir becomes zero, is calculated for each of three phases. This value is calculated to understand that how much reactive component leakage current Ic flows at the secondary side of the zero-phase current converter  10  if the active component leakage current Ir generated by the insulation resistors becomes zero. 
     Describing it easily, it is to make active component leakage currents generated by the insulation resistors Ir be balanced in all three phases. An active component leakage current value of R-phase is calculated first to find out the phase of active component zero leakage current Ir′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   rr  zero, i.e., the value of Equation 2 becomes zero. 
     Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Irr′, Irs′, and Irt′ values are calculated so that the in-phase active component leakage current of R-phase becomes zero, i.e., the value of Equation 15 becomes zero. If the Irr′, Irs′, and Irt′ values are put into Equation 16 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 19. If the Irr′, Irs′, and Irt′ values are put into Equation 15 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 20. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Irr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Icr 
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     73.8 
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         + 
                         
                           
                             
                               3 
                             
                              
                             
                               Irs 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             
                               3 
                             
                              
                             
                               Irt 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     19 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lrr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Irr 
                           - 
                           
                             Irr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       
                         - 
                         19.5 
                       
                       + 
                       
                         Irr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irs 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irt 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     20 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 20 becomes zero when Irr′=19.5, Irs′=−39, and Irt′=−39, if the values are put into Equation 19 respectively, the zero-phase leakage currents become 0+j73.8, 0+j40, and 0+j107.6, and the Irr′, Irs′, and Irt′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The active component leakage current value of S-phase becomes zero when Irr′=147.3, Irs′=−73.7, and Irt′=147.3, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0−j20, and 0+j107.5, and the Irr′, Irs′, and Irt′ values of S-phase are stored in the memory unit  86 . 
     The active component leakage current value of T-phase becomes zero when Irr′=−108.3, Irs′=−108.3, and Irt′=54.1, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0+j40.0, and 0−j53.8, and the Irr′, Irs′, and Irt′ values of T-phase are stored in the memory unit  86 . 
     Next, if the step of verifying the calculated data  160  is performed, active component zero leakage current Ir′ values calculated and stored in the memory unit  86  in the step of calculating an in-phase component zero value of each phase  150 , with which the active component leakage current Ir of each phase becomes zero, are read. Then, using the zero-phase leakage current for R-phase voltage Io=19.5+j73.8 mA, the zero-phase leakage current for S-phase voltage Io=73.6−j20.0 mA, and the zero-phase leakage current for T-phase voltage Io=−54.1−j53.8 mA, which are calculated in the step of calculating an in-phase and a phase-shifted components of the leakage current component Io 1  of each phase  130 , it is examined that an in-phase component and a 90° phase-shifted component of which phase is large or small. Examining a large value first, since the largest in-phase component is +73.6 of S-phase, if the in-phase component of S-phase is added + (plus), the active component leakage current of S-phase Irs or the reactive component leakage current of R-phase Icr is larger than the active component leakage current of T-phase Irt, the active component leakage current of R-phase Irr, or the reactive component leakage current of T-phase Ict. Since the largest 90° phase-shifted component is +73.8 of R-phase, if the 90° phase-shifted component of R-phase is added + (plus), the reactive component leakage current of R-phase Irr or the active component leakage current of S-phase Irs is larger than the reactive component leakage current of S-phase Ics, the reactive component leakage current of T-phase Ict, or the active component leakage current of T-phase Irt. Therefore, this case satisfies two conditions, and thus although a small value are examined, among the zero-phase leakage current components flowing between the power line  3  and the ground, active component leakage current generated by insulation resistors is largest in S-phase, and reactive component leakage current generated by electrostatic capacitance is largest in R-phase. Accordingly, in the cases where active component leakage current becomes zero, a condition corresponding to R-phase where 90° phase-shifted component leakage current of R-phase is plus + is Irs′=39 with which the active component leakage current of R-phase becomes zero. At this point, the 90° phase-shifted reactive component leakage current of R-phase is +40. Accordingly, it is understood that actual reactive component leakage current of R-phase flowing between the power line  3  and the ground due to the electrostatic capacitance is larger than those of the other phases by +40 mA, and actual active component leakage current of S-phase flowing between the power line  3  and the ground due to the insulation resistors is larger than those of the other phases by +39 mA. 
     3) The step of calculating a 90° phase-shifted component zero value of each phase  140  and the step of calculating an in-phase component zero value of each phase  150  will be described. 
     A. The Step of Calculating a 90° Phase-Shifted Component Zero Value of Each Phase  140   
     A reactive leakage current value generated by electrostatic capacitance with which a reactive component leakage current value becomes zero is calculated for each of three phases. This value is calculated to understand that active component leakage current of which component flows at the secondary side of the zero-phase current converter  10  if the reactive component leakage current Ic generated by the electrostatic capacitance becomes zero. Describing it easily, it is to make reactive component leakage currents generated by the electrostatic capacitances be balanced in all three phases. A reactive component leakage current value of R-phase is calculated first to find out the phase and magnitude of reactive component zero leakage current Ic′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   cr  zero, i.e., the value of Equation 20 becomes zero. Io 1   rr  and Io 1   cr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Icr′, Ics′, and Ict′ values are calculated so that the 90° phase-shifted reactive component leakage current of R-phase becomes zero, i.e., the value of Equation 16 becomes zero. If the Icr′, Ics′, and Ice values are put into Equation 15 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 21. If the Icr′, Ics′, and Ict′ values are put into Equation 16 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 22. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Icr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Irr 
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       - 
                       19.5 
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ics 
                             ′ 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             Ict 
                             ′ 
                           
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     21 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lcr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Icr 
                           + 
                           
                             Icr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     
                       73.8 
                       + 
                       
                         Icr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lcs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           
                             ( 
                             
                               Ics 
                               + 
                               
                                 Ics 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ics 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lct 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Ict 
                               + 
                               
                                 Ict 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             Ict 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     22 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 22 becomes zero when Icr′=−73.8. Ics′=147.6, and Ict′=147.6, if the values are put into Equation 21 respectively, the zero-phase leakage currents become −19.5+j0, −147.3+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The reactive component leakage current value of S-phase becomes zero when Icr′=−40.0, Ics′=20.0, and Ict′=−40.0, and if the values are put as described above, the zero-phase leakage currents become 39.9+j0, 73.6+j0, and 108.3+j0, and the Icr′, Ics′, and Ict′ values of S-phase are stored in the memory unit  86 . 
     The reactive component leakage current value of T-phase becomes zero when Icr′=−107.5, Ics′=−107.5, and Ict′=53.8, and if the values are put as described above, the zero-phase leakage currents become 39.0+j0, −147.3+j0, and −54.1+j0, and the Icr′, Ics′, and Ict′ values of T-phase are stored in the memory unit  86 . 
     B. The Step of Calculating an In-Phase Component Zero Value of Each Phase  150   
     The step of calculating an in-phase component zero value of each phase  150  of  FIG. 25  is described. In a method almost similar to the step of calculating a 90° phase-shifted component zero value of each phase  140 , an active component leakage current generated by insulation resistors, with which the active component leakage current value becomes zero, is calculated for each of three phases. This value is calculated to understand that how much reactive component leakage current flows at the secondary side of the zero-phase current converter  10  if the active component leakage current generated by the insulation resistors becomes zero. Describing it easily, it is to make active component leakage currents generated by the insulation resistors Ir be balanced in all three phases. An active component leakage current value of R-phase is calculated first to find out the phase of active component zero leakage current Ir′ that should be additionally flown through the primary winding of the zero-phase current transformer  10  to make Io 1   rr  zero, i.e., the value of Equation 11 becomes zero. Io 1   rr  and Io 1   rr  of R-phase, Io 1   rs  and Io 1   cs  of S-phase, and Io 1   rt  and Io 1   ct  of T-phase described in the above example and stored in the memory unit  86  are read. Then, Irr′, Irs′, and Irt′ values are calculated so that the in-phase active component leakage current becomes zero, i.e., the value of Equation 15 becomes zero. If the Irr′, Irs′, and Irt′ values are put into Equation 16 and a 90° phase-shifted component leakage current value Io 1   cr ′ of R-phase is calculated, Io 1   cr ′ becomes as shown in Equation 23. If the Irr′, Irs′, and Irt′ values are put into Equation 15 and an in-phase component leakage current value Io 1   rr ′ of R-phase is calculated, Io 1   rr ′ becomes as shown in Equation 24. 
     
       
         
           
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       Irr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       Icr 
                       - 
                       
                         Ics 
                         2 
                       
                       - 
                       
                         Ict 
                         2 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irs 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Irt 
                       
                     
                     = 
                     73.8 
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irt 
                         
                       
                       = 
                       
                         73.8 
                         + 
                         
                           
                             
                               3 
                             
                              
                             
                               Irs 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                         
                             
                         
                          
                         
                             
                         
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Icr 
                         - 
                         
                           Ics 
                           2 
                         
                         - 
                         
                           Ict 
                           2 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Irs 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                         
                       
                       = 
                       
                         73.8 
                         - 
                         
                           
                             
                               3 
                             
                              
                             
                               Irt 
                               ′ 
                             
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     23 
                   
                   &gt; 
                 
               
             
             
               
                 
                   
                     If 
                      
                     
                         
                     
                      
                     
                       lrr 
                       ′ 
                     
                      
                     
                         
                     
                      
                     is 
                      
                     
                         
                     
                      
                     applied 
                   
                   , 
                 
               
               
                 
                     
                 
               
             
             
               
                 
                   
                     
                       
                         ( 
                         
                           Irr 
                           - 
                           
                             Irr 
                             ′ 
                           
                         
                         ) 
                       
                       - 
                       
                         Irs 
                         2 
                       
                       - 
                       
                         Irt 
                         2 
                       
                       - 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ics 
                       
                       + 
                       
                         
                           
                             3 
                           
                           2 
                         
                          
                         Ict 
                       
                     
                     = 
                     
                       
                         - 
                         19.5 
                       
                       + 
                       
                         Irr 
                         ′ 
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrs 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           
                             ( 
                             
                               Irs 
                               + 
                               
                                 Irs 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           Irt 
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irs 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     
                       If 
                        
                       
                           
                       
                        
                       
                         lrt 
                         ′ 
                       
                        
                       
                           
                       
                        
                       is 
                        
                       
                           
                       
                        
                       applied 
                     
                     , 
                     
                       
 
                     
                      
                     
                       
                         Irr 
                         - 
                         
                           Irs 
                           2 
                         
                         - 
                         
                           
                             ( 
                             
                               Irt 
                               + 
                               
                                 Irt 
                                 ′ 
                               
                             
                             ) 
                           
                           2 
                         
                         - 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ics 
                         
                         + 
                         
                           
                             
                               3 
                             
                             2 
                           
                            
                           Ict 
                         
                       
                       = 
                       
                         
                           - 
                           19.5 
                         
                         + 
                         
                           
                             Irt 
                             ′ 
                           
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   &lt; 
                   
                     Equation 
                      
                     
                         
                     
                      
                     24 
                   
                   &gt; 
                 
               
             
           
         
       
     
     Since the value of Equation 24 becomes zero when Irr′=19.5, Irs′=−39, and Irt′=−39, if the values are put into Equation 23 respectively, the zero-phase leakage currents become 0+j73.8, 0+j40, and 0+j107.6, and the Irr′, Irs′, and Irt′ values of R-phase are stored in the memory unit  86 . 
     Next, if zero-phase leakage currents of S-phase and R-phase are respectively calculated in the same manner as that of R-phase, following results will be obtained. 
     The active component leakage current value of S-phase becomes zero when Irr′=147.3, Irs′=−73.7, and Irt′=147.3, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0−j20, and 0+j107.5, and the Irr′, Irs′, and Irt′ values of S-phase are stored in the memory unit  86 . 
     The active component leakage current value of T-phase becomes zero when Irr′=−108.3, Irs′=−108.3, and Irt′=54.1, and if the values are put as described above, the zero-phase leakage currents become 0−j147.5, 0+j40.0, and 0−j53.8, and the Irr′, Irs′, and Irt′ values of T-phase are stored in the memory unit  86 . 
     Next, the step of verifying the calculated data  160  is performed. Reactive component zero leakage current Ic′ values calculated and stored in the memory unit  86  in the step of calculating a 90° phase-shifted component zero value of each phase  140 , with which the reactive component leakage current of each phase becomes zero, are combined by each case with Ir′ values calculated and stored in the memory unit  86  in the step of calculating an in-phase component zero value of each phase  150 , with which the active component leakage current of each phase becomes zero, and a combination where an Io value corresponding to the zero-phase leakage current becomes zero is searched. The Ic′ values and the Ir′ values are combined case by case, and the recalculated Io value becomes zero. In this case, since the leakage current generated by the insulation resistor of S-phase is the largest, a selected combination is Irs′=−39.0 and Icr′=−40.0. Analyzing the meaning of the result, the active component leakage current generated by the insulation resistor of S-phase is larger than those of R-phase and T-phase by about 39 mA, and the reactive component leakage current generated by the electrostatic capacitance of R-phase is larger than those of S-phase and T-phase by about 40 mA. That is, it is understood that the insulation resistance of S-phase is the lowest, which means poor insulation, and the electrostatic capacitance of R-phase to the ground is the highest. 
     The step of displaying and outputting a variety of data shown in  FIGS. 23 to 25   170  is described. 
     This is a step of displaying the combinations recalculated in the step of verifying the calculated data  160  and a result of performing the step of detecting Io 1 , and Vf of each phase  120 , in which the detected data, such as the active component leakage current Ior=39 mA, reactive component leakage current Ioc=40 mA, zero-phase leakage current Io=76.3 mA, information on a phase where insulation is the most poor (e.g., S-phase in the example described above), information on a phase where the reactive component leakage current generated by electrostatic capacitance is the highest (e.g., T-phase in the example described above), or the like, is displayed on the display unit  84 . In addition, insulation resistance R for the active component leakage current Ior=39 mA or electrostatic capacitance C for the reactive component leakage current Ioc=40 mA of the zero-phase voltage between the power line  3  and the ground detected in the step of detecting Io 1 , and Vf of each phase  120  can also be displayed. 
     Here, the insulation resistance R can be expressed as shown in Equation 13, and the electrostatic capacitance C can be expressed as shown in Equation 14. In Equations 13 and 14, the voltage amplification coefficient is an amplification-related coefficient of the voltage detecting means  30 , and it is assumed that the amplification-related co-efficient of the leakage current detecting means  40  including the zero-phase current transformer  10  is one. 
     In addition, using various types of communication methods (RS-232, RS-485, RS-422, code division multiple access (CDMA), or power line communication), the data described above can be outputted to outside through the communication unit  90 . 
     In addition, a variety of data described above is compared with an alarm setting value that is previously stored in the memory unit  86 , inputted through the input unit  82 , or received through the communication unit  90 . If the alarm setting value is larger than the active leakage current Ior or Ir, or smaller than the insulation resistance R, warning alarm can be displayed on the display unit  84  or outputted through the communication unit  90 . 
     Other Embodiments 
     Next, other embodiments shown in  FIGS. 3 to 7  describe that the leakage detecting apparatus of the present invention can be embodied in a variety of forms, i.e., the leakage detecting apparatus can be embodied in a system where the secondary side connection of the transformer  1  is a Y-connection or a delta-connection, in a three-phase three-wire or three-phase four-wire system, in a non-grounding system, or in a system where the voltage component of the power line  3  is implemented by detecting phase voltage or line voltage. 
     The difference between the connection diagram of the second embodiment of the insulation detecting apparatus according to the present invention shown in  FIG. 3  and the connection diagram of the first embodiment shown in  FIG. 2  is that the zero-phase current transformer  10 , such as ZCT for detecting zero-phase leakage current Io, is installed in the middle of the power line  3  to detect the insulation state of the power line  3  including the load  4  in the embodiment of  FIG. 2 , whereas the zero-phase current transformer  10  is installed in the middle of the ground line  5  of the transformer  1  in the embodiment of  FIG. 3 , and the other aspects of the two embodiments are the same. 
     Next, the connection diagram of the third embodiment of the insulation detecting apparatus according to the present invention shown in  FIG. 4  is a connection diagram for describing that the connection diagram of the third embodiment is almost the same as that of the first embodiment shown in  FIG. 2  and the insulation detecting apparatus can be embodied in a three-phase four-wire system where the neutral phase (N-phase) is used, and the operational flow and detecting method of the insulation detecting apparatus  20  is the same as those described above. 
     Next, the connection diagram of the fourth embodiment of the insulation detecting apparatus according to the present invention shown in  FIG. 5  is a connection diagram for describing that the insulation detecting apparatus can be embodied in a power line  3  where the secondary side connection of the transformer  1  is a delta connection and a phase of the delta connection is grounded, and the operational flow and detecting method of the insulation detecting apparatus  20  is the same as those described above. 
     Next, the connection diagram of the fifth embodiment of the insulation detecting apparatus according to the present invention shown in  FIG. 6  is a connection diagram for describing that the insulation detecting apparatus can be embodied in a non-grounding system where the secondary side connection of the transformer  1  is a delta connection that is not grounded, and the operational flow and detecting method of the insulation detecting apparatus  20  is the same as those described above. 
     Next, the connection diagram of the sixth embodiment of the insulation detecting apparatus according to the present invention is a connection diagram for describing that the insulation state of the power line  3  can be monitored by detecting line voltage, and the operational flow and detecting method of the insulation detecting apparatus  20  is the same as those described above. 
     The leakage current detecting means  40  used in the aforementioned embodiments of the present invention shown in  FIGS. 9 and 11  is configured in the sequence of detecting a leakage current component through the zero-phase current transformer  10 , converting the detected leakage current component to a voltage component in the current-to-voltage conversion unit  41 , amplifying a value converted to a voltage component through the amplification unit  42 , and filtering the amplified value through the current filter unit  43 . However, another embodiment of the leakage current detecting means shown in  FIG. 22  is configured in the sequence of detecting a leakage current component through the zero-phase current converter  10 , converting the detected leakage current component to a voltage component in the current-to-voltage conversion unit  41 , filtering a value converted to a voltage component in the current filter unit  43 , and then amplifying the filtered value through the amplification unit  42 . 
     Although the voltage detecting unit  31  of the power line  3  is configured inside of the insulation detecting apparatus  20  to detect a voltage component in the embodiments of the present invention, there can be other embodiments where a part having a function of the voltage detecting unit  31  is configured outside of the insulation detecting apparatus  20 . Although the current-to-voltage conversion unit  41  is configured inside of the insulation detecting apparatus  20  to detect leakage current, if an electronic component element such as a resistor is directly connected to the secondary side of the zero-phase current converter  10 , there can be other embodiments where a part having a function of the current-to-voltage conversion unit  41  is configured outside of the insulation detecting apparatus  20 . There also can be other embodiments where the amplification unit  42  is configured to have a function of the current-to-voltage conversion unit  41 . 
     Although resistance or capacitance of the resistors or capacitors used in the voltage detecting unit  31  is the same for all three phases in the embodiments of the present invention, there can be another embodiment where previously known different values can be applied to three phases respectively. 
     There can be another embodiment where phase angles and leakage current of each phase are measured using a power factor meter, leakage current meter, or wattmeter, which has the same functions as those of the embodiments of the present invention, together with capability of measuring phase angles and leakage current. 
     Although the present invention has been described with reference to several preferred embodiments, the description is illustrative of the invention and is not to be construed as limiting the invention. Various modifications and variations may occur to those skilled in the art, without departing from the scope of the invention as defined by the appended claims.