# Fault Location in Double Circuit Medium Power Distribution Networks Using an Impedance-Based Method
## 2 The Proposed Method  
The distribution network consists of different parts. In this article, the line is considered as the part between two buses. By using a \(\pi\) line model, Figure 1 shows a single-line diagram of each section for the positive, negative, and zero sequences of a double circuit network for a normal fault in point \(F\). Notations in Figure 1 are as follows:  
Figure 1: (**a**) Single-line diagram for each section of a power distribution system; (**b**) Positive; (**c**) negative; and (**d**) zero-sequence networks of a double-circuit distribution line during a usual fault. This table is an extended version of the one in [18]:  
\[\left\{ \begin{array}{cccc}V_{S_{1}}&V_{S_{2}}&V_{S_{0}}&\text{The positive, negative, and zero}\\ I_{S_{1-I}}&I_{S_{2-I}}&I_{S_{0-I}}&\text{sequence of voltage and current}\\ I_{S_{1-II}}&I_{S_{2-II}}&I_{S_{0-II}}&\text{during fault at node $S$}\end{array} \right.,\]  
\[\left\{ \begin{array}{cccc}V_{F_{1-I}}&V_{F_{2-I}}&V_{F_{0-I}}&\text{The positive, negative, and zero}\\ V_{F_{1-II}}&V_{F_{2-II}}&V_{F_{0-II}}&\text{sequence of voltage and current}\\ I_{F_{1-I}}&I_{F_{2-I}}&I_{F_{0-I}}&\text{during fault at node $F$}\end{array} \right.,\]  
\[\left\{ \begin{array}{cccc}&\text{Positive, zero, and mutual sequence}\\ Z_{1}&Z_{0}&Z_{m0}&\text{of impedance series and parallel}\\ Y_{1}&Y_{0}&Y_{m0}&\text{admittance of line}\end{array} \right.,\]  
\[x\]  
Unknown distance of error from reference node  
According to Figure 1b-d and by defining by (1) to simplify by (3), positive, negative, and zero sequence voltages in fault point \(F\) are obtained, and by using [PERSON]'s transformation matrix (by (2)) as (_abc_) by (3) is produced:  
\[W=1+0.5x^{2}y_{0}z_{0}+0.5x^{2}y_{0}z_{m_{0}}, \tag{1}\]  
\[A=\left[ \begin{array}{cccc}1&1&1\\ 1&a^{2}&a\\ 1&a&a^{2}\end{array} \right], \tag{2}\]  
\[\left[ \begin{array}{cccc}V_{F_{2}}\\ V_{F_{3}}\\ V_{F_{c}}\end{array} \right]=A\left[ \begin{array}{cccc}W&0&0\\ 0&1+0.5x^{2}y_{1}z_{1}&a\\ 0&0&1+0.5x^{2}y_{1}z_{1}\end{array} \right]A^{-1}\left[ \begin{array}{cccc}V_ {S_{a}}\\ V_{S_{b}}\\ V_{S_{c}}\end{array} \right] \tag{3}\]  
Now, by defining \(E\), \(G\), \(H\), and \(L\), \(V_{F_{abc}}\) can be obtained from Equation (5):  
\[\left\{ \begin{array}{cccc}&E=ACA^{-1}V_{S_{abc}}\\ &G=AZA^{-1}I_{S_{abc-I}}\\ &H=AZ_{M}A^{-1}I_{S_{abc-II}}\\ &L=ABA^{-1}V_{S_{abc}}\end{array} \right., \tag{4}\]  
\[V_{F_{abc}}=0.5x^{2}E-x(G+H)+L. \tag{5}\]  
The most general fault model is shown in Figure 2, which models faults such as single phase to the ground, two phases to the ground, and three phases to the ground. In this section, a quadratic equation is suggested, and it is proven that it can determine the distance of ground faults with a suitable accuracy. By using the equation below, the voltage at the point of the fault is calculated:  
\[\left[ \begin{array}{cccc}V_{F_{a}}\\ V_{F_{b}}\\ V_{F_{c}}\end{array} \right]=\left[ \begin{array}{cccc}Z_{F_{a}}+Z_{F_{g}}&Z_{ F_{g}}&Z_{F_{g}}\\ Z_{F_{g}}&Z_{F_{b}}+Z_{F_{g}}&Z_{F_{g}}\\ Z_{F_{g}}&Z_{F_{g}}&Z_{F_{c}}+Z_{F_{g}}\end{array} \right]\left[ \begin{array}[ ]{cccc}I_{F_{a}}\\ I_{F_{b}}\\ I_{F_{c}}\end{array} \right], \tag{6}\]where \(a\), \(b\), and \(c\) are the phases of the faults impedance \(Z_{F_{k},h,c,g}\), as shown in Figure 2. The above equation is only suitable for fault phases that have a current not equal to zero. By inserting Equation (6) in to Equation (5) for each \(k\)-phase, we will have:  
\[Z_{F_{k}}I_{F_{k}}+Z_{F_{g}}I_{F}=0.5x^{2}\;E-x(G+H)+L=M, \tag{7}\]  
where \(Z_{F_{k}}\) is the impedance and \(I_{F_{k}}\) is the current of the \(k\)-phase. As shown in Figure 2, \(I_{F}\) is the fault's current. Considering the fault's impedance as a pure resistance and by separating Equation (7) into two parts (real and imaginary), Equation (8) is obtained:  
\[\left\{