## 1 Introduction
### Use of Source and Load Neutral Points as Voltage Reference Points  
Let us consider the three-phase four-wire power system illustrated in Figure 2. If at least one of its subsystems, the source or load, is unbalanced and/or distorted and the neutral conductor impedance is not negligible, a voltage drop could occur between the source and load neutral points (N and n), the value of which can be determined using [PERSON]'s second law. This voltage is the neutral-point displacement voltage [3] and is responsible for the variation of the measured source and load instantaneous powers with the voltage reference point.  
The load instantaneous power of the system in Figure 2 can be expressed as follows:  
\[p_{l}(t)=v_{An}i_{A}+v_{Bn}i_{B}+v_{Cn}i_{C}. \tag{1}\]  
The value given by Equation (1) would be registered by a power analyzer placed at the PCC (A, B, or C) of the system in Figure 2 if the load neutral point (n) is used as the voltage reference. According to [PERSON]'s second law, the line-to-neutral instantaneous voltages in the load phases (\(v_{An},\ v_{Bn},\) and \(v_{Cn}\)) can be expressed as functions of the source line-to-neutral instantaneous voltages (\(v_{AN},\ v_{BN},\) and \(v_{CN}\)) and the neutral-point displacement voltage (\(v_{nN}\)), as follows:  
\[\begin{array}{l}v_{An}=v_{AN}-v_{nN},\\ v_{Bn}=v_{BN}-v_{nN},\\ v_{Cn}=v_{CN}-v_{nN}.\end{array} \tag{2}\]  
From Equations (1) and (2), we obtain  
\[p_{l}(t)=v_{AN}i_{A}+v_{BN}i_{B}+v_{CN}i_{C}-v_{nN}\cdot(i_{A}+i_{B}+i_{C}). \tag{3}\]  
According to [PERSON]'s theorem [31], the neutral-point displacement instantaneous voltage can be expressed as the difference between the zero-sequence line-to-neutral instantaneous voltages of the source (\(v_{AN0}\)) and load (\(v_{An0}\)),  
\[v_{nN}=v_{AN0}-v_{An0}, \tag{4}\]  
both for the fundamental frequency and harmonics, i.e.,  
\[v_{AN0}=\frac{1}{3}\sum_{x=1}^{\infty}(v_{ANx}+v_{BNx}+v_{CNx})\quad v_{An0}= \frac{1}{3}\sum_{y=1}^{\infty}\Bigl{(}v_{Any}+v_{Bny}+v_{Cny}\Bigr{)}. \tag{5}\]  
The load instantaneous power can, thus, be expressed as  
\[p_{l}(t)=v_{AN}i_{A}+v_{BN}i_{B}+v_{CN}i_{C}+(v_{An0}-v_{AN0})\cdot(i_{A}+i_{B }+i_{C})=p_{s}(t)+p_{n}(t). \tag{6}\]  
Figure 2: Power analyzer with the load neutral point as the voltage reference point.  
The first three terms in Equation (6) represent the source instantaneous power.  
\[p_{s}(t)=v_{AN}i_{A}+v_{BN}i_{B}+v_{CN}i_{C}. \tag{7}\]  
The value given by Equation (6) would be registered by a power analyzer placed at a PCC of the system in Figure 2 if the source neutral point (\(N\)) is used as the voltage reference point.  
Because the voltage \(v_{nN}\), given by Equation (4), determines the displacements of the source and load neutral points [3], the term in Equation (6),  
\[p_{n}(t)=(v_{An0}-v_{AN0})\cdot(i_{A}+i_{B}+i_{C})=p_{l}(t)-p_{s}(t), \tag{8}\]  
should be the neutral-displacement instantaneous power, which defines the load and source instantaneous power differences. This power is necessary to satisfy the conservation of energy principle and identifies the phenomenon associated with the variation of the apparent power measured by a power analyzer with the voltage reference point (n or N).  
From Equation (8), two components of the neutral-displacement instantaneous power can be defined, namely, the load component,  
\[p_{nl}(t)=v_{An0}\cdot(i_{A}+i_{B}+i_{C}), \tag{9}\]  
and the source component,  
\[p_{ns}(t)=v_{AN0}\cdot(i_{A}+i_{B}+i_{C}). \tag{10}\]  
These two components of the neutral-displacement instantaneous power satisfy the equation  
\[p_{n}(t)=p_{nl}(t)-p_{ns}(t). \tag{11}\]  
Hence, the instantaneous power difference between the source and load of the system in Figure 2 is exclusively due to the different neutral-displacement instantaneous powers of the two subsystems.