## 2 Unbalanced Power Analysis in a Three-Phase Linear System  
Figure 1 displays a linear unbalanced load connected to a three-phase system with three-wire unbalanced line-to-line voltages.  
The instantaneous power p of the system is determined by (1), where \(v_{ab}\), \(v_{bc}\) and \(v_{ca}\) are the line-to-line voltages on the bus, and \(i_{a}\), \(i_{b}\) and \(i_{c}\) are the line currents.  
\[p=v_{ab}i_{a}-v_{bc}i_{c}=v_{bc}i_{b}-v_{ca}i_{a}=v_{ca}i_{c}-v_{ab}i_{b} \tag{1}\]  
Figure 1: Three-phase three-wire system with an unbalanced load.  
For example, considering the first equality of Equation (1) and expressing these values in complex form, we can determine the active power \(P\), reactive power \(Q\), and apparent power of the system \(S\) using Equations (2)-(4), respectively.  
\[P=V_{ab}I_{a}cos(\alpha_{ab}-\beta_{a})-\ V_{bc}I_{c}\ cos(\alpha_{bc}-\beta_{c}) \tag{2}\]  
\[Q=V_{ab}I_{a}sin(\alpha_{ab}-\beta_{a})-\ V_{bc}I_{c}\ sin(\alpha_{bc}-\beta_{c}) \tag{3}\]  
\[S=\sqrt{p^{2}+Q^{2}} \tag{4}\]  
Using the Fortescue transformation matrix, the line-to-line voltages and line currents in terms of symmetric components are expressed in Equations (5) and (6). As it is a three-wire system, the zero-sequence line-to-line voltage and line current are zero.  
\[\frac{V_{ab}}{V_{ab}}=\frac{V_{ab+}}{V_{ab-}}\ \frac{V_{bc}}{V_{bc}}=\frac{V_{ bc+}}{V_{bc-}}\ \frac{V_{ac}}{V_{ac+}}=\frac{V_{ac+}}{V_{ac-}} \tag{5}\]  
\[\frac{I_{a}}{V_{ab+}}=\underline{I_{a+}}+\underline{I_{a-}}\ \underline{I_{b}}= \underline{I_{b+}}+\underline{I_{b-}}\ \underline{I_{c}}=\underline{I_{c+}}+\underline{I_{c-}} \tag{6}\]  
Given that \(\underline{V_{ab+}}=V_{ab+}e^{i\alpha_{ab+}}\) and \(\underline{V_{ab-}}=V_{ab-}e^{i\alpha_{ab-}}\), the positive-sequence line-to-neutral voltage \(\underline{V_{a+}}\) and negative-sequence line-to-line voltage \(\underline{V_{a-}}\) are determined by Equations (7) and (8), respectively, where \(\alpha_{ab+}\) and \(\alpha_{ab-}\) are the angles referred to Phase \(A\) of both line-to-line voltages and \(\underline{a}=e^{i120}\), where \(\underline{a}\) is complex phasor rotation operator.  
\[\underline{V_{a+}}=\frac{V_{ab+}}{\sqrt{3}}e^{i(\alpha_{ab+}-30)}\ \ \underline{V_{b+}}=\underline{a}^{2}\underline{V_{a+}}\ \underline{V_{c+}}=\underline{a}\underline{V_{a+}} \tag{7}\]  
\[\underline{V_{a-}}=\frac{V_{ab-}}{\sqrt{3}}e^{i(\alpha_{ab+}-30)}\ \ \underline{V_{b-}}= \underline{a}\underline{V_{a-}}\ \underline{V_{c-}}=\underline{a}^{2}\underline{V_{a-}} \tag{8}\]  
At the measuring point (PCC or bus) and under these conditions, the total apparent power of Buchholz \(S_{T}\) is determined by Equation (9). In balanced systems, \(S\) has the same value as \(S_{T}\), while in unbalanced systems, \(S\) does not reflect the real apparent power of the system, and has a different value than \(S_{T}\).  
\[S_{T}=3\sqrt{\Big{(}V_{+}^{2}+V_{-}^{2}\Big{)}\Big{(}\underline{I_{+}^{2}+ \underline{I_{-}^{2}}}\Big{)}} \tag{9}\]  
Developing Equation (9), we obtain Equation (10).  
\[S_{T}=3\sqrt{{V_{+}}^{2}{I_{+}}^{2}+{V_{-}}^{2}{I_{-}}^{2}+{V_{+}}^{2}{I_{-}}^ {2}+{V_{-}}^{2}{I_{+}}^{2}}=\sqrt{{S_{+}}^{2}+{S_{-}}^{2}+{S_{u+}^{-}}^{2}+{S_ {u-}^{+}}^{2}} \tag{10}\]  
Here  
* \(S_{+}\) indicates the positive Substituting Equation (11) into Equation (10) and considering Equation (4), we obtain Equation (12).  
\[S_{T}=\sqrt{P^{2}+Q^{2}-2P_{+}P_{-}-2Q_{+}Q_{-}+{S_{u+}^{-}}^{2}+{S_{u-}^{+}}^{2}} \tag{12}\]