## 3 Compensation of Unbalanced Three-Phase 4-Wire Linear Systems
### ZSCC Compensator with Unbalanced Voltages  
If the voltages in the bus in Figure 3 are unbalanced, then in this case, two situations with different behaviors are possible:  
* When the modules of the line-to-neutral voltages are different, but their phase angles are offset \(\pm 120\) degrees from each other. * When the phase angles are not offset \(\pm 120\) degrees from each other. When the voltage imbalance is only because the modules of the line-to-neutral voltages in each phase are not equal, the behavior of the ZSCC compensator is identical to that set out above for balanced voltages. The decomposition of currents \(\underline{l_{AW}}\), \(\underline{l_{BW}}\) and \(\underline{l_{CW}}\) into symmetric components, gives rise to a system of zero-sequence currents and a system of negative-sequence currents. As with balanced voltages, the ZSCC compensator does not consume positive-sequence currents. Therefore, the reactive power in each phase consumed by the ZSCC compensator will also be equal to twice the reactive power caused by the line-to-neutral voltage and zero-sequence current of the load, but of the opposite sign. Hence, the values of the compensator reactances calculated from Equation (25) are perfectly valid for this type of situation. On the contrary, when the phase angles of the line-to-neutral voltages are not offset \(\pm 120\) degrees, regardless of the RMS value of the voltages, when decomposing the currents \(\underline{l_{AW}}\), \(\underline{l_{BW}}\) and \(\underline{l_{CW}}\), in addition to the system of zero-sequence currents and from the negative-sequence current system, a positive-sequence current system is obtained that is not null. Therefore, the currents \(\underline{l_{zW}}\) consumed by the compensator are determined by Equation (26). All this implies that Equation (18) is not valid when the phase angles of the voltages are not compensated \(\pm 120\) degrees from each other, and as a consequence, Equation (25) is also not valid. Moreover, if we calculate the compensator reactances at unbalanced voltages using Equation (25) and analyze the system, we observe that the value of the zero-sequence current consumed by the ZSCC compensator is different from that necessary to compensate for the load. A new zero-sequence current different from the desired one appears and must be taken into account to perform the calculation again. This procedure will be performed iterating indefinitely until a valid solution is obtained. The number of iterations required will depend on the degree of voltage imbalance and the accuracy that is desired. \[\underline{l_{zW}}=\underline{l_{zW+}}+\underline{l_{zW-}}+\underline{l_{zW 0}} \tag{26}\]  
To solve this problem, the authors present an exact calculation procedure that makes it unnecessary to use iterative methods. To do this, we will use the zero-sequence currents consumed by the load with the opposite sign \(-\underline{l_{zL0}}\) and the zero-sequence currents consumed by the ZSCC compensator obtained from the values of the reactances according to Equation (25), which we will call \(\underline{l_{zW0(25)}}\) for \(z=\{a,b,c\}\). If we define the phasor of deviation of the zero-sequence current \(\underline{f_{W0}}\) as the quotient between the phasors of both currents, Equation (27) is obtained; where, \(A\) is the real part of \(\underline{f_{W0}}\) and \(B\) is the imaginary part of \(\underline{f_{W0}}\). Obviously, \(\underline{f_{W0}}\) is identical in each of the phases. When the angles of the line-to-neutral voltages are \(\pm 120\) degrees out of phase, then \(A=1\) and \(B=0\). \[\underline{f_{W0}}=\frac{-\underline{l_{zL0}}}{\underline{l_{zW0(25)}}}=\frac {-\underline{l_{zL0}}}{\underline{l_{zW0(25)}}}=\frac{-\underline{l_{zL0}}}{ \underline{l_{CW0(25)}}}=A+jB \tag{27}\]where \(I_{aW0(25)}=I_{bW0(25)}=I_{cW0(25)}\) is given by Equation (28). Here, \(\underline{X_{aW(25)}}\), \(\underline{X_{aW(25)}}\) and \(\underline{X_{aW(25)}}\) are the ZSCC compensator reactances calculated from Equation (25).