## 4 Unbalanced Power Vectors Proposed  
In a linear three-phase system balanced in voltages and currents, the negative sequence and zero sequence currents and voltages are null. Under these conditions, the total apparent power of the system is determined exclusively from the positive-sequence currents and voltages. This apparent power is called the apparent positive-sequence power \(S_{+}\). Positive-sequence active power \(P_{+}\) and positive-sequence reactive power \(Q_{+}\) are calculated from (16) and (17). If we consider these powers in each phase, the following conditions are satisfied:  
\[P_{a+}^{a+}=P_{b+}^{b+}=P_{c+}^{c+}=\frac{P_{+}}{3} \tag{33}\]  
\[Q_{a+}^{a+}=Q_{b+}^{b+}=Q_{c+}^{c+}=\frac{Q_{+}}{3} \tag{34}\]  
In an unbalanced linear three-phase system [10], the only apparent power that is not due to the imbalance of the voltages and currents is the positive sequence apparent power \(S_{+}\); the rest of the apparent powers are caused by imbalances, representing the unbalanced apparent power. As in the previous section, unbalanced power is defined from three vectors: vector of unbalanced apparent power of positive-sequence voltage \(\overrightarrow{S_{uv+}}\), vector of unbalanced apparent power of negative-sequence voltage \(\overrightarrow{S_{uv-}}\), and vector of unbalanced apparent power of zero-sequence voltage \(\overrightarrow{S_{uv0}}\). The vectors \(\overrightarrow{S_{uv-}}\) and \(\overrightarrow{S_{uv0}}\), according to (29) and (30), represent only the unbalanced powers; therefore, \(\overrightarrow{S_{uv-}}=\overrightarrow{S_{p-}}\) and \(\overrightarrow{S_{uv0}}=\overrightarrow{S_{v0}}\). The vector \(\overrightarrow{S_{v+}}\) includes \(\overrightarrow{S_{+}}\), which is not an unbalanced power; therefore, to obtain \(\overrightarrow{S_{uv+}}\), the corresponding values of \(P_{+}\) and \(Q_{+}\), which according to (33) and (34) are equal to one-third of its value, must be subtracted in each of its components, as in (35), where \(z=\{a,b,c\}\). \[\overrightarrow{S_{uv+}}=\sqrt{3}\left[\sum\left(P_{z+}^{z}-\frac{P_{+}}{3} \right)\overrightarrow{u_{pz+}}+\sum\left(Q_{z+}^{z}-\frac{Q_{+}}{3}\right) \overrightarrow{u_{qz+}}\right] \tag{35}\]  
The modulus of unbalanced apparent power is given by  
\[S_{uv+}=\sqrt{3}\sqrt{\sum\left(P_{z+}^{z}-\frac{P_{+}}{3}\right)^{2}+\sum \left(Q_{z+}^{z}-\frac{Q_{+}}{3}\right)^{2}} \tag{36}\]The total unbalanced apparent power \(S_{u}\) according to (37) is calculated from the vector moduli \(\overrightarrow{S_{uv+}},\overrightarrow{S_{uv-}}\), and \(\overrightarrow{S_{uv0}}\). \[S_{u}=\sqrt{S_{uv+}{}^{2}+S_{uv-}{}^{2}+S_{uv0}{}^{2}} \tag{37}\]  
Therefore, the apparent total power is given by (38). \[S_{1}=\sqrt{S_{+}{}^{2}+S_{uv+}{}^{2}+S_{uv-}{}^{2}+S_{uv0}{}^{2}} \tag{38}\]  
Additionally, if we express the line currents in symmetric components, the total apparent power is determined by (39). \[S_{1}=\sqrt{S_{+}{}^{2}+S_{+}{}^{2}+S_{+}{}^{0}{}^{2}+S_{-}{}^{2}+S_{-}{}^{+}{ }^{2}+S_{0}{}^{2}+S_{0}{}^{+}{}^{2}+S_{0}{}^{-}{}^{2}} \tag{39}\]  
Following the same procedure as in previous sections, the apparent powers of any of the nine addends of the square root can be expressed by means of its vector expression according to (40) and its moduli from (41). Here \(w=\{+,-,0\}\) indicates the symmetric component of the voltage and \(x=\{+,-,0\}\) indicates the symmetric component of the current. When the symmetrical components of voltage and current are the same in (39), we have indicated it only with the subscript, therefore in (40) and (41); \(S_{+}=S_{+}^{+}\), \(S_{-}=S_{-}^{-}\) and \(S_{0}=S_{0}^{0}\). On the other hand, \(\overrightarrow{u_{pxx}^{\pm}}\) and \(\overrightarrow{u_{qxx}^{\pm}}\) are the unit vectors associated with each active and reactive component, respectively.