## 2 BCC--From Load Balancing to Power Flow Balancing
\begin{cases}\begin{split}&\underline{Im}\left(\underline{I}^{s+} \right)=0\\ &\underline{Re}\left(\underline{I}^{s-}\right)=0\\ &\underline{Im}\left(\underline{I}^{s-}\right)=0\end{split}\end{cases}  \tag{5}\]  
It results in several forms of expression of the relations for calculating the values of the three susceptances, depending on the quantities known in the section controlled by the BCC (Figure 1), of which the following are noted here [26]:  
* depending on the values of the three equivalent susceptances of the load:  
Figure 1: BCC (BRC) for load balancing in a three-phase three-wire network. \[ \begin{bmatrix}B_{ab}^{compp}\\ B_{bc}^{Bump}\\ B_{ca}^{Bump}\end{bmatrix} = \begin{bmatrix}-1&0&0\\ 0&-1&0\\ 0&0&-1\end{bmatrix} \cdot \begin{bmatrix}B_{ab}^{load}\\ B_{bc}^{load}\\ B_{ca}^{load}\end{bmatrix} +\frac{1}{\sqrt{3}} \begin{bmatrix}0&1&-1\\ -1&0&1\\ 1&-1&0\end{bmatrix} \cdot \begin{bmatrix}G_{ab}^{load}\\ G_{bc}^{load}\\ G_{ca}^{load}\end{bmatrix}  \tag{6}\]  
* depending on the reactive components of the currents on the load phases: \[ \begin{bmatrix}B_{ab}^{compp}\\ B_{bc}^{Bump}\\ B_{ca}^{Bump}\end{bmatrix} =\frac{1}{3U} \begin{bmatrix}-1&-1&1\\ 1&-1&-1\\ -1&1&-1\end{bmatrix}\begin{bmatrix}I_{tx}^{load}\\ I_{tx}^{load}\end{bmatrix} \] (7)
* depending on the reactive power components in the load phases: \[ \begin{bmatrix}B_{ab}^{compp}\\ B_{ca}^{Bump}\end{bmatrix} =\frac{1}{3U^{2}}\cdot \begin{bmatrix}-1&-1&1\\ 1&-1&-1\\ -1&1&-1\end{bmatrix} \cdot \begin{bmatrix}Q_{a}^{load}\\ Q_{b}^{load}\\ Q_{c}^{load}\end{bmatrix} \] (8)
* depending on the real and imaginary parts of the sequence components of the load currents: \[ \begin{bmatrix}B_{ab}^{compp}\\ B_{bc}^{Bump}\\ B_{ca}^{Bump}\end{bmatrix} =\frac{1}{3U}\cdot \begin{bmatrix}1&-\sqrt{3}&1\\ 1&0&-2\\ 1&\sqrt{3}&1\end{bmatrix} \cdot \begin{bmatrix}Im\left(\underline{I}^{load+} \right)\\ \operatorname{Re}\left(\underline{I}^{load-}\right)\\ Im\left(\underline{I}^{load-}\right)\end{bmatrix} \] (9)  
In (6)-(9) with \(U\), the rms value of the phase voltage belonging to the symmetrical three-phase set provided by the source was noted. The way in which the BCC, containing only reactive circuit elements, achieves the balancing of the active power on the network phases can be deduced from the expressions of the active powers flowing on the BCC phases, written according to the active powers on the load phases:  
\[ \begin{bmatrix}p_{a}^{compp}\\ p_{b}^{comp}\end{bmatrix} =\frac{1}{3}\cdot \begin{bmatrix}-2&1&1\\ 1&-2&1\\ 1&1&-2\end{bmatrix} \cdot \begin{bmatrix}P_{a}^{load}\\ P_{b}^{load}\end{bmatrix}  \tag{10}\]  
According to [26], if it is noted with \(P_{av}^{load}\) the arithmetic mean of the active powers on the load phases,  
\[p_{av}^{load}=\frac{1}{3}\Big{(}P_{a}^{load}+P_{b}^{load}+P_{c}^{load}\Big{)} \tag{11}\]  
(10) can be written in the form:  
\[ \begin{split} P_{a}^{comp}=&\ P_{av}^{load}-P_{a}^{load}\\ P_{b}^{comp}=&\ P_{av}^{load}-P_{b}^{load}\\ P_{c}^{comp}=&\ P_{av}^{load}-P_{c}^{load}\end{split}  \tag{12}\]  
It is observed that:  
\[P_{a}^{comp}+P_{b}^{comp}+P_{c}^{comp}=0 \tag{13}\]  
which means that BCC does not change the active power flow on the whole of the three phases. But on one of the phases (or on two phases) BCC takes active power (\(P>0\)) from the network, which it then feeds back into the network on the other two phases (or on the third phase). The value of the active power taken or discharged on each BCC phase is equal to the algebraic difference between the value of the active power of the load on the respective phase and the average value of the active power on the load phases (12). This is how the active powers are equalized in the phases of the load-BCC assembly:  
\[ \begin{split} P_{a}^{s}&=P_{a}^{load}+P_{a}^{comp}=P_{av }^{load}\\ P_{b}^{s}&=P_{b}^{load}+P_{b}^{comp}=P_{av}^{load}\\ P_{c}^{s}&=P_{c}^{load}+P_{c}^{comp}=P_{av}^{load} \end{split}