## 1 Introduction
### Deduction for Space 1  
The \(i_{R2}=2i_{F1}\) in Equation (4), i.e., the total currents of N are two times that of F. According to Table 1, considering that the total current of R1, R2, R3, R4, PW1, EW1, PW2, EW2, E is \(i_{R2}\) and the current of FW1 is \(i_{F1}\), it is reasonable that \(\sum\limits_{i=1}^{9}i_{i}+\sum\limits_{j=11}^{19}i_{j}=i_{R2}\) and \(i_{10}+i_{20}=i_{F1}\). In the synthesis of above relations, Equation (6) can be gained. \[\sum\limits_{i=1}^{9}i_{i}+\sum\limits_{j=11}^{19}i_{j}-2i_{10}-2i_{20}\ =\ 0 \tag{6}\]  
Meanwhile, owing to the fact that the unit-length voltage drop between JW1 and FW1 is equal to the unit-length voltage drop between CW1 and FW1, this can be formulated as:  
\[\Delta u_{10}\ =\ \Delta u_{20} \tag{7}\]Then, by introducing Equations (6) and (7), the subtraction of \(\Delta u_{10}\), \(\Delta u_{20}\) is performed and the current ratio of N and F is introduced in Equation (5). Equation (5) is transformed into Equation (8), where \(a_{i}=l_{10:i}-l_{20:i}\) (\(i=1\), \(2\), \(\cdots\), \(20\)) in the tenth row of inductance matrix. \[\left[ \begin{array}{c}\Delta u_{1}\\ \vdots\\ \Delta u_{9}\\ 0\\ \Delta u_{11}\\ \vdots\\ \Delta u_{19}\\ 0\\ \end{array} \right]\ =\ j\omega\left[ \begin{array}{cccccccc}l_{1-1}&\cdots&l_{1-9}&l_ {1-10}&l_{1-11}&\cdots&l_{1-19}&l_{1-20}\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ l_{9-1}&\cdots&l_{9-9}&l_{9-10}&l_{9-11}&\cdots&l_{9-19}&l_{9-20}\\ a_{1}&\cdots&a_{9}&a_{10}&a_{11}&\cdots&a_{19}&a_{20}\\ l_{11-1}&\cdots&l_{11-9}&l_{11-10}&l_{11-11}&\cdots&l_{11-19}&l_{11-20}\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ l_{19-1}&\cdots&l_{19-9}&l_{19-10}&l_{19-11}&\cdots&l_{19-19}&l_{19-20}\\ 1&\cdots&1&-2&1&\cdots&1&-2\\ \end{array} \right]\left[ \begin{array}{c}i_{1}\\ \vdots\\ i_{9}\\ i_{10}\\ i_{11}\\ \vdots\\ i_{19}\\ i_{20}\\ \end{array} \right] \tag{8}\]  
Assuming that the total current of transmission conductors or return conductors is \(i_{0}\) and the current distribution coefficient of each loop circuit of Space 1 is \(k_{i}\) (\(i\)=1, 2... 20), \(i_{i}=k_{i}i_{0}\) is gained. \(\Delta u_{1}\), \(\Delta u_{2}\)... \(\Delta u_{9}\), \(\Delta u_{11}\), \(\Delta u_{12}\)... \(\Delta u_{19}\) are the unit-length potential difference between T and N and they can be seen as equal to each other. In view of this, \(\Delta u_{1}=\ldots\ =\Delta u_{9}=\Delta u_{11}=\ldots\ =\Delta u_{19}=c_{1}\) (\(c_{1}\) is arbitrary constant) is set. Then, focusing on Equation (8), by performing an inverse matrix transformation and introducing \(i_{i}=k_{i}i_{0}\) and \(\Delta u_{1}=\ldots\ =\Delta u_{9}=\Delta u_{11}=\ldots\ =\Delta u_{19}=c_{1}\), Equation (8) is transformed into Equation (9). \[\left[ \begin{array}{c}k_{1}i_{0}\\ \vdots\\ k_{9}i_{0}\\ k_{10}i_{0}\\ k_{11}i_{0}\\ \vdots\\ k_{19}i_{0}\\ k_{20}i_{0}\\ \end{array} \right]\ =\ \frac{1}{j\omega}\left[ \begin{array}{cccccccc}l_{1-1}& \cdots&l_{1-9}&l_{1-10}&l_{1-11}&\cdots&l_{1-19}&l_{1-20}\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ l_{9-1}&\cdots&l_{9-9}&l_{9-10}&l_{9-11}&\cdots&l_{9-19}&l_{9-20}\\ a_{1}&\cdots&a_{9}&a_{10}&a_{11}&\cdots&a_{19}&a_{20}\\ l_{11-1}&\cdots&l_{11-9}&l_{11-10}&l_{11-11}&\cdots&l_{11-19}&l_{11-20}\\ \vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ l_{19-1}&\cdots&l_{19-9}&l_{19-10}&l_{19-11}&\cdots&l_{19-19}&l_{19-20}\\ 1&\cdots&1&-2&1&\cdots&1&-2\\ \end{array} \right]^{-1}\left[ \begin{array}{c}c_{1}\\ \vdots\\ c_{1}\\ 0\\ c_{1}\\ \vdots\\ c_{1}\\ 0\\ \end{array} \right] \tag{9}\]  
For Equation (9), after calculating the coefficients of the inductance matrix and eliminating \(i_{0}\) in the first column, the values of \(k_{1}\), \(k_{2}\)\(\ldots\)\(k_{20}\), are deduced. Subsequently, by introducing \(i_{i}=k_{i}i_{0}\) into Equation (5), it can be found that,  
\[\left[ \begin{array}{c}\Delta u_{1}\\ \Delta u_{2}\\ \vdots\\ \Delta u_{i}\\ \vdots\\ \Delta u_{20}\\ \end{array} \right]\ =\ \left[