# Leakage Detection Method of Coal Mine Power Supply System Based on 3 AC
## 3 Phase Selection of Leakage Fault Based on 3 AC Detection Method  
The leakage protection devices currently in use do not have the function of fault phase selection, which does not facilitate efficient fault repair. The leakage protection method based on the 3 AC detection method can provide a criterion for selecting the fault phase. Figure 5: The simulation results of phase A leakage current. Figure 6: Relationship between amplitude and phase of \(\hat{I}_{\text{C0}}\). We replaced the \(\dot{U}_{\rm CO}\) in the case of the phase A leakage fault into the \(\dot{U}_{0}\) formula under different leakage conditions and substituted different insulation resistance values, which are taken as 1000 \(\Omega\), 2000 \(\Omega\), 10 k\(\Omega\), 20 k\(\Omega\), 24 k\(\Omega\), and 100 M\(\Omega\), respectively. Among them, the 100 M\(\Omega\) is equivalent to the case of infinite insulation resistance, that is, no insulation degradation occurs. Using the \(\dot{U}_{0}\) calculation formula of the phase A insulation drop, the \(\dot{I}_{\rm C0}\) is calculated, and then it is substituted into the calculation formula of the seven fault types, and the real and imaginary parts of \(Rd\) (phase A leakage) are obtained as shown in Tables 1 and 2. It can be observed that in all simulated leakage cases, only when phase A leakage occurs, the imaginary part of the complex result calculated by the formula is the smallest. Under the current numerical accuracy, the result is close to zero, so the absolute value of the imaginary part in this case is the smallest, and the size of the real part is consistent with the preset leakage resistance value. This is accurate feedback to the mathematical model, which verifies the effectiveness and accuracy of this method. That is, in the actual detection process, the leakage fault of a certain phase can be quickly and accurately located by finding the measurement results that have a very small imaginary part (which can be approximated to zero), and the real part matches the expected leakage resistance value. From this, we conclude the following: \(\dot{U}_{0}\) is calculated by the measured \(\dot{I}_{\rm C0}\) because the \(\dot{I}_{\rm C0}\) has been deduced in detail to obtain its calculation formula:  
\[\dot{I}_{\rm C0}=\dot{\rm z}\omega C_{2}(\dot{U}_{A}+4\dot{U}_{0}) \tag{13}\]  
therefore,  
\[\dot{U}_{0}=\frac{\dot{I}_{\rm C0}}{\dot{\rm z}\omega 4C_{2}}-\frac{\dot{U}_{A}} {4} \tag{14}\]  
Using the measured \(\dot{I}_{\rm C0}\), \(\dot{U}_{0}\) is calculated, and then the \(\dot{U}_{0}\) is substituted into the function of seven kinds of \(\dot{U}_{0}\) and insulation resistance, and the insulation resistance in the complex form is inversely derived. Comparing the 'complex resistance' of the seven calculation results, since the real resistance cannot be complex, shows that only the real part \begin{table}
\begin{tabular}{l l l l l l l l} \hline \hline \multirow{2}{*}{**Insulation Resistance**} & \multicolumn{4}{c}{**Current Leakage**} \\ \cline{2-9}  & **A Phase** & **B Phase** & **C Phase** & **AB Phase** & **BC Phase** & **AC Phase** & **ABC Phase** \\ \hline
1000 \(\Omega\) & 1000 & 3730 & 2395 & 1326 & 667 & 1697 & 111 \\
2000 \(\Omega\) & 2000 & 4397 & 1728 & 1198 & 1334 & 1864 & 222 \\
10 k\(\Omega\) & 10 \(\times\) 10\({}^{3}\) & 9730 & 3605 & 134 & 6667 & 3197 & 1111 \\
20 k\(\Omega\) & 20 \(\times\) 10\({}^{3}\) & \(1.6\times 10^{4}\) & 10,271 & 1801 & 13,334 & 4865 & 2222 \\
24 k\(\Omega\) & 24 \(\times\) 10\({}^{3}\) & \(1.9\times 10^{4}\) & 12,938 & 2468 & 16,000 & 5530 & 2667 \\
100 M\(\Omega\) & 10 \(\times\) 10\({}^{7}\) & \(66\times 10^{6}\) & \(66\times 10^{6}\) & \(16\times 10^{6}\) & \(66\times 10^{6}\) & \(16\times 10^{6}\) & \(11\times 10^{6}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Calculate the real part of \(Rd\) (phase A leakage).