## 2 Current Vector Analysis of IMs with ITF  
### Mathematical Model for ITF  
Figure 1 shows the mathematical model for an induction motor with an ITF in Phase A. Since the fault resistance is connected to the fault phase parallelly, the voltage equation is expressed as follows [34]:  
\[ \begin{array}{l}[v_{s}]=[R_{s}][i_{s}]+\frac{d}{dt}[\lambda_{s}]-[\mu R_{s} \ 0\ 0]^{T}\ i_{f}\\ R_{f}i_{f}=\mu R_{s}\Big{(}i_{as}-i_{f}\Big{)}+\frac{d}{dt}\lambda_{as2}\\ 0=[R_{r}][i_{r}]+\frac{d}{dt}[\lambda_{r}]\end{array}  \tag{1}\]  
where  
\[ \begin{array}{l}[v_{s}]=[v_{as}\ v_{bs}\ v_{cs}]^{T},\ [i_{s}]=[i_{as}\ i_{bs}\ i_{cs}]^{T}\\ [\lambda_{s}]=[L_{ss}][i_{s}]+[L_{sr}][i_{r}]+\mu[-(L_{ls}+L_{ms})\ L_{ms}/2\ L_{ms}/2]^{T}i_{f},\\ [\lambda_{r}]=[L_{rs}][i_{s}]+[L_{rr}][i_{r}]-\mu L_{ms}[\cos\theta_{e}\ \cos(\theta_{e}+2\pi/3)\ \cos(\theta_{e}-2\pi/3)]^{T}i_{f},\\ \lambda_{as2}=-\mu[-(L_{ls}+L_{ms})\ L_{ms}/2\ L_{ms}/2][i_{s}]\\ +\mu L_{ms}[\cos\theta_{r}\ \cos(\theta_{r}-2\pi/3)\ \cos(\theta_{r}+2\pi/3)][ i_{r}]-\mu(L_{ls}+\mu L_{ms})i_{f},\end{array} \]  
\(v_{as}\), \(v_{bs}\), and \(v_{cs}\) are the phase voltages; \(i_{as}\), \(i_{bs}\), and \(i_{cs}\) are the phase currents; \(i_{s}\) is the rotor current; \(i_{f}\) is the fault current; \(\lambda_{s}\) and \(\lambda_{r}\) are the flux linkages of the stator and rotor, respectively; \(L_{ss}\) and \(L_{rr}\) are the self-inductances of the stator and rotor, respectively; \(L_{sr}\) and \(L_{rs}\) are the mutual inductances between the stator and rotor; \(R_{s}\) and \(R_{r}\) are the resistances of the stator and rotor, respectively; \(L_{ls}\) and \(L_{ms}\) are the leakage and magnetize inductance, respectively; \(\mu=N_{fault}/N_{ph}\) is the ratio of the number of turns in the fault winding; and \(\theta_{e}\) is the electrical angle. Figure 1: Configuration of ITF in IM with \(\Delta\) connection. To analyze the current vector pattern in the stationary coordinate frame (the \(\alpha\beta\)-axis plane), the coordinate transformation is considered as illustrated in Figure 2. From Figure 2, the coordinate transformation can be expressed as follows:  
\[\left[C_{abc\to a\beta}\right]=\frac{2}{3} \begin{bmatrix}1&-\frac{1}{2}&-\frac{1 }{2}\\ 0&\frac{\sqrt{3}}{2}&-\frac{\sqrt{3}}{2}\end{bmatrix} ,\ \ \left[C_{a\beta\to abc}\right]= \begin{bmatrix}1&-\frac{1}{2}&-\frac{1}{2}\\ 0&\frac{\sqrt{3}}{2}&-\frac{\sqrt{3}}{2}\end{bmatrix} ^{T} \tag{2}\]  
where \(\left[C_{abc\to a\beta}\right]\) is the coordinate transformation from abc to \(\alpha\beta\), and \(\left[C_{a\beta\to abc}\right]\) is the coordinate transformation from \(\alpha\beta\) to abc. From Equation (2), three-phase components are transformed into stationary coordinate frame as shown in Equation (3). \[\left[f_{a\beta}\right]=\left[C_{abc\to a\beta}\right]\left[f_{abc}\right],\ \left[f_{abc}\right]=\left[C_{a\beta\to abc}\right]\left[f_{a\beta}\right] \tag{3}\]  
where \(\left[f_{abc}\right]\) is the physical value, such as the voltage and current, in \(abc\), and \(\left[f_{a\beta}\right]\) is the physical value, such as the voltage and current, in \(\alpha\beta\). By subtracting Equation (3) from Equation (1), the voltage equations in the stationary coordinate frame can be obtained using the stationary coordination transformations:  
\[ \begin{array}{l}\left[v_{a\beta s}\right]=\left[R_{s}\right]\left[i_{a\beta s }\right]+\frac{d}{dt}\left[\lambda_{a\beta s}\right]-\frac{2}{3}\mu R_{s}if \\ 0=\left[R_{r}\right]\left[i_{a\beta r}\right]+\frac{d}{dt}\left[\lambda_{a\beta r }\right]-j\omega_{e}\left[\lambda_{a\beta r}\right]\end{array}  \tag{4}\]  
where  
\[\left[\lambda_{a\beta s}\right]=\left[L_{s}\right]\left[i_{a\beta s}\right]+ \left[L_{m}\right]\left[i_{a\beta r}\right]-\frac{2}{3}\mu\left[ \begin{array}[ ]{c}L_{s}\\ 0\end{array} \right]if_{r},\ \left[\lambda_{a\beta r}\right]=\left[L_{m}\right]\left[i_{a \beta s}\right]+\left[L_{r}\right]\left[i_{a\beta r}\right]-\frac{2}{3}\mu \left[ \begin{array}{c}L_{m}\\ 0\end{array}