## 2 Preliminaries and Basic Idea
### Circuit Analysis Single--Phasing Fault  
When one of the three phases is opened, as shown in Figure 1, single phasing occurs. In electrical analysis, sequence components are typically used to analyze unbalanced conditions [2]. The terminal voltages are resolved into positive, negative, and zero sequence components. As a result, the phase voltages at the three-phase induction motor's terminals are given in matrix form, as shown in Equation (1).  
\[\begin{bmatrix}V_{an}\\ V_{bn}\\ V_{cn}\end{bmatrix}=\begin{bmatrix}1&1&1\\ 1&a^{2}&a\\ 1&a&a^{2}\end{bmatrix}\cdot\begin{bmatrix}V_{a0}\\ V_{a1}\\ V_{a2}\end{bmatrix};\begin{aligned} a&=1\angle 120^{\circ}\\ a^{2}&=1\angle 240^{\circ}\end{aligned} \tag{1}\]  
where \(\mathrm{V_{an}}\), \(\mathrm{V_{bn}}\), \(\mathrm{V_{cn}}\) are line to neutral voltages, and \(\mathrm{V_{a0}}\), \(\mathrm{V_{a1}}\), \(\mathrm{V_{a2}}\) are the phase voltage (zero, positive, negative) sequences, respectively. Rewriting Equation (1) using the symmetrical sequence components as follows,  
\[\begin{bmatrix}V_{a0}\\ V_{a1}\\ V_{a2}\end{bmatrix}=\frac{1}{3}\cdot\begin{bmatrix}1&1&1\\ 1&a^{2}&a\end{bmatrix}\cdot\begin{bmatrix}V_{an}\\ V_{bn}\\ V_{cn}\end{bmatrix};\begin{aligned} a&=1\angle 120^{\circ}\\ a^{2}&=1\angle 240^{\circ}\end{aligned} \tag{2}\]  
Figure 1: Single phasing in three-phase induction motor.  
For a single-phasing fault in Line-\(a\), the line currents must satisfy the following:  
\[I_{a}=0,\text{ and }I_{c}=-I_{b} \tag{3}\]  
Therefore, the matrix form of the symmetrical sequence components of the phase-\(a\) currents becomes the following:  
\[\begin{bmatrix}Ia_{0}\\ Ia_{1}\\ Ia_{2}\end{bmatrix}=\frac{1}{3}\cdot\begin{bmatrix}1&1&1\\ 1&a&a^{2}\\ 1&a^{2}&a\end{bmatrix}\cdot\begin{bmatrix}0\\ I_{b}\\ -I_{b}\end{bmatrix}=\frac{1}{3}\cdot\begin{bmatrix}0\\ (a-a^{2})\cdot I_{b}\\ (a^{2}-a)\cdot I_{b}\end{bmatrix}=\begin{bmatrix}0\\ j\frac{1}{\sqrt{3}}\cdot I_{b}\\ j\frac{1}{\sqrt{3}}\cdot I_{b}\end{bmatrix};\begin{aligned} a=1\angle 120^{\circ}\\ a^{2}=1\angle 240^{\circ}\end{aligned} \tag{4}\]  
where \(\mathrm{I}_{a}\), \(\mathrm{I}_{b}\), \(\mathrm{I}_{c}\) are the three-phase currents and \(\mathrm{Ia}_{0},\mathrm{Ia}_{1},\mathrm{Ia}_{2}\) are the phase current (zero, positive, negative) sequences, respectively. Equation (4) shows that the zero-sequence current will not flow through a faulty phase, and \(Ia_{1}=-Ia_{2}\). As a result of the analysis in Figure 2, the input voltages of the positive and negative sequences under single-phasing conditions (5) and (6) can be calculated.  
In Figure 2, the sequence input voltages are given by the following:  
\[V_{in}=Va_{1}-Va_{2} \tag{5}\] \[Va_{1}=\frac{1}{3}\cdot(V_{an}+a\cdot V_{bn}+a^{2}\cdot V_{cn})\] \[Va_{2}=\frac{1}{3}\cdot(V_{an}+a^{2}\cdot V_{bn}+a\cdot V_{cn})\] \[V_{in}=\frac{1}{3}\cdot(V_{an}-V_{an}+(a-a^{2})\cdot V_{bn}+(a^{ 2}-a)\cdot V_{cn})\] \[V_{in}=j\frac{V_{bc}}{\sqrt{3}} \tag{6}\]  
where \(\mathrm{V}_{in}\) is the input voltage. Equation (6) depicts a direct relationship between the induction motor's input voltage and the actual line-to-line voltage of the electric grid.  
Based on Figure 2, the sequence currents \(\mathrm{I}_{a1}\) and \(\mathrm{I}_{a2}\) can be found as follows [2]:  
\[Z_{\mathrm{eq},\mathrm{i}}=\mathrm{Rs}+\mathrm{j}\mathrm{Xs}+\frac{\mathrm{Rr }/\mathrm{s}+\mathrm{j}\mathrm{Xr}}{\mathrm{j}\mathrm{Xm}+\mathrm{Rr}/\mathrm{ s}+\mathrm{j}\mathrm{Xr}} \tag{7}\]  
where \(\mathrm{i}=1\), \(2\) for positive and negative sequences, respectively.  
The stator sequence currents are as follows:  
\[I_{s,1}=\frac{Vin}{Z_{eq,1}+Z_{eq,2}},\quad I_{s,2}=-I_{s,1}\;,\;\;\;\;\;I_{s, 0}=0\]  
Figure 2: Positive and negative sequence network for single phasing.  
And the rotor sequence currents are as follows:  
\[I_{r,i}=I_{s,1}\times\frac{jXm}{j\,Xm+\mathrm{Rr}/\mathrm{s}+\mathrm{jXr}} \tag{8}\]