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Discuss LL fault in detail
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Line to Line fault

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Current equation:

Phase Current relation:

$$ I_{b}=I_{f}=-I_{c} \quad I_{a}=0 $$ $\frac{\text { Symmetrical Current relation: }}{I_{b}+I_{c}=0}$ $$ \begin{aligned} &I_{b}+I_{c}=0 \\ &I_{a 0}+a^{2} I_{a 1}+a I_{a 2}+I_{a 0}+a I_{a 1}+a^{2} I_{a 2}=0 \\ &2 I_{a 0}+\left(a^{2}+a\right) I_{a 1}+\left(a^{2}+a\right) I_{a 2}=0 \\ &2 I_{a 0}-I_{a 1}-I_{a 2}=0 \quad 2 I_{a 0}=I_{a 1}+I_{a 2} \ldots .(1) \\ &I_{a}=0 \quad I_{a 0}+I_{a 1}+I_{a 2}=0 \\ &3 I_{a 0}=0 \quad I_{a 0}=0 \end{aligned} $$ Fault Current relation: $$ \begin{gathered} I_{b}=I_{f}=I_{a 0}+a^{2} I_{a 1}+a I_{a 2}=\left(a^{2}-a\right) I_{a 1} \\ I_{f}=\left(a^{2}-a\right) I_{a 1} \end{gathered} $$

$\underline{\text { Voltage equation: }}$ $$ \begin{aligned} &V_{b}=V_{c}+I_{f} Z_{f} \\ &V_{a 0}+a^{2} V_{a 1}+a V_{a 2}=V_{a 0}+a V_{a 1}+a^{2} V_{a 2}+I_{f} Z_{f} \\ &\left(a^{2}-a\right) V_{a 1}=\left(a^{2}-a\right) V_{a 2}+I_{f} Z_{f} \\ &\left(a^{2}-a\right)\left[V_{a 1}-V_{a 2}\right]=\left(a^{2}-a\right) I_{a 1} Z_{f} \\ &V_{a 1}=V_{a 2}+I_{a 1} Z_{f} \quad V_{a 0}=0 \quad \text { since } \end{aligned} $$ $V_{a 0}=0$ Since $I_{a 0}=0$

Equivalent circuit:

From the current and Voltage equation

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$$ I_{a 1}=\frac{E_{b}}{Z_{1}+Z_{2}+Z_{f}} \quad I_{f}=\frac{\left(a^{2}-a\right) E_{b}}{Z_{1}+Z_{2}+Z_{f}} $$

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