The positive, negative and zero-sequence impedances of a three-phase generator are $Z_1$, $Z_2$, and $Z_0$, respectively. For a line-to-line fault with fault impedance $Z_f$, the fault current is $I_{f1} = k I_f$, where $I_f$ is the fault current with zero fault impedance. The

Question

The positive, negative and zero-sequence impedances of a three-phase generator are $Z_1$, $Z_2$, and $Z_0$, respectively. For a line-to-line fault with fault impedance $Z_f$, the fault current is $I_{f1} = k I_f$, where $I_f$ is the fault current with zero fault impedance. The relation between $Z_f$ and $k$ is

Accepted Answer

$Z_f = (Z_1 + Z_2)(1-k)/k$. In a line-to-line fault, the positive and negative sequence networks are in series, so the fault current depends on $Z_1 + Z_2 + Z_f$. If the current with fault impedance is $k$ times the zero-impedance fault current, equating the two expressions gives the required relation. Solving yields $Z_f = (Z_1 + Z_2)(1-k)/k$.

Solution

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