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A 3-phase transmission line is shown in the figure. The voltage drop across the transmission line is given by the following equation: $\begin{bmatrix}\Delta V_a\\ \Delta V_b\\ \Delta V_c\end{bmatrix}=\begin{bmatrix}Z_s&Z_m&Z_m\\ Z_m&Z_s&Z_m\\ Z_m&Z_m&Z_s\end{bmatrix}\begin{bmatrix}I_a\\ I_b\\ I_c\end{bmatrix}$. Shunt capacitance of the line can be neglected. If the line has positive-sequence impedance of $15\,\Omega$ and zero-sequence impedance of $48\,\Omega$, then the values of $Z_s$ and $Z_m$ will be

1. $Z_s = 31.5\,\Omega;\ Z_m = 16.5\,\Omega$
2. $Z_s = 26\,\Omega;\ Z_m = 11\,\Omega$
3. $Z_s = 16.5\,\Omega;\ Z_m = 31.5\,\Omega$
4. $Z_s = 11\,\Omega;\ Z_m = 26\,\Omega$

Correct answer: $Z_s = 26\,\Omega;\ Z_m = 11\,\Omega$

Solution

For a transposed 3-phase line, the phase impedance matrix has diagonal terms $Z_s$ and mutual terms $Z_m$. The sequence impedances satisfy $Z_1 = Z_s - Z_m$ and $Z_0 = Z_s + 2Z_m$. Solving $Z_s - Z_m = 15$ and $Z_s + 2Z_m = 48$ gives $Z_m = 11\,\Omega$ and $Z_s = 26\,\Omega$.

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