Question

The series impendance matrix of a short three-phase transmission line in phase coordinates is $\left[\begin{array}{ccc}{Z}_s& Z_m& Z_m\\ Z_m& Z_s& Z_m\\ Z_m& Z_m& Z_s\end{array}\right]$. If the positive sequence impedance is (1 + j10)$\Omega$, and the zero sequence impedance is (4 + j31)$\Omega$, then the imaginary part of Z${}_m$(in $\Omega$) is (upto to 2 decimal places).

• A. 7

Answer 1

The correct option is A 7
Given that,
$Z_1$ = (1 + j10)$\Omega$

$Z_0$ = (4 + j31)$\Omega$

We know, $Z_1$ = $Z_s$ - $Z_m$

$Z_0$ = $Z_s$ + $2Z_m$

$Z_1$ - $Z_0$ = $-3Z_m$

Z${}_m$ = $\frac{Z_0-Z_1}{3}$

$\frac{4+j31-1-j10}{3}=\frac{3+j21}{3}$

= $\frac{3+j21}{3}$ = (1 + j7)

The imaginary part of Z${}_m$ is 7.00