GATE 2018 EE Q.20: solution and answer

GATE 2018 · EEQ.20 · 1 mark · NAT

The series impedance matrix of a short three-phase transmission line in phase coordinates is [[Z_s, Z_m, Z_m],[Z_m, Z_s, Z_m],[Z_m, Z_m, Z_s]]. If the positive sequence impedance is (1 + j10) Ω, and the zero sequence is (4 + j31) Ω, then the imaginary part of Z_m (in Ω) is ______ (up to 2 decimal places).

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Answer and solution

ANSWER

The imaginary part of $Z_{m}$ is $7.00 Ω$ .

Official answer: 7.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

The matrix has $Z_{s}$ on the diagonal and $Z_{m}$ in every off-diagonal position, so the line is fully symmetric (balanced mutual coupling). Its sequence impedances are found by the symmetrical-component transformation $Z_{012} = A^{- 1} Z_{ab c} A$ , which gives a diagonal matrix.
Positive and negative sequence: $Z_{1} = Z_{2} = Z_{s} - Z_{m} = (1 + j 10) Ω$ .
Zero sequence: $Z_{0} = Z_{s} + 2 Z_{m} = (4 + j 31) Ω$ .
Subtract the first from the second so that $Z_{s}$ cancels: $Z_{0} - Z_{1} = 3 Z_{m}$ .
$Z_{m} = \frac{( 4 + j 31 ) - ( 1 + j 10 )}{3} = \frac{3 + j 21}{3} = (1 + j 7) Ω$
The imaginary part of $Z_{m}$ is $7.00 Ω$ .

KEY IDEA

For a symmetric line matrix with $Z_{s}$ on the diagonal and $Z_{m}$ off it, $Z_{1} = Z_{2} = Z_{s} - Z_{m}$ and $Z_{0} = Z_{s} + 2 Z_{m}$ . Hence $Z_{m} = (Z_{0} - Z_{1}) /3$ , and $Z_{s}$ cancels.

WATCH OUT FOR

Writing $Z_{0} = Z_{s} + Z_{m}$ , or $Z_{1} = Z_{s} + Z_{m}$ , instead of $Z_{s} + 2 Z_{m}$ and $Z_{s} - Z_{m}$ . Another slip is dividing the difference by 2 rather than 3, or taking the real part instead of the imaginary part.

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GATE 2018 EE Q.20: solution and answer

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