GATE 2018 EE Q.28: solution and answer

GATE 2018 · EEQ.28 · 2 marks · MCQ

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

(A)
Z_f = (Z_1+Z_2)(1-k)/k
(B)
Z_f = (Z_1+Z_2)(1+k)/k
(C)
Z_f = (Z_1+Z_2)k/(1-k)
(D)
Z_f = (Z_1+Z_2)k/(1+k)

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Answer and solution

ANSWER

Option (A): $Z_{f} = \frac{( Z _{1} + Z _{2} ) ( 1 - k )}{k}$

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

For a line-to-line fault between phases $b$ and $c$ through a fault impedance $Z_{f}$ , the zero-sequence current is zero, so $Z_{0}$ plays no part. The positive- and negative-sequence networks are connected in series with $Z_{f}$ : $I_{a 1} = - I_{a 2} = \frac{E}{Z _{1} + Z _{2} + Z _{f}}$
The faulted-phase current is $I_{b} = - I_{c} = - j 3 I_{a 1}$ . This is a fixed multiple of $I_{a 1}$ whatever the value of $Z_{f}$ . So the ratio of the fault current with $Z_{f}$ to the fault current with $Z_{f} = 0$ equals the ratio of the corresponding positive-sequence currents, and $I_{f 1} = k I_{f}$ can be read as a ratio of $I_{a 1}$ values.
With $Z_{f} = 0$ : $I_{a 1} = \frac{E}{Z _{1} + Z _{2}}$
Take the ratio: $k = \frac{E / ( Z _{1} + Z _{2} + Z _{f} )}{E / ( Z _{1} + Z _{2} )} = \frac{Z _{1} + Z _{2}}{Z _{1} + Z _{2} + Z _{f}}$
Cross-multiply: $k (Z_{1} + Z_{2}) + k Z_{f} = Z_{1} + Z_{2}$ , so $k Z_{f} = (Z_{1} + Z_{2}) (1 - k)$ .
Hence $Z_{f} = \frac{( Z _{1} + Z _{2} ) ( 1 - k )}{k}$ which is option (A). Check: for $k = 1$ this gives $Z_{f} = 0$ , and as $k \to 0$ , $Z_{f} \to \infty$ , both as expected.

KEY IDEA

In a line-to-line fault the sequence networks $Z_{1}$ , $Z_{2}$ and $Z_{f}$ are in series and $Z_{0}$ is not involved. So $k = \frac{Z _{1} + Z _{2}}{Z _{1} + Z _{2} + Z _{f}}$ , and solving for $Z_{f}$ gives the answer.

WATCH OUT FOR

Putting $Z_{0}$ into the fault loop, or using the line-to-ground connection (all three sequence networks in series, with $3 Z_{f}$ ). A line-to-line fault has no zero-sequence path. Another slip is a rearrangement error that gives $(1 + k)$ or $k / (1 - k)$ in place of $(1 - k) / k$ .

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GATE 2018 EE Q.28: solution and answer

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