GATE 2020 EE Q.8: solution and answer

Official answer: DSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Relay reach is 80 % of the line, so the reach impedance is $Z_r=0.8\times j0.2=j0.16\text{pu}$.

2. The relay is at the generator bus, and the generator terminal voltage is $1\angle 0^{\circ }$ pu. The relay therefore sees $V_r=1\angle 0^{\circ }$ pu. The generator is treated as an ideal 1 pu source because the question gives no source impedance.

3. A distance relay operates when the apparent impedance $V_r/I_r$ is at or below its reach. The threshold current is the current for which the apparent impedance equals the reach, which is a solid three-phase fault at the 80 % point.

4. For a solid fault at that point, the voltage at the relay equals the drop across the line section up to the fault: $V_r=I_f{Z}_r$. The line is lossless and uniformly distributed, so the impedance up to the fault is $j0.16$ pu.

5. $$I_{th}=\frac{V_r}{|Z_r|}=\frac{1}{0.16}=6.25\text{pu}$$

6. A fault nearer than 80 % gives a smaller impedance and a larger current, so the relay operates. A fault beyond 80 % gives a smaller current, so the relay does not operate. The threshold is $6.25$ pu, option (D).