ISRO EC 2023 Q.66: solution and answer

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

1. For a uniform line of length $l$ with propagation constant $\gamma$, the open-circuit and short-circuit input impedances are $Z_{oc}=Z_0\coth \gamma l$ and $Z_{sc}=Z_0\tanh \gamma l$. The given $Z_{oc}$ has a real part, so the line is lossy and the general form with $\gamma$ is the right one.

2. Multiplying the two gives $Z_{oc}{Z}_{sc}=Z_0^2\coth \gamma l\tanh \gamma l=Z_0^2$, so $Z_{sc}=\frac{Z_0^2}{Z_{oc}}$.

3. Substitute: $Z_{sc}=\frac{50^2}{100+j150}=\frac{2500}{100+j150}\Omega$.

4. Rationalise: $Z_{sc}=\frac{2500(100-j150)}{100^2+150^2}=\frac{2500(100-j150)}{32500}=\frac{100-j150}{13}$.

5. So $Z_{sc}\approx 7.69-j11.54\Omega$.

6. Check against the other options: (d) repeats $Z_{oc}$, which would need $Z_{oc}^2=Z_0^2$. (c) $50\Omega$ would need $Z_{oc}=50\Omega$. For (b), $(100+j150)(2-j50)=7700-j4700\ne 2500$. Only (a) satisfies $Z_{oc}{Z}_{sc}=Z_0^2$.