GATE 2025 EC Q.64: solution and answer

Official answer: 6.3 to 6.5Source: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Reflection coefficient at the load, with $Z_0=50\Omega$ and $Z_L=(50-j75)\Omega$: $${\Gamma }_L=\frac{Z_L-Z_0}{Z_L+Z_0}=\frac{(50-j75)-50}{(50-j75)+50}=\frac{-j75}{100-j75}$$

2. Power depends only on the magnitude of ${\Gamma }_L$: $$|{\Gamma }_L{|}^2=\frac{75^2}{100^2+75^2}=\frac{5625}{15625}=0.36(|{\Gamma }_L|=0.6)$$

3. On a lossless line the reflected average power is $P_{ref}=|{\Gamma }_L{|}^2{P}_{inc}=0.36\times 10=3.6\text{mW}$.

4. The line absorbs nothing, so power balance gives the power delivered to the load: $$P_L=P_{inc}-P_{ref}=P_{inc}\left(1-|{\Gamma }_L{|}^2\right)=10\times (1-0.36)=6.4\text{mW}$$

5. Check with the load resistance directly: $$1-|{\Gamma }_L{|}^2=\frac{4R_L{Z}_0}{|Z_L+Z_0{|}^2}=\frac{4\times 50\times 50}{100^2+75^2}=\frac{10000}{15625}=0.64$$ which gives the same $P_L=6.4\text{mW}$.