GATE 2025 EE Q.32: solution and answer

Official answer: 0.975 to 0.985Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. For a short feeder with series impedance $R+jX$ and a load current $I$ at a lagging angle $\varphi$, the approximate voltage drop from the sending end to the receiving end is $$\Delta V\approx I(R\cos \varphi +X\sin \varphi ).$$

2. The load is fixed at 350 kVA, so with the receiving-end voltage held fixed the current magnitude $I$ does not change. Only $\varphi$ varies, so maximise $f(\varphi )=R\cos \varphi +X\sin \varphi$.

3. Set $\frac{df}{d\varphi }=-R\sin \varphi +X\cos \varphi =0$, which gives $\tan \varphi =\frac{X}{R}$. The second derivative is $-f(\varphi )<0$ at this point, so it is a maximum, and $f_{\max }=\sqrt{R^2+X^2}$. The load angle equals the feeder impedance angle.

4. Given $R/X=5$, so $\tan \varphi =\frac{X}{R}=\frac{1}{5}=0.2$.

5. Take $R=5$ and $X=1$ in any common unit. Then $\cos \varphi =\frac{R}{\sqrt{R^2+X^2}}=\frac{5}{\sqrt{26}}=\frac{5}{5.0990}=0.98058$.

6. The power factor is $\cos \varphi \approx 0.981$ (lagging).