ISRO EC 2025 Q.65: solution and answer

Official answer: BSource: ISRO's final answer key of 15 May 2026

1. The resistor and inductor are in series, so the total impedance is $Z=R+j{X}_L=10+j10\Omega$.

2. Magnitude: $|Z|=\sqrt{10^2+10^2}=10\sqrt{2}\Omega \approx 14.14\Omega$.

3. Phase angle: $\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)={\tan }^{-1}(1)=45^{\circ }$.

4. Power factor $=\cos \varphi =\frac{R}{|Z|}=\frac{10}{10\sqrt{2}}=\frac{1}{\sqrt{2}}\approx 0.707$.

5. An inductive load makes the current lag the voltage, since $\varphi =+45^{\circ }$ is the angle of the voltage ahead of the current. The power factor is therefore lagging.

6. Hence the power factor is $0.707$ lagging. Option (a), $0.5$, would need $\varphi =60^{\circ }$, that is $X_L=R\sqrt{3}$. Option (c) needs $X_L=0$. Option (d) is leading, which only a net capacitive circuit gives, and $0.866$ would need $\varphi =30^{\circ }$.