GATE 2025 EE Q.27: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Average power comes only from terms of the same frequency in $v$ and $i$. The voltage has only the fundamental, so the $3\omega$ and $5\omega$ current terms give zero average power. $P=V_{rms}{I}_{1,rms}\cos {\varphi }_1$.

2. $V_{rms}=\frac{300}{\sqrt{2}}=212.13\text{V}$ and $I_{1,rms}=\frac{10}{\sqrt{2}}=7.071\text{A}$. The fundamental current is $10\sin (\omega t-\pi /6)$ against $300\sin (\omega t)$, so the current lags by ${\varphi }_1=30^{\circ }$.

3. $$P=212.13\times 7.071\times \cos 30^{\circ }=1500\times 0.8660=1299.0\text{W}$$

4. The rms current includes every harmonic, because the rms of a sum of sinusoids of different frequencies is the root of the sum of the squared rms values. $$I_{rms}=\sqrt{\frac{10^2+2^2+1^2}{2}}=\sqrt{52.5}=7.246\text{A}$$

5. Apparent power: $S=V_{rms}{I}_{rms}=212.13\times 7.246=1537.0\text{VA}$.

6. $$\text{PF}=\frac{P}{S}=\frac{1299.0}{1537.0}\approx 0.845$$ As a check, the distortion factor is $\frac{7.071}{7.246}=0.976$ and the displacement factor is $\cos 30^{\circ }=0.866$. Their product is $0.845$. This is option (A).