GATE 2022 EE Q.33: solution and answer

Official answer: 0.43 to 0.47Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. The supply voltage is a pure sinusoid at $\omega$, so only the fundamental of the current can deliver average power. The 3rd and 5th harmonic currents act at different frequencies from the voltage, and their average products with it are zero. Hence $P=V_{rms}{I}_{1,rms}\cos {\varphi }_1$, with $V_{rms}=\frac{230\sqrt{2}}{\sqrt{2}}=230\text{V}$.

2. The fundamental current is $10\sin \left(\omega t-\frac{\pi }{3}\right)$. So $I_{1,rms}=\frac{10}{\sqrt{2}}\text{A}$. The current lags the voltage by ${\varphi }_1=\frac{\pi }{3}=60^{\circ }$, so the displacement factor is $\cos {\varphi }_1=0.5$.

3. The rms value of the distorted current is the root of the sum of the squares of the harmonic rms values: $$I_{rms}=\sqrt{\frac{10^2+4^2+3^2}{2}}=\sqrt{\frac{125}{2}}\approx 7.906\text{A}$$

4. The apparent power is $S=V_{rms}{I}_{rms}$. So $$\text{PF}=\frac{P}{S}=\frac{V_{rms}{I}_{1,rms}\cos {\varphi }_1}{V_{rms}{I}_{rms}}=\underset{\text{distortion factor}}{\underbrace{\frac{I_{1,rms}}{I_{rms}}}}\times \underset{\text{displacement factor}}{\underbrace{\cos {\varphi }_1}}$$

5. Substituting: $$\text{PF}=\frac{10/\sqrt{2}}{\sqrt{125/2}}\times 0.5=\frac{10}{\sqrt{125}}\times 0.5=0.8944\times 0.5=0.447\approx 0.45\text{(lagging)}$$