GATE 2022 EE Q.32: solution and answer

Official answer: 795 to 805Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. The supply is 200 V, 50 Hz single phase (taken as the rms value, the usual convention), so $V_m=200\sqrt{2}\text{V}$. With ideal diodes and no gating, the full-bridge output is $v_o(t)=|v_s(t)|=V_m\left|\sin \omega t\right|$. The firing angle is $\alpha =0^{\circ }$ (diodes conduct naturally) and there is no switching, so no duty ratio applies; the load carries current continuously.

2. The load is purely resistive, so the active power is $P=\frac{V_{o,rms}^2}{R}$. It depends on the rms output voltage, not the average.

3. The rms of the full-wave rectified sine, over one output period $\pi$: $$V_{o,rms}=\sqrt{\frac{1}{\pi }{\int }_0^{\pi }(V_m\sin \theta {)}^{2}d\theta }=\sqrt{\frac{V_m^2}{2\pi }{\int }_0^{\pi }(1-\cos 2\theta )d\theta }=\frac{V_m}{\sqrt{2}}=200\text{V}$$

4. $$P=\frac{V_{o,rms}^2}{R}=\frac{200^2}{50}\text{W}=800\text{W}$$

5. Check on the average value: $V_{dc}=\frac{2V_m}{\pi }\approx 180.1\text{V}$. Using it wrongly would give $\frac{180.1^2}{50}\approx 648\text{W}$, which is not the active power. The correct value is $800\text{W}$.