GATE 2025 EE – Question 48
In the following circuit, the average voltage
$$V_o=400\left(1+\frac{\cos\alpha}3\right)\ \text{V},$$
where $\alpha$ is the firing angle. If the power dissipated in the resistor is 64 W, then the closest value of $\alpha$ in degrees is
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Correct answer: (A) 35.9
Explanation
The resistor dissipates $I^2R=64$ W with $R=1\ \Omega$, so $I=8$ A. In steady state the inductor has no average voltage, so $V_o=E+IR=500+8\times1=508$ V. Then $400\left(1+\dfrac{\cos\alpha}{3}\right)=508$ gives $\cos\alpha=0.81$ and $\alpha=35.9^\circ$.