GATE 2026 EE Q.34: solution and answer

Official answer: 32.3 to 32.5Source: Official answer key, GATE 2026 EE (IIT Guwahati)

1. This is a single-phase half-wave rectifier feeding an R-L load, with no freewheeling diode. The diode is an uncontrolled (plain) diode, so the firing angle is $\alpha =0$. There is no chopping, so no duty ratio $D$ applies.

2. Load reactance: $X_L=\omega L=2\pi (50)(500\times 10^{-6})=0.1571\Omega$.

3. Load impedance: $Z=\sqrt{R^2+X_L^2}=\sqrt{100+0.0247}=10.0012\Omega$. Load angle: $\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)={\tan }^{-1}(0.01571)=0.0157\text{rad}$.

4. While the diode conducts, with $\theta =\omega t$ and $\alpha =0$, the current is $$i(\theta )=\frac{V_m}{Z}\left[\sin (\theta -\varphi )+\sin \varphi e^{-\theta /\tan \varphi }\right].$$

5. The time constant is very small: $\tau =L/R=50\mu \text{s}$, so $\omega \tau =\tan \varphi =0.0157\text{rad}$. By $\theta \approx \pi /2$ the transient term has fallen to about $e^{-100}\approx 0$. The current is therefore effectively the steady-state sinusoid $\frac{V_m}{Z}\sin (\theta -\varphi )$.

6. The peak occurs where $\frac{di}{d\theta }=0$, that is at $\theta =\frac{\pi }{2}+\varphi$. The diode is still conducting there, because conduction lasts until about $\theta \approx \pi +\varphi$. At this point $I_{peak}=\frac{V_m}{Z}$.

7. $$I_{peak}=\frac{325}{10.0012}=32.496\text{A}\approx 32.5\text{A}.$$