GATE 2025 EE Q.47: solution and answer

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

1. Source: $V_m=230\sqrt{2}\approx 325.27\text{V}$, $\omega =2\pi \times 50=314.16\text{rad/s}$, so $\omega L=314.16\times 0.1=31.416\Omega$.

2. The SCR is forward biased from $\omega t=0$, when the source voltage turns positive. With firing angle $\alpha =0^{\circ }$ it is triggered at $\omega t=0$ and conducts for as long as the current stays positive.

3. While the SCR conducts, the inductor obeys $L\frac{di}{dt}=V_m\sin \omega t$ with $i(0)=0$.

4. Integrating: $$i(\omega t)=\frac{V_m}{\omega L}\left(1-\cos \omega t\right)$$

5. Since $1-\cos \omega t\ge 0$, the current is never negative. It returns to zero only at $\omega t=2\pi$, so the SCR conducts for the full cycle (conduction angle $360^{\circ }$). The negative half-cycle of the source only pulls the current back down to zero, not below it.

6. The current is largest where $\cos \omega t=-1$, that is at $\omega t=\pi$: $$I_{peak}=\frac{2V_m}{\omega L}$$

7. $$I_{peak}=\frac{2\times 325.27}{31.416}\approx 20.71\text{A}$$ This is option (A).