GATE 2025 EE – Question 47
An ideal sinusoidal voltage source $v(t)=230\sqrt2\sin(2\pi\times50t)$ V feeds an ideal inductor $L$ through an ideal SCR with firing angle $\alpha=0^\circ$. If $L=100$ mH, then the peak of the inductor current, in ampere, is closest to
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Correct answer: (A) 20.71
Explanation
With $\alpha=0$ the SCR turns on at $\omega t=0$ and the current is $i(t)=\dfrac{V_m}{\omega L}(1-\cos\omega t)$, which continues until it returns to zero at $\omega t=2\pi$. Its peak is $\dfrac{2V_m}{\omega L}=\dfrac{2\times325.27}{2\pi\times50\times0.1}=\dfrac{650.5}{31.42}=20.71$ A.