Question

An ideal sinusoidal voltage source \( v(t) = 230\sqrt{2} \sin(2\pi \times 50t) \, \text{V} \) feeds an ideal inductor \( L \) through an ideal SCR with firing angle \( \alpha = 0^\circ \). If \( L = 100 \, \text{mH} \), then the peak of the inductor current, in ampere, is closest to:

• A. 20.71
• B. 0
• C. 10.35
• D. 7.32

Hint:

When a pure inductor is supplied with a sinusoidal voltage and triggered via an SCR at \( \alpha = 0^\circ \), use: \[ i_{peak} = \frac{V_m}{L \omega}(1 - \cos(\omega t)) \] and for full half-cycle: \( t = \frac{\pi}{\omega} \Rightarrow i_{peak} = \frac{2V_m}{L \omega} \).

Answer 1

The Correct Option is A

Given: \[ v(t) = 230\sqrt{2} \sin(\omega t), \quad \omega = 2\pi \times 50 = 100\pi, \quad L = 100 \, {mH} = 0.1 \, {H} \] The peak current through an inductor with SCR firing at \( \alpha = 0^\circ \) is given by: \[ i(t) = \frac{1}{L} \int_0^t v(\tau) \, d\tau = \frac{230\sqrt{2}}{L} \int_0^t \sin(\omega \tau) \, d\tau \] The integral becomes: \[ i(t) = \frac{230\sqrt{2}}{0.1 \cdot \omega} (1 - \cos(\omega t)) \] Maximum current occurs at \( \omega t = \pi \Rightarrow t = \frac{\pi}{\omega} \): \[ i_{max} = \frac{230\sqrt{2}}{0.1 \cdot \omega} (1 - \cos(\pi)) = \frac{230\sqrt{2}}{0.1 \cdot 100\pi} (1 + 1) \] \[ i_{max} = \frac{230\sqrt{2} \cdot 2}{10\pi} \approx \frac{325.27 \cdot 2}{10\pi} \approx \frac{650.54}{31.4159} \approx 20.71 \, {A} \] Therefore, the peak current through the inductor is: \[ \boxed{20.71 \, {A}} \]