Question

As shown in the figure an inductor of inductance $200mH$ is connected to an AC source of emf $220\text{V}$ and frequency $50\text{Hz}$. The instantaneous voltage of the source is $0\text{V}$ when the peak value of current is $\frac{\sqrt{a}}{\pi }\text{A}$. The value of $a$ is

Answer 1

Current in the circuit is:
$I_{\text{rms}}=\frac{V_{\text{rms}}}{z}$
So, imoedence if circuit is:
$z=X_L=\omega L$
$=2\pi \times 50\times \frac{200}{1000}$
$=20\pi$
$\therefore I_{\text{rms}}=\frac{220}{20\pi }=\frac{11}{\pi }$
$\therefore I_{\text{peak}}=\sqrt{2}\times \frac{11}{\pi }$ ($I_{peak}=\sqrt{2}{I}_0$)
$=\frac{\sqrt{2\times 121}}{\pi }$
$=\frac{\sqrt{242}}{\pi }$