An alternating voltage

(4) As $V(t)=220 \sin 100 \pi t$

so, $I(t)=\frac{220}{50} \sin 100 \pi \mathrm{t}$

i.e., $I=I_{m}=\sin (100 \pi \mathrm{t})$

For $I=I_{m}$

$t_{1}=\frac{\pi}{2} \times \frac{1}{100 \pi}=\frac{1}{200} \mathrm{sec}$

and for $I=\frac{I_{m}}{2}$

$\Rightarrow \frac{I_{m}}{2}=I_{m} \sin \left(100 \pi t_{2}\right)$

$\Rightarrow \frac{\pi}{6}=100 \pi t_{2} \Rightarrow t_{2}=\frac{1}{600} s$

$\therefore t_{\text {req }}=\frac{1}{200}-\frac{1}{600}=\frac{2}{600}=\frac{1}{300} s=3.3 \mathrm{~ms}$