In an AC circuit, $I=100 \sin 200 \pi t$, the time required for the current to achieve its peak value is:

$\frac{1}{400} s$

The correct answer is Option (4) → $\frac{1}{400} s$

$I=100 \sin (200 \pi t)$ [given]

$I=I_0\sin)(kx-ωt)$ [general form]

Now,

$I_0=100A$

$ω=200\pi\,rad/s=2\pi f$

for peak condition,

$\sin(200 \pi t)=1$

$200 \pi t=\frac{\pi}{2}$

$t=\frac{1}{400}s$