In an AC circuit, $I=100 \sin 200 \pi t$, the time required for the current to achieve its peak value is:
Answer & explanation
Correct answer: option 4
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,
$\sin(200 \pi t)=1$
$200 \pi t=\frac{\pi}{2}$
$t=\frac{1}{400}s$