An alternating voltage V(t) = 220 sin 100 πt volt is applied to a purely

Question

An alternating voltage V(t) = 220 sin 100 \(\pi t\) volt is applied to a purely resistive load of \(50\,\Omega\). The time taken for the current to rise from half of the peak value to the peak value is

(1) 5 ms 

(2) 3.3 ms 

(3) 7.2 ms 

(4) 2.2 ms

Answer 1

(2) 3.3 ms 

Rising half to peak

\(t =\frac {T}{6}\)

\(t =\frac {2\pi}{6\omega}=\frac {\pi}{3\omega}=\frac {\pi}{300 \pi}=\frac {1}{300}= 3.33 ms\)