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:

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YESPYQ · Accepted Answer

Correct answer: c) $3.3 ms$. $i(t)={i}_{0}\sin 100\pi t$ $\frac{{i}_{0}}{2}={i}_{0}\sin 100\pi {t}_{1}$ $\sin 100\pi {t}_{1}=\sin \frac{\pi }{6}$ $100\pi {t}_{1}=\frac{\pi }{6}$ ${t}_{1}=\frac{1}{600}\text{sec}$ Similarly ${t}_{2}=\frac{1}{200}\text{sec}$ $t={t}_{2}−{t}_{1}$ $=\frac{1}{200}−\frac{1}{600}$ $t=\frac{3−1}{600}=\frac{1}{300}$ t = 3.3 ms

An alternating voltage \(V(t)=220 \sin 100 \pi t\) volt is applied to a purely resistive load of \(50 \Omega\). The time…

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