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:
Solution
<p>Rising half to peak</p>
<p>$$\begin{aligned}
& \mathrm{t}=\mathrm{T} / 6 \\
& \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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