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  4. An alternating voltage v(t) = 220 sin 100 nt volt is applied to a purely resistance load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is :

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Q. An alternating voltage $v(t) = 220$ $sin\, 100\, nt$ volt is applied to a purely resistance 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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Solution:

$V(t) = 220 \; \sin (100 \pi t) volt$
time taken,
$t = \frac{\theta}{\omega} = \frac{\frac{\pi}{3}}{100\pi} = \frac{1}{300}\sec $
$= 3.3 \,ms $

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