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(A) The instantaneous current in an $AC$ circuit is given by $i = i_{0} \cos(\omega t)$,where $i_{0}$ is the peak current.At $t = 0$,the current is at

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$2.5\, ms$

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(A) The instantaneous current in an $AC$ circuit is given by $i = i_{0} \cos(\omega t)$,where $i_{0}$ is the peak current.At $t = 0$,the current is at its maximum value,$i = i_{0}$.The $rms$ value of the current is $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.We need to find the time $t$ when $i = \frac{i_{0}}{\sqrt{2}}$.Setting $\cos(\omega t) = \frac{1}{\sqrt{2}}$,we get $\omega t = \frac{\pi}{4}$.Substituting $\omega = 2\pi f$,we have $2\pi f t = \frac{\pi}{4}$.Solving for $t$,$t = \frac{1}{8f}$.Given $f = 50\, Hz$,$t = \frac{1}{8 	imes 50} = \frac{1}{400}\, s$.$t = 0.0025\, s = 2.5\, ms$.

An $AC$ source rated $220\, V, 50\, Hz$ is connected to a resistor. The time taken by the current to change from its maximum value to the $rms$ value is:

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