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(D) The instantaneous current in an $A.C.$ circuit is given by $i = i_0 \sin(\omega t + \phi)$.At maximum value,$i = i_0$,which occurs at $\omega t_1

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$2.5 ms$

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(D) The instantaneous current in an $A.C.$ circuit is given by $i = i_0 \sin(\omega t + \phi)$.At maximum value,$i = i_0$,which occurs at $\omega t_1 = \frac{\pi}{2}$.At $R.M.S.$ value,$i = i_{R.M.S.} = \frac{i_0}{\sqrt{2}}$,which occurs at $\omega t_2 = \frac{\pi}{4}$ (or $\frac{3\pi}{4}$).The time interval $\Delta t = t_1 - t_2$ is given by $\omega \Delta t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.Since $\omega = 2 \pi f$,we have $2 \pi f \Delta t = \frac{\pi}{4}$.Substituting $f = 50 Hz$,we get $2 \pi (50) \Delta t = \frac{\pi}{4}$.$100 \pi \Delta t = \frac{\pi}{4} \Rightarrow \Delta t = \frac{1}{400} s$.$\Delta t = 0.0025 s = 2.5 ms$.

$A$ resistance of $20 \Omega$ is connected to an alternating current source of $110 V$. If the frequency of the $A.C.$ source is $50 Hz$,then the time taken by the current to change from its maximum value to the $R.M.S.$ value is

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