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(A) The instantaneous current in an $AC$ circuit is given by $I = I_0 \sin(\omega t)$.At maximum value,the phase $\omega t_1 = \frac{\pi}{2}$.At $rms$

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$2.5 \,ms$

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(A) The instantaneous current in an $AC$ circuit is given by $I = I_0 \sin(\omega t)$.At maximum value,the phase $\omega t_1 = \frac{\pi}{2}$.At $rms$ value,the current is $I = \frac{I_0}{\sqrt{2}}$,so $\sin(\omega t_2) = \frac{1}{\sqrt{2}}$. The first instance after the peak is $\omega t_2 = \frac{3\pi}{4}$.The phase difference is $\Delta \phi = \omega t_2 - \omega t_1 = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.Given frequency $f = 50 \,Hz$,the angular frequency $\omega = 2\pi f = 2\pi 	imes 50 = 100\pi \,rad/s$.The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\pi / 4}{100\pi} = \frac{1}{400} \,s$.Converting to milliseconds: $t = 0.0025 \,s = 2.5 \,ms$.

$A$ resistance of $40 \,\Omega$ is connected to a source of alternating current rated $220 \,V, 50 \,Hz$. Find the time taken by the current to change from its maximum value to its $rms$ value.

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