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Question

(A) The instantaneous current in an $AC$ circuit is given by $i = i_0 \sin(\omega t)$.At maximum value,$i = i_0$,so $i_0 = i_0 \sin(\omega t_1) \Right

Vedclass · Accepted Answer

$2.5$

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(A) The instantaneous current in an $AC$ circuit is given by $i = i_0 \sin(\omega t)$.At maximum value,$i = i_0$,so $i_0 = i_0 \sin(\omega t_1) \Rightarrow \omega t_1 = \frac{\pi}{2}$.At rms value,$i = \frac{i_0}{\sqrt{2}}$,so $\frac{i_0}{\sqrt{2}} = i_0 \sin(\omega t_2) \Rightarrow \omega t_2 = \frac{\pi}{4}$.The time taken to change from maximum value to rms value is $\Delta t = t_1 - t_2$.$\Delta t = \frac{\pi}{2\omega} - \frac{\pi}{4\omega} = \frac{\pi}{4\omega}$.Since $\omega = 2\pi f$,we have $\Delta t = \frac{\pi}{4(2\pi f)} = \frac{1}{8f}$.Given $f = 50\, Hz$,$\Delta t = \frac{1}{8 	imes 50} = \frac{1}{400}\, s$.Converting to milliseconds: $\Delta t = \frac{1}{400} 	imes 1000\, ms = 2.5\, ms$.

$A$ $10\, \Omega$ resistance is connected across $220\, V - 50\, Hz$ $AC$ supply. The time taken by the current to change from its maximum value to the rms value is $....\, ms$.

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