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(D) The instantaneous current is given by $ I = I_0 \sin(\omega t) $.At the $ rms $ value, $ I = I_{rms} = \frac{I_0}{\sqrt{2}} $.So, $ \frac{I_0}{\sq

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$ 2.5 	imes 10^{-3} \,s $

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(D) The instantaneous current is given by $ I = I_0 \sin(\omega t) $.At the $ rms $ value, $ I = I_{rms} = \frac{I_0}{\sqrt{2}} $.So, $ \frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = \frac{1}{\sqrt{2}} $.This gives $ \omega t_1 = \frac{\pi}{4} $.Since $ \omega = \frac{2\pi}{T} $, we have $ \frac{2\pi}{T} t_1 = \frac{\pi}{4} \Rightarrow t_1 = \frac{T}{8} $.The peak value occurs at $ t_2 = \frac{T}{4} $.The time taken to reach the peak value from the $ rms $ value is $ \Delta t = t_2 - t_1 = \frac{T}{4} - \frac{T}{8} = \frac{T}{8} $.Given frequency $ f = 50 \,Hz $, the time period is $ T = \frac{1}{f} = \frac{1}{50} \,s = 0.02 \,s $.Therefore, $ \Delta t = \frac{0.02}{8} = 0.0025 \,s = 2.5 	imes 10^{-3} \,s $.Thus, option $ D $ is correct.

The frequency of an alternating current is $ 50 \,Hz $. What is the minimum time taken by the current to reach its peak value from its $ rms $ value?

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