In an ac circuit I = 100 sin(200 pi t). The t | vedclass

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(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 100 	ext{ A}$ and $\omega = 200 \pi 	ext{ rad/s}$.To reach the peak valu

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$\frac{1}{400} 	ext{ s}$

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(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 100 	ext{ A}$ and $\omega = 200 \pi 	ext{ rad/s}$.To reach the peak value,the phase $\omega t$ must be equal to $\frac{\pi}{2}$.So,$200 \pi t = \frac{\pi}{2}$.Solving for $t$: $t = \frac{\pi}{2 	imes 200 \pi} = \frac{1}{400} 	ext{ s}$.Alternatively,the time taken to reach the peak value is $\frac{T}{4}$,where $T$ is the time period.Since $\omega = \frac{2 \pi}{T} = 200 \pi$,we get $T = \frac{1}{100} 	ext{ s}$.Therefore,$t = \frac{1}{4} 	imes \frac{1}{100} = \frac{1}{400} 	ext{ s}$.

In an $ac$ circuit $I = 100 \sin(200 \pi t)$. The time required for the current to achieve its peak value will be

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