In an a.c. circuit I=100 sin 200 pi t. The ti | vedclass

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(D) The given equation for the alternating current is $I = 100 \sin(200 \pi t)$.Comparing this with the standard form $I = I_0 \sin(\omega t)$,the pea

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

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(D) The given equation for the alternating current is $I = 100 \sin(200 \pi t)$.Comparing this with the standard form $I = I_0 \sin(\omega t)$,the peak value $I_0$ is $100 \ A$.The current reaches its peak value when $\sin(200 \pi t) = 1$.This occurs when the argument of the sine function is equal to $\frac{\pi}{2}$.So,$200 \pi t = \frac{\pi}{2}$.Dividing both sides by $\pi$,we get $200 t = \frac{1}{2}$.Therefore,$t = \frac{1}{2 	imes 200} = \frac{1}{400} \ s$.

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

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