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In an AC circuit, \(I = 100\sin(200\pi t)\). The time required for the current to achieve its peak value is

1. \(\frac{1}{100}\) s
2. \(\frac{1}{200}\) s
3. \(\frac{1}{300}\) s
4. \(\frac{1}{400}\) s

Correct answer: \(\frac{1}{400}\) s

Solution

For \(I = I_0\sin(\omega t)\), the first peak occurs when \(\omega t = \pi/2\). Here \(\omega = 200\pi\), so \(t = \frac{\pi/2}{200\pi} = \frac{1}{400}\) s.

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