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In an circuit $I = 250 \sin 250 π t$ . The time required for the current to achieve its peak value will be

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$\frac{1}{200} \sec$

$\frac{1}{500} \sec$

$\frac{1}{400} \sec$

$\frac{1}{100} \sec$

answer is C.

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Detailed Solution

$ω = 2 π f \Rightarrow 250 π = 2 π f \Rightarrow f = 125$

$t = \frac{T}{4} = \frac{1}{4 f} = \frac{1}{500} \sec$ .

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