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Q.

The frequency of an alternating current is 50 Hz. The minimum time taken by it in reaching from zero to peak value is ____ ms

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answer is 5.

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Detailed Solution

$\begin{aligned} T = \frac{1}{t} = \frac{1}{50} \sec \\ t = \frac{T}{4} = \frac{1}{200} \sec \\ t = 5 \times 10^{- 3} = 5 ms \end{aligned}$

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