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An alternating voltage $V (t) = 220 \sin 100 πt$ volt is applied to a purely resistive load of $50 Ω$ . The time
taken for the current to rise from half of the peak value to the peak value is:

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5 ms

3.3 ms

7.2 ms

2.2 ms

answer is B.

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Detailed Solution

Rising half to peak

$t = T / 6$

$t = \frac{2 π}{6 ω} = \frac{π}{3 ω} = \frac{π}{300 π} = \frac{1}{300} = 3.33 ms$

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