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An alternating voltage v(t) = 220 sin(100*pi*t) volt is applied across a purely resistive load of 50 ohm. Find the time taken for the current to rise from half its peak value to its peak value.

1. 2.2 ms
2. 5 ms
3. 3.3 ms
4. 7.2 ms

Correct answer: 3.3 ms

Solution

Current i = i0 sin(omega t) with omega = 100*pi. i = i0/2 when omega t = pi/6; i = i0 when omega t = pi/2. The phase change is pi/2 - pi/6 = pi/3. Time = (pi/3)/omega = (pi/3)/(100*pi) = 1/300 s about 3.3 ms.

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