The peak value of an alternating emf E given byE=E0 cosωt is 10 V and frequency is 50 HZ. At time t = (1/600) s, the instantaneous value of emf is

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10 V

53V

5 V

1 V

answer is B.

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Detailed Solution

Given that E0=10 V,t=1600s∴E=E0cos 2πft=10 cos2π×50×1600=10 cos π/6 = 103/2=53 V

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