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The peak value of an alternating emf E given by E = `underset(o)(E) ` cos `omega`t is 10 V and frequency is 50 Hz . At time t = (1/600) s, the instantaneous value of emf isA. 10 VB. 5`sqrt 3` VC. 5 VD. 1 V |
Answer» Correct Answer - B E = 10 cos(`2pi`ft) = 10 cos (`2pi xx 50 xx 1/600`) = 10 cos`pi`/6 = `5sqrt 3`V |
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