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The peak value of an alternating emf E g...

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 is

A

10 V

B

5`sqrt 3` V

C

5 V

D

1 V

Text Solution

Verified by Experts

The correct Answer is:
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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