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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

10V

B

`5sqrt(3)V`

C

`5 V`

D

1 V

Text Solution

Verified by Experts

The correct Answer is:
B

Given that `E_(0) =10 V, t=1/600 s`
`:. E=(E_0) cos 2 pi f t `
`=10 cos [ 2 pi xx 50 xx 1/600]`
`=10 cos (pi//6)=10 (sqrt(3)///2)=5 sqrt(3)V`.
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