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The peak value of an alternating emf E given by `E=(E_0) cos omega t` is 10V and freqency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf isA. 10VB. `5sqrt(3)V`C. `5 V`D. 1 V |
Answer» Correct Answer - 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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