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The peak value of an alternating emf E given by `E=(E_0) cos omega t` is 10V and frequency is 50 Hz. At time `t=(1//600)s` the instantaneous value of emf isA. `10V`B. `5sqrt(3)V`C. `5V`D. `1V` |
Answer» Correct Answer - B `E=E_(0)cos omegat=E_(0)cos (2pit)/T` `=10cos.(2pixx50xx1)/600=10cos(pi/6)=5sqrt(3)`volt. |
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