1.

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.


Discussion

No Comment Found

Related InterviewSolutions