1.

If the emf of an ac circuit be `E=E_0` sinomegatand current ` I=I_("cosomegat")``, what is the power dissipated in the circuit?

Answer» `E=E_0sinomegat,I=I_0cosomegat=I_0sin(omegat+90^@)`
So, phase difference , `theta=90^@`
Therefore , power factor`=costheta=cos90^@=0`
i.e., power dispated=0


Discussion

No Comment Found

Related InterviewSolutions