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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 |
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