InterviewSolution
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InterviewSolution
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Prove that an ideal capacitor in an a.c. circuit does not dissipate power. |
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Answer» In a circuit containing inductor L, current I lags behind the voltage E be a phase angle of `90^(@) or (pi)/(2)` `therefore " "E=E_(0) sin omega"t then I"=I_(0)sin(omegat-(pi)/(2))` `therefore " "I=-I_(0)cos omegat` Work done in one complete cycle is `W=underset(0)overset(T)int EI dt=underset(0)overset(T)int E_(0) sin omegat(-I_(0) cos omegat) dt=E_(0)I_(0) underset(0)overset(T)int sin omegat cos omegat` `=-E_(0)I_(0) underset(0)overset(T)intsin (sin 2omegat)/(2)=-(E_(0)I_(0))/(2)[-(cos2 omegat)/(2omega)]_(0)^(T)=-(E_(0)I_(0))/(2)[(cos 2 omegaT)/(2omega)-(cos 0)/(2omega)]=(E_(0)I_(0))/(2)[(cos 2.(2pi)/(T).T)/(2omega)-(1)/(2omega)] ` `=(E_(0)I_(0))/(2)[-(cos4pi)/(2 omega)-(1)/(2omega)]=(E_(0)I_(0))/(2)[(1)/(2omega)-(1)/(2omega)]=0` Average power `=(W)/(T)=(0)/(T)=0`. Hence, average power supplied to an ideal capacitor by the source over a complete cycle of a,c cycle is zero. |
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