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Obtain an expression for average power of AC over a cycle. Discuss its special cases. |
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Answer» SOLUTION :i. POWER of a circuit is defined as the rate of consumption of electric energy in that circuit. It is given by the PRODUCT of the voltage and current. In an AC circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle. ii. The alternating voltage and altelrnating current in the series RLC circuit at an instant are given by `v=V_(m)sinomegat" and "i=I_(m)=(omega+phi)` iii. where `phi` is the phase angle between v and i. The instantaneous power is then written as P = vi `=V_(m)I_(m)sinomegatsin(omegat+phi)` `=V_(m)I_(m)sinomegatsinomegat` `[sinomegatcosphi-cosomegatdsinphi]` `P=V_(m)I_(m)` `[cosphisin^(2)omegat-sinomegatcosomegatsinphi]` iv. Here the average of `sin^(2)` over a cycle is `(1)/(2)` and that of `sinomegat cosomegat`zero. Substituting these values, we obtain average power over a cycle. `P_(av)=V_(m)I_(m)cosphixx(1)/(2)` `=(V_(m)I_(m))/(sqrt(2)sqrt(2))cosphi` `P_(av)=V_(RMS)I_(RMS)cosphi` v. where `V_(RMS)I_(RMS)` is called apparent power and `cosphi` is power factor. The average power of an AC circuit is also KNOWN as the true power of the circuit. Special Cases i. For a purely resistive circuit, the phase angle between voltage and current is zero and `cosphi=1.` `thereforeP_(av)=V_(RMS)I_(RMS)` ii. For a purely inductive or CAPACITIVE circuit, the phase angle is `pm(pi)/(2)andcos(pm(pi)/(2))=0thereforeP_(av)=0` iii. For series RLC circuit, the phase angle `phi=tan^(-1)((X_(L)-X_(C))/(R))` `thereforeP_(av)=V_(RMS)I_(RMS)cosphi=1.` iv. For series RLC circuit at resonance, the phase angle is zero and `cosphi=1.` `thereforeP_(av)=V_(RMS)I_(RMS)` |
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