Derive an experssion for the average power in LCR circuit connected to a.c. supply. Hence define power factor. Show that average power cousumed per cycle in an a.c. circuit conataining an ideal capacitor is zero.

Explanation

Power in `LCR` circuit /`AC` circuit: the voltage `V=V_(m) sin omegat` is applied to a series `RLC` circuit.Then the current in the circuit is given by `I=I_(m)sin (omegat+phi)`
where `I_(m)=V_(m)/Z` and `phi=tan^(-1)((X_(L)-X_(C))/R)`
The instaneous power `P` supplied to the source is:
`P=VI`
`P=V_(m)sin omegat.I_(m)sin(omegat+phi)`
`=(V_(m)I_(m))/2 . 2sin omegat sin(omegat+phi)`
`=(V_(m)I_(m))/2 [cos(omegat-omegat-phi)]-cos(omegat+omegat+phi)`
`[because 2 sin A sin B=cos (A-B)-cos (A+B)]`
`=(V_(m)I_(m))/2[cos phi-cos (2omegat+phi)]`...(i)
`because` The average power over a cycle is given by the average of the two terms in `RHS` of equation (i).The second terms in `RHS` is time dependent.Its average is zero
[`because` Positive half of the cosine cancles the negative half]
Therefore, average power
`barP=(V_(m)I_(m))/2cos phi=V_(m)/sqrt2I_(m)/sqrt2cosphi`
`barP=V_(rms)I_(rms)cos phi`
Where `cosphi` is called power factor.It depends on voltage, current and phase angle `phi` CaseI:For resistive circuit:If circuit contains only resistance `R`
`therefore phi=0`, Therefore
`P=V_(rms)I_(rms)cos 0=V_(rms)I_(rms)`
Therefore, power disispation is maximum
Case II: For pure inductive or capacitive circuit, the phase difference between voltage and current is `pi/2`
`thereforephi=pi/2 rArrcos phi="cos" pi/2=0`
`therefore barP=V_(rms)I_(rms)cos phi=0`
Therefore, power dissipation is zero. hence, no power is dissipated even through a current is flowing in the circuit, the current is called wattless current.