Derive an expression for average power of an AC (alternating current) circuit.

Explanation

Let instantaneous emf and the current in an a.c. circuit are given by
`E = E_0 sin omegat`
and `I=I_0sin(omegat+theta)`
where `theta =` Angle by which current leads the voltage
Power at instant.s, t `(dW)/(dt) = EI`
`(dW)/(dt)=E_0sinomegat xxI_0sin(omegat-theta)`
`=E_0I_0sinomegat(sinomegatcos theta-cosomegatsin theta)`
`=E_0I_0sin^2omegatcos theta-E_0I_0sinomegatcosomegatsin theta`
`=E_0I_0sin^2omegatcos theta-(E_0I_0)/2(2sinomegatcosomegat)sin theta`
`=E_0I_0sin^2omegatcos theta-(E_0I_0)/2sin2omegatsin theta`
If this instantaneous power is assumed to remain constant for a small time dt , then small amount of work done in this time is
`dW=[E_0I_0sin^2omegatcos theta-(E_0t_0)/2 sin2omegasin theta]dt`
Total work done over a complete cycle is
`W=int_0^TE_0I_0sin^2omegatcostheta-int_0^T(E_0I_0)/2 sin 2omegatsin thetadt`.
`=E_0I_0costhetaint_0^Tsin^2omegatdt-(E_0I_0sintheta)/(2)int_0^Tsin2omegatdt`
`=E_0I_0costheta(X)-(E_0I_0sintheta)/2xx(Y)" "...(3)`
Now, `X=int_0^Tsin^2omegatdt=int_0^T((1-cos2omegat)/2)dt =int_0^T1/2dt-int_0^T(cos2omegat)/2dt`
`=1/2|t|_0^T-1/2|(sin2omegat)/(2omega)|_0^T=T/2-1/(4omega)(sin2omegaT -sin0)`
`=T/2-1/(4omega)(sin2(2pi)/(T)T-sintheta)`
`=T/2-1/(4omega)(sin4pi-sin0)" "( :. sin4pi=sin0^@="zero")`
`Y=int_0^Tsin2omegatdt=|(-cos2omegat)/(2omega)|_0^T=(-1)/(2omega)(cos2omegaT-cos2omegaxx0)`
`=(-1)/(2omega)(cos2xx(2pi)/(T)T-cos0)=(-1)/(2omega)(cos4pi-cos0)`
`=(-1)/(2omega)(1-1)`= zero
Putting value of X and Y in eq. (3)
`W=E_0I_0cos theta(T/2)-E_0I_0(sintheta)/2(0)`
`W = E_0I_0cos theta(T//2)`
So average power in the circuit over a complete cycle is given by
`P=W/T=(E_0I_0costheta)/(T)(T/2)`
`P=(E_0I_0costheta)/2=E_0/(sqrt2)I_0/sqrt2costheta`
`P=E_vI_vcostheta`
Average power of an a.c. circuit is also called true power of the circuit. The product `E_vI_v` is apparent power and cos O is known as power factor.
So, True power = Virtual power `xx` Power factor