Question

Obtain an expression for average power dissipated in a purely resistive A.C. circult. 

Define power factor of the circuit and state its value for purely resistive A.C. circult.

Answer 1

Expression for average power in purely resistive circuit:-

i. Let, e = e₀ sinωt be applied e.m.f across a resistor of resistance ‘R’ as shown in figure. At certain instant, current I = I₀ sin ωt is flowing through the resistor. In this case both ‘e’ and ‘I’ are in phase.

ii. Instantaneous power in circuit is given by,

P = eI = (e₀ sinωt) (I₀ sin ωt)

∴ P = e₀I₀ sin² ωt                                                            ..........................(1)

iii. Average power for a complete cycle can be obtained by integrating equation (1).

 `thereforeP_"av"="work done in one cycle"/"time for one cycle"`

          `=(int_0^TPdt)/T`

          `=(int_0^Te_0I_0sin^2omegatdt)/T`

`therefore P_"av"=(e_0I_0)/T(int_0^Tsin^2omegatdt)`                                               ...............(2)

iv. `"But"int_0^Tsin^2omegatdt=int_0^T((1-cos2omegat)/2)dt`

                             `=1/2[int_0^Tdt-int_0^Tcos2omegatdt]`

                              `=1/2[t-(sine2omegat)/(2omega)]_0^T`

                              `=1/2[T-(sin2omegaxx(2pi)/omega)/(2omega)]`

                              `=1/2[T-(sin4pi)/(2omega)]`

                              `=T/2" ........................................."[becausesin4pi=0]`

From equation (2),

`thereforeP_"av"=(e_0I_0)/T(T/2)`

          `=(e_0I_0)/2`

          `=(1/2)e_0I_0`

`thereforeP_"av"=e_0/sqrt2xxI_0/sqrt2`

           = e_rms x I_rms

∴ P_av = e_rms x I_rms

Power factor:-

Power factor of an a.c circuit is defined as the cosine of the phase difference between the applied voltage and the circuit current.

For purely resistive circuit, power factor = 1