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प्रश्न
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.
उत्तर
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`
= erms x Irms
∴ Pav = erms x Irms
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
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