In the series LCR circuit, the power delivered by the AC circuit is defined as the product of current and emf.
Mathematically,
Power (P) = I∗E .......... (i) (∵ P = I * IR = I2R)The instantaneous value of current and emf are,
I=I0SinωtE=E0Sin(ωt+θ)} .......... (ii)
Then,
Power (P) = I0 Sin ωt.E0 Sin (ωt+θ)
= I0.E0 Sin ωt (Sin ωt Cos θ + Cos ωt Sin θ)
= I0.E0 (Cos θ . Sin2ωt + Sinθ . Cos ωt . Sin ωt)
Now the small amount of work done (dW) is given by,
dW=P.dt
or, dW=I0.E0 (Cosθ . Sin2ωt + Sinθ . Cos ωt . Sin ωt) .......... (iii)
Hence, the total amount of work done (W) can be obtained by integrating the above equation (iii) from 0 to T, then
or, ∫dW=∫T0I0.E0 (Cosθ . Sin2ωt + Sinθ . Cos ωt . Sin ωt)
By Integrating,
W=I0.E02T.Cosθ
Therefore, the average Power (Pav) in the AC circuit is,
Pav=WT=I0.E02∗TT.Cosθ
=I0.E02Cosθ
=I0√2∗E0√2∗Cosθ
⇒ Pav=Irms∗Erms∗Cosθ .......... (iv)
Here Irms and Erms are called apparent power and Cosθ is called power factor.
Special Cases:
Special Cases:
Case I: Power Consumption across R, If θ=0; Then,
⇒ Pav=Irms∗Erms
Case II: Power Consumption across L, If θ=π2; Then,
⇒ Pav=Irms∗Erms∗Cos(π2)=0
Case III: Power Consumption across C, If θ=−π2; Then,
⇒ Pav=Irms∗ErmsCos(−π2)=0
Case IV: Power Consumption across RL, If θ=RZ; Then,
⇒ Pav=Irms∗Erms(R√XL2+R2)
Case V: Power Consumption across RC, If θ=RZ; Then,
⇒ Pav=Irms∗Erms(R√XC2+R2)
Case VI: Power Consumption across LCR, If θ=RZ; Then,
⇒ Pav=Irms∗Erms(R√XL−XC)2+R2)
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