T6.8 Power in LCR Circuit:


In the series LCR circuit, the power delivered by the AC circuit is defined as the product of current and emf.
Mathematically,
Power (P) = IE  .......... (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.E02TT.Cosθ
             =I0.E02Cosθ
             =I02E02Cosθ
  Pav=IrmsErmsCosθ .......... (iv)

Here Irms and Erms are called apparent power and Cosθ is called power factor.


Special Cases:
Case I: Power Consumption across R, If θ=0; Then, 
Pav=IrmsErms
Case II: Power Consumption across L, If θ=π2; Then, 
Pav=IrmsErmsCos(π2)=0
Case III: Power Consumption across C, If θ=π2; Then, 
Pav=IrmsErmsCos(π2)=0
Case IV: Power Consumption across RL, If θ=RZ; Then, 
  Pav=IrmsErms(RXL2+R2)
Case V: Power Consumption across RC, If θ=RZ; Then, 
  Pav=IrmsErms(RXC2+R2)
Case VI: Power Consumption across LCR, If θ=RZ; Then, 
Pav=IrmsErms(RXLXC)2+R2)

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