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(a) Use the results of part (a) of Exercise 31.21 to show that the average power delivered by the source in an L-R-C series circuit is given byPavg=(Irms)2R.(b) An L-R-C series circuit has R = 96.0 鈩, and the amplitude of the voltage across the resistor is 36.0 V. What is the average power delivered by the source?

Short Answer

Expert verified

The average power delivered by the source in an L-R-C series circuit is given byPavg=(Irms)2Rand the average power delivered by the source is 6.75W.

Step by step solution

01

Step-1: Formula of average power delivered

Pavgis the average power dissipated by the circuit.

Pavg=VrrmsIrms肠辞蝉蠒

VrmsandIrms are the rms voltage and current respectively.

颁辞蝉蠒=Rz,Irms=Vrmszvrms=v2

02

Step-2: Calculations for average power delivered by the source

Pavg=VrmsIrmscos=VrmsIrms(RZ)=(VrmsZ)IrmsR=Irms2R=V2rmsR

Therefore, the average power delivered by the source is I2rmsR.

03

Step-3: Calculations for average power dissipated by the source

Vrms=36V2=25.45V

Now, we plug the values ofVrmsand R in the formula to get the average power.

Pav=25.45V296=6.75W

Therefore the average power delivered by the source in an L-R-C series circuit is given byPavg=Irms2Rand the average power delivered by the source is 6.75W.

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