Chapter 4: Q22E (page 1045)

(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 by $P_{a v g} = {(I_{r m s})}^{2} R$ .(b) An L-R-C series circuit has R = 96.0 Ω, 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 by $P_{a v g} = (I_{r m s})^{2} R$ and the average power delivered by the source is 6.75W.

Step by step solution

Step-1: Formula of average power delivered

$P_{a v g}$ is the average power dissipated by the circuit.

$P_{avg} = V_{rrms} I_{rms} cosϕ$

$V_{r m s}$ and $I_{r m s}$ are the rms voltage and current respectively.

$Cosϕ = \frac{R}{z}, I_{rms} = \frac{V_{rms}}{z} v_{rms} = \frac{v}{\sqrt{2}}$

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

$P_{a v g} = V_{r m s} I_{r m s} \cos ϕ = V_{r m s} I_{r m s} (\frac{R}{Z}) = (\frac{V_{r m s}}{Z}) I_{r m s} R = {(I_{r m s})}^{2} R = \frac{{V^{2}}_{r m s}}{R}$

Therefore, the average power delivered by the source is ${I^{2}}_{rms} R$ .

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

$V_{r m s} = \frac{36 V}{\sqrt{2}} = 25.45 V$

Now, we plug the values of $V_{r m s}$ and R in the formula to get the average power.

$P_{av} = \frac{{(25.45 V)}^{2}}{96 Ω} = 6.75 W$

Therefore the average power delivered by the source in an L-R-C series circuit is given by $P_{a v g} = {(I_{r m s})}^{2} R$ and the average power delivered by the source is 6.75W.