An L-R-C series circuit is connected to a 120-Hz ac
source that has. The circuit has a resistance of
75.0Ω and an impedance at this frequency of 105 Ω. What aver-
age power is delivered to the circuit by the source?

$\mathit{\varphi }$is the power factor of the circuit.

$\mathrm{cos\varphi }=\frac{\mathrm{R}}{\mathrm{Z}}$

${\mathbf{I}}_{\mathbf{rms}}$represents the d.c. current that dissipates the same amount of power as the average power dissipated by the alternating current/voltage.

$I_{rms}=\frac{V_{rms}}{Z}$

${\mathit{P}}_{\mathbf{a}\mathbf{v}\mathbf{g}}$is the average power dissipated by the circuit.

$P_{avg}=V_{rms}{I}_{rms}\cos \varphi$

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

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$$\begin{gathered} Z=105\Omega \\ {V}_{rms}=75V \\ \cos \varphi =\frac{75V}{105\Omega } \\ =0.714 \end{gathered}$$

$$\begin{gathered} I_{rms}=\frac{80V}{105\Omega } \\ =0.762V \end{gathered}$$

Finally,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\left(80.00\mathrm{V}\right)\left(0.762\mathrm{A}\right)\left(0.714\right) \\ =43.5\mathrm{W} \end{gathered}$$

Therefore, the average power delivered to the circuit by the source is 43.5W.