A series ac circuit contains a 250-Ω resistor, a 15-mH
inductor, a 3.5-mF capacitor, and an ac power source of voltage
amplitude 45 V operating at an angular frequency of$360\frac{\mathrm{rad}}{\mathrm{s}}$.
(a) What is the power factor of this circuit? (b) Find the average
power delivered to the entire circuit. (c) What is the average power
delivered to the resistor, to the capacitor, and to the inductor?

$\mathit{\varphi }$is the phase difference between voltage and current and $\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }$is called the power factor of the circuit.

$\varphi is{\tan }^{-1}\left(\frac{\omega L-{\displaystyle \frac{1}{\omega c}}}{R}\right)$where$\varphi$is the phase difference between voltage and current, $\omega$is the angular frequency of voltage supply, L is the value of inductance of the inductor, C is the capacitance, R is the resistance.

Inductance reactance is the resistance provided to alternating current due to the inductor.

${\mathrm{X}}_1=\mathrm{wL}$

Capacitive reactance is the resistance provided to alternating current due to the capacitor.

${\mathrm{X}}_{\mathrm{c}}=\frac{1}{\mathrm{wC}}$

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.

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$, where Z is the impedance

localid="1664179953769" ${\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$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 which is effectively the power dissipated by the resistor in one cycle.

$$\begin{gathered} P_{avg}=V_{rms}{I}_{rms}\cos \varphi \\ ={\left(I_{rms}\right)}^{2}R \end{gathered}$$

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

$\mathrm{w}=360\frac{\mathrm{rad}}{\mathrm{s}}\varphi$

$$\begin{gathered} \mathrm{L}=15\mathrm{mH} \\ \mathrm{C}=3.5\mathrm{\mu F} \\ {\mathrm{X}}_1=\left(360\right)\left(15\times {10}^{-3}\right) \\ =5.4\mathrm{\Omega } \end{gathered}$$

$$\begin{gathered} X_c=\frac{1}{\left(360\right)\left(3.5\times {10}^{-6}\right)} \\ =794\Omega \end{gathered}$$

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{5.4\Omega -794\Omega }{250\Omega }\right) \\ =-72.41° \end{gathered}$$

$$\begin{gathered} \cos \varphi =\cos \left(-72.41°\right) \\ =0.302 \end{gathered}$$

Therefore, the power factor of the circuit is 0.302.

$$\begin{gathered} Z=\sqrt{{250}^2+{\left(5.4-794\right)}^2\Omega } \\ =827\Omega \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_{\mathrm{rms}}=\frac{31.8\mathrm{V}}{827\mathrm{\Omega }} \\ =0.145\mathrm{A} \\ {\mathrm{P}}_{\mathrm{avg}}=\left(31.8\mathrm{V}\right)\left(0.0385\mathrm{A}\right)\left(0.302\right) \\ =0.37\mathrm{W} \end{gathered}$$

Therefore, the average power delivered to the entire circuit is

The average power delivered to the entire circuit is equal to the power consumed in the resistor.

${\mathrm{P}}_{\mathrm{av}}=(0.0385\mathrm{A}{)}^2(250\mathrm{\Omega })=0.37\mathrm{W}$.

All the power is dissipated in the resistor only. So, no power is consumed in the capacitor and the inductor.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{c}}=0 \\ {\mathrm{P}}_{\mathrm{L}}=0 \end{gathered}$$

Therefore, the power factor of the circuit is 0.302, the average power delivered to the entire circuit is 0.37W and the average power delivered to the resistor is 0.37W, to the capacitor is 0 and to the inductor is 0W.