In an $RLC$ circuit, energy is supplied by the emf and dissipated by the resistor. But an $RLC$ circuit is unlike a purely resistive circuit in that the current is not in phase with the potential difference of the emf.

We found in Equation 32.22 that the instantaneous current in an $RLC$ circuit is $i=I\text{cos}(\omega t-\varphi )$, where $\varphi$ is the angle by which the current lags the emf. Thus the instantaneous power supplied by the emf is

We can use the expression $\text{cos}(x-y)=\text{cos}\left(x\right)\text{cos}\left(y\right)+\text{sin}\left(x\right)\text{sin}\left(y\right)$ to write the power as

In our analysis of the power of a resistor and a capacitor, we found that the average of ${\text{cos}}^2\omega t$ is $\frac{1}{2}$ and the average of $\text{sin}\omega t\text{cos}\omega t$ is zero. Thus we can immediately write that the average power supplied by the emf is

The rms values, you will recall, are $I\text{/}\sqrt{2}$ and ${\epsilon }_{\text{}0}\text{/}\sqrt{2}$.

The term $\text{cos}\varphi$, called the power factor, arises because the current and the emf in a series $RLC$ circuit are not in phase. Because the current and the emf aren’t pushing and pulling together, the source delivers less energy to the circuit.

We’ll leave it as a homework problem for you to show that the peak current in an $RLC$ circuit can be written $I=I_{\text{max}}\cos \varphi$, where $I_{\text{max}}={\epsilon }_{\text{}0}\text{/}R$ was given in Equation 32.31. In other words, the current term in Equation 32.44 is a function of the power factor. Consequently, the average power is

where $P_{\text{max}}=\frac{1}{2}{I}_{\text{max}}{\epsilon }_{\text{}0}$ is the maximum power the source can deliver to the circuit.

The source delivers maximum power only when $\text{cos}\varphi =1$. This is the case when $X_{\text{L}}-X_{\text{C}}=0$, requiring either a purely resistive circuit or an $RLC$ circuit operating at the resonance frequency ${\omega }_0$. The average power is zero for a purely capacitive or purely inductive load with, respectively, $\varphi =-90\text{°}$ or $\varphi =+90\text{°}$, as found above.

Motors of various types, especially large industrial motors, use a significant fraction of the electric energy generated in industrialized nations. Motors operate most efficiently, doing the maximum work per second, when the power factor is as close to 1 as possible. But motors are inductive devices, due to their electromagnet coils, and if too many motors are attached to the electric grid, the power factor is pulled away from 1. To compensate, the electric company places large capacitors throughout the transmission system. The capacitors dissipate no energy, but they allow the electric system to deliver energy more efficiently by keeping the power factor close to 1.

Finally, we found in Equation 32.28 that the resistor’s peak voltage in an $RLC$ circuit is related to the emf peak voltage by $V_{\text{R}}={\epsilon }_{\text{}0}\text{ cos }\varphi$ or, dividing both sides by $\sqrt{2}\text{,}{V}_{\text{rms}}={\epsilon }_{\text{rms}}\cos \varphi$. We can use this result to write the energy loss in the resistor as

But this expression is $P_{\text{source}}$, as we found in Equation 32.44. Thus we see that the energy supplied to an $RLC$ circuit by the emf is ultimately dissipated by the resistor.