Power in an AC Circuit

A circuit element dissipates or produces power according to $P=IV,$ where I is the current through the element and V is the voltage across it. Since the current and the voltage both depend on time in an ac circuit, the instantaneous power $p\left(t\right)=i\left(t\right)v\left(t\right)$ is also time dependent. A plot of p(t) for various circuit elements is shown in CNX_UPhysics_32_04_acpower. For a resistor, i(t) and v(t) are in phase and therefore always have the same sign (see CNX_UPhysics_32_02_acresistor). For a capacitor or inductor, the relative signs of i(t) and v(t) vary over a cycle due to their phase differences (see CNX_UPhysics_32_02_accap and CNX_UPhysics_32_02_acinduct). Consequently, p(t) is positive at some times and negative at others, indicating that capacitive and inductive elements produce power at some instants and absorb it at others.

Because instantaneous power varies in both magnitude and sign over a cycle, it seldom has any practical importance. What we’re almost always concerned with is the power averaged over time, which we refer to as the average power. It is defined by the time average of the instantaneous power over one cycle:

where $T=2\pi \text{/}\omega$ is the period of the oscillations. With the substitutions $v\left(t\right)=V_0\text{sin}\omega t$ and $i\left(t\right)=I_0\text{sin}(\omega t-\varphi ),$ this integral becomes

Using the trigonometric relation $\text{sin}(A-B)=\text{sin}A\text{cos}B-\text{sin}B\text{cos}A,$ we obtain

Evaluation of these two integrals yields

and

Hence, the average power associated with a circuit element is given by

In engineering applications, $\text{cos}\varphi$ is known as the power factor, which is the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase. For a resistor, $\varphi =0,$ so the average power dissipated is

A comparison of p(t) and $P_{\text{ave}}$ is shown in CNX_UPhysics_32_04_acpower(d). To make $P_{\text{ave}}=\left(1\text{/}2\right)I_0{V}_0$ look like its dc counterpart, we use the rms values $I_{\text{rms}}\text{and}{V}_{\text{rms}}$ of the current and the voltage. By definition, these are

where

With $i\left(t\right)=I_0\text{sin}(\omega t-\varphi )\text{and}v\left(t\right)=V_0\text{sin}\omega t,$ we obtain

We may then write for the average power dissipated by a resistor,

This equation further emphasizes why the rms value is chosen in discussion rather than peak values. Both equations for average power are correct for fs-id1172102049240, but the rms values in the formula give a cleaner representation, so the extra factor of 1/2 is not necessary.

Alternating voltages and currents are usually described in terms of their rms values. For example, the 110 V from a household outlet is an rms value. The amplitude of this source is $110\sqrt{2}\text{V}=\text{156 V}\text{.}$ Because most ac meters are calibrated in terms of rms values, a typical ac voltmeter placed across a household outlet will read 110 V.

For a capacitor and an inductor, $\varphi =\pi \text{/}2\text{and}-\pi \text{/}2\text{rad,}$ respectively. Since $\text{cos}\pi \text{/}2=\text{cos}\left(\text{−}\pi \text{/}2\right)=0,$ we find from fs-id1172101736047 that the average power dissipated by either of these elements is $P_{\text{ave}}=0.$ Capacitors and inductors absorb energy from the circuit during one half-cycle and then discharge it back to the circuit during the other half-cycle. This behavior is illustrated in the plots of CNX_UPhysics_32_04_acpower, (b) and (c), which show p(t) oscillating sinusoidally about zero.

The phase angle for an ac generator may have any value. If $\text{cos}\varphi >0,$ the generator produces power; if $\text{cos}\varphi <0,$ it absorbs power. In terms of rms values, the average power of an ac generator is written as

For the generator in an RLC circuit,

and

Hence the average power of the generator is

This can also be written as

which designates that the power produced by the generator is dissipated in the resistor. As we can see, Ohm’s law for the rms ac is found by dividing the rms voltage by the impedance.

Power Output of a Generator

is connected to an RLC circuit for which $L=2.00\times {10}^{\mathrm{-3}}\text{H}$, $C=4.00\times {10}^{\mathrm{-6}}\text{F}$, and $R=5.00\text{Ω}$. (a) What is the rms voltage across the generator? (b) What is the impedance of the circuit? (c) What is the average power output of the generator?

Strategy

Solution

1. Since $V_0=4.00\text{V,}$ the rms voltage across the generator is $V_{\text{rms}}=\frac{1}{\sqrt{2}}\left(4.00\text{V}\right)=2.83\text{V}\text{.}$
2. The impedance of the circuit is $\begin{array}{cc}Z& =\sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ & ={\left\{{\left(5.00\text{Ω}\right)}^2+{\left[\left(1.00\times {10}^4\text{rad/s}\right)\left(2.00\times {10}^{\mathrm{-3}}\text{H}\right)-\frac{1}{\left(1.00\times {10}^4\text{rad/s}\right)\left(4.00\times {10}^{\mathrm{-6}}\text{F}\right)}\right]}^2\right\}}^{1\text{/}2}\\ & =7.07\text{Ω}\text{.}\end{array}$
3. From fs-id1172102033252, the average power transferred to the circuit is $P_{\text{ave}}=\frac{V_{\text{rms}}^{2}R}{Z^2}=\frac{{\left(2.83\text{V}\right)}^2\left(5.00\text{Ω}\right)}{{\left(7.07\text{Ω}\right)}^2}=0.801\text{W}\text{.}$

Significance