Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
University Physics Volume 2

15.4 Power in an AC Circuit

University Physics Volume 215.4 Power in an AC Circuit

Learning Objectives

By the end of this section, you will be able to:

  • Describe how average power from an ac circuit can be written in terms of peak current and voltage and of rms current and voltage
  • Determine the relationship between the phase angle of the current and voltage and the average power, known as the power factor

A circuit element dissipates or produces power according to P=IV,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(t)=i(t)v(t)p(t)=i(t)v(t) is also time dependent. A plot of p(t) for various circuit elements is shown in Figure 15.16. For a resistor, i(t) and v(t) are in phase and therefore always have the same sign (see Figure 15.5). For a capacitor or inductor, the relative signs of i(t) and v(t) vary over a cycle due to their phase differences (see Figure 15.7 and Figure 15.9). 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.

Figures a through d show sine waves on graphs of P versus t. All have the same amplitude and frequency. Figure a is labeled resistor. P bar is equal to half I0 V0. The sine wave is above the x axis, with the minimum y value being 0. It starts from a trough. Figure b is labeled capacitor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a positive slope. Figure c is labeled inductor. P bar is equal to 0. The equilibrium position of the sine wave is along the x axis. It starts at equilibrium with a negative slope. Figure d is labeled AC source. P bar is equal to half I0 V0 cos phi. The equilibrium position of the sine wave is above the x axis, with the minimum y-value of the wave being negative.
Figure 15.16 Graph of instantaneous power for various circuit elements. (a) For the resistor, Pave=I0V0/2,Pave=I0V0/2, whereas for (b) the capacitor and (c) the inductor, Pave=0.Pave=0. (d) For the source, Pave=I0V0(cosϕ)/2,Pave=I0V0(cosϕ)/2, which may be positive, negative, or zero, depending on ϕ.ϕ.

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=2π/ωT=2π/ω is the period of the oscillations. With the substitutions v(t)=V0sinωtv(t)=V0sinωt and i(t)=I0sin(ωtϕ),i(t)=I0sin(ωtϕ), this integral becomes


Using the trigonometric relation sin(AB)=sinAcosBsinBcosA,sin(AB)=sinAcosBsinBcosA, we obtain


Evaluation of these two integrals yields




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


In engineering applications, cosϕcosϕ 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, ϕ=0,ϕ=0, so the average power dissipated is


A comparison of p(t) and PavePave is shown in Figure 15.16(d). To make Pave=(1/2)I0V0Pave=(1/2)I0V0 look like its dc counterpart, we use the rms values IrmsandVrmsIrmsandVrms of the current and the voltage. By definition, these are



iave2=1T0Ti2(t)dtand vave2=1T0Tv2(t)dt.iave2=1T0Ti2(t)dtand vave2=1T0Tv2(t)dt.

With i(t)=I0sin(ωtϕ)andv(t)=V0sinωt,i(t)=I0sin(ωtϕ)andv(t)=V0sinωt, 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 Equation 15.13, 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 1102V=156 V.1102V=156 V. 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, ϕ=π/2andπ/2rad,ϕ=π/2andπ/2rad, respectively. Since cosπ/2=cos(π/2)=0,cosπ/2=cos(π/2)=0, we find from Equation 15.12 that the average power dissipated by either of these elements is Pave=0.Pave=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 Figure 15.16, (b) and (c), which show p(t) oscillating sinusoidally about zero.

The phase angle for an ac generator may have any value. If cosϕ>0,cosϕ>0, the generator produces power; if cosϕ<0,cosϕ<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,




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.

Example 15.3

Power Output of a Generator

An ac generator whose emf is given by

is connected to an RLC circuit for which L=2.00×10−3HL=2.00×10−3H, C=4.00×10−6FC=4.00×10−6F, and R=5.00ΩR=5.00Ω. (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?


The rms voltage is the amplitude of the voltage times 1/21/2. The impedance of the circuit involves the resistance and the reactances of the capacitor and the inductor. The average power is calculated by Equation 15.14, or more specifically, the last part of the equation, because we have the impedance of the circuit Z, the rms voltage VrmsVrms, and the resistance R.


  1. Since V0=4.00V,V0=4.00V, the rms voltage across the generator is
  2. The impedance of the circuit is
  3. From Equation 15.14, the average power transferred to the circuit is


If the resistance is much larger than the reactance of the capacitor or inductor, the average power is a dc circuit equation of P=V2/R,P=V2/R, where V replaces the rms voltage.

Check Your Understanding 15.4

An ac voltmeter attached across the terminals of a 45-Hz ac generator reads 7.07 V. Write an expression for the emf of the generator.

Check Your Understanding 15.5

Show that the rms voltages across a resistor, a capacitor, and an inductor in an ac circuit where the rms current is IrmsIrms are given by IrmsR,IrmsXC,andIrmsXL,IrmsR,IrmsXC,andIrmsXL, respectively. Determine these values for the components of the RLC circuit of Equation 15.12.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.


This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at
Citation information

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.