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15.4 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(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, whereas for (b) the capacitor and (c) the inductor, Pave=0. (d) For the source, Pave=I0V0(cosϕ)/2, which may be positive, negative, or zero, depending on 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:

Pave=1T0Tp(t)dt,

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

Pave=I0V0T0Tsin(ωtϕ)sinωtdt.

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

Pave=I0V0cosϕT0Tsin2ωtdtI0V0sinϕT0Tsinωtcosωtdt.

Evaluation of these two integrals yields

1T0Tsin2ωtdt=12

and

1T0Tsinωtcosωtdt=0.

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

Pave=12I0V0cosϕ.

(15.12)

In engineering applications, cosϕ 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, so the average power dissipated is

Pave=12I0V0.

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

Irms=iave2andVrms=vave2,

where

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

With i(t)=I0sin(ωtϕ)andv(t)=V0sinωt, we obtain

Irms=12I0andVrms=12V0.

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

Pave=12I0V0=IrmsVrms=Irms2R.

(15.13)

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. 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, respectively. Since 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. 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, the generator produces power; if cosϕ<0, it absorbs power. In terms of rms values, the average power of an ac generator is written as

Pave=IrmsVrmscosϕ.

For the generator in an RLC circuit,

tanϕ=XLXCR

and

cosϕ=RR2+(XLXC)2=RZ.

Hence the average power of the generator is

Pave=IrmsVrmscosϕ=VrmsZVrmsRZ=Vrms2RZ2.

(15.14)

This can also be written as

Pave=Irms2R,

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.

Summary

  • The average ac power is found by multiplying the rms values of current and voltage.
  • Ohm’s law for the rms ac is found by dividing the rms voltage by the impedance.
  • In an ac circuit, there is a phase angle between the source voltage and the current, which can be found by dividing the resistance by the impedance.
  • The average power delivered to an RLC circuit is affected by the phase angle.
  • The power factor ranges from –1 to 1.

Conceptual Questions

For what value of the phase angle ϕ between the voltage output of an ac source and the current is the average power output of the source a maximum?

Discuss the differences between average power and instantaneous power.

The instantaneous power is the power at a given instant. The average power is the power averaged over a cycle or number of cycles.

The average ac current delivered to a circuit is zero. Despite this, power is dissipated in the circuit. Explain.

Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?

The instantaneous power can be negative, but the power output can’t be negative.

The power rating of a resistor used in ac circuits refers to the maximum average power dissipated in the resistor. How does this compare with the maximum instantaneous power dissipated in the resistor?

Problems

The emf of an ac source is given by v(t)=V0sinωt, where V0=100V and ω=200πrad/s. Calculate the average power output of the source if it is connected across (a) a 20-μF capacitor, (b) a 20-mH inductor, and (c) a 50-Ω resistor.

Calculate the rms currents for an ac source is given by v(t)=V0sinωt, where V0=100V and ω=200πrad/s when connected across (a) a 20-μF capacitor, (b) a 20-mH inductor, and (c) a 50-Ω resistor.

a. 0.89 A; b. 5.6A; c. 1.4 A

A 40-mH inductor is connected to a 60-Hz AC source whose voltage amplitude is 50 V. If an AC voltmeter is placed across the inductor, what does it read?

For an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively; R=500Ω; and L=0.20H. Find the average power dissipated in the resistor for the following values for the capacitance: (a) C=2.0μF and (b) C=0.20μF.

a. 5.3 W; b. 2.1 W

An ac source of voltage amplitude 10 V delivers electric energy at a rate of 0.80 W when its current output is 2.5 A. What is the phase angle ϕ between the emf and the current?

An RLC series circuit has an impedance of 60Ω and a power factor of 0.50, with the voltage lagging the current. (a) Should a capacitor or an inductor be placed in series with the elements to raise the power factor of the circuit? (b) What is the value of the reactance across the inductor that will raise the power factor to unity?

a. inductor; b. XL=52Ω