For a resistor, power is the rate at which energy is dissipated. This is

We can use $i_{\text{R}}=I_{\text{R}}\cos \omega t$ to write the resistor’s instantaneous power as

FIGURE 32.20 shows the instantaneous power graphically. You can see that, because the cosine is squared, the power oscillates twice during every cycle of the emf. The energy dissipation peaks both when $i_{\text{R}}=I_{\text{R}}$ and when $i_{\text{R}}=-I_{\text{R}}$.

FIGURE 32.20 The instantaneous current and power of a resistor.

In practice, we’re more interested in the average power than in the instantaneous power. The average power $P$ is the total energy dissipated per second. We can find $P_{\text{R}}$ for a resistor by using the identity ${\text{cos}}^2\left(x\right)=\frac{1}{2}(1+\text{cos}2x)$ to write

The $\text{cos}2\omega t$ term oscillates positive and negative twice during each cycle of the emf. Its average, over one cycle, is zero. Thus the average power dissipation of a resistor is

It is useful to write Equation 32.25 as

where the quantity

is called the root-mean-square current, or rms current, $I_{\text{rms}}$. Technically, an rms quantity is the square root of the average, or mean, of the quantity squared. For a sinusoidal oscillation, the rms value turns out to be the peak value divided by $\sqrt{2}$.

The rms current allows us to compare Equation 32.36 directly to the energy dissipated by a resistor in a DC circuit: $P=I^{2}R$. You can see that the average power of a resistor in an AC circuit with $I_{\text{rms}}=1\text{ A}$ is the same as in a DC circuit with $I=1\text{A}$. As far as power is concerned, an rms current is equivalent to an equal DC current.

Similarly, we can define the root-mean-square voltage and emf:

The resistor’s average power in terms of the rms quantities is

and the average power supplied by the emf is

The single-resistor circuit that we analyzed in Section 32.1 had $V_{\text{R}}=\epsilon$ or, equivalently, $V_{\text{rms}}={\epsilon }_{\text{rms}}$. You can see from Equations 32.39 and 32.40 that the power of the resistor (rate of energy dissipation) exactly matches the power of the emf (rate of energy supplied). This must be the case in order to conserve energy.

EXAMPLE 32.7 Drying your hair

A 1600 W hair dryer is plugged into a $120\text{V}\text{/}60\text{Hz}$ outlet. What is the resistance of the heater element? What is the peak current through the hair dryer?

MODEL The dryer’s heater element acts as a resistor.

VISUALIZE FIGURE 32.21 is a simple one-resistor circuit.

FIGURE 32.21 An AC circuit with a heater element as a resistor.

FIGURE 32.21 Full Alternative Text

SOLVE An appliance labeled 1600 W is designed to dissipate an average 1600 W at $V_{\text{rms}}=120\text{V}$. We can use Equation 32.39 to find

$$R={\displaystyle \frac{(V_{\text{rms}}{)}^2}{P_{\text{R}}}}={\displaystyle \frac{(120\text{V}{)}^2}{1600\text{W}}}=9\text{Ω}$$

The rms current is then found from

$$I_{\text{rms}}={\displaystyle \frac{P_{\text{R}}}{V_{\text{rms}}}}={\displaystyle \frac{1600\text{W}}{120\text{V}}}=13.3\text{A}$$

The peak current is $I_{\text{R}}=\sqrt{2}{I}_{\text{rms}}=19\text{A}$.

REVIEW Calculations with rms values are just like the calculations for DC circuits.