28.3.1 Energy Dissipation in Resistors

$P_{bat}$ is the energy transferred per second from the battery’s store of chemicals to the moving charges that make up the current. But what happens to this energy? Where does it end up? FIGURE 28.12, a section of a current-carrying resistor, reminds you of our microscopic model of conduction. The electrons accelerate in the electric field, transforming potential energy into kinetic, then collide with atoms in the lattice. The collisions transfer the electron’s kinetic energy to the thermal energy of the lattice. The potential energy was acquired in the battery, so the entire energy-transfer process looks like

E_{chem} \to U \to K \to E_{th}

The net result is that the battery’s chemical energy is transferred to the thermal energy of the resistors, raising their temperature.

The figure illustrates the dissipation of energy in a current-carrying resistor.

FIGURE 28.12 Full Alternative Text

Consider a charge $q$ that moves all the way through a resistor with a potential difference $Δ V_{R}$ between its two ends. The charge loses potential energy $Δ U = - q Δ V_{R}$ , and, after a vast number of collisions, all that energy is transformed into thermal energy. Thus the resistor’s increase in thermal energy due to this one charge is

Δ E_{th} = q Δ V_{R}

(28.9)

The rate at which energy is transferred from the current to the resistor is then

P_{R} = \frac{d E_{th}}{d t} = \frac{d q}{d t} Δ V_{R} = I Δ V_{R}

(28.10)

Power—so many joules per second—is the rate at which energy is dissipated by the resistor as charge flows through it. The resistor, in turn, transfers this energy to the air and to the circuit board on which it is mounted, causing the circuit and all its surroundings to heat up.

From our analysis of the basic circuit, in which a single resistor is connected to a battery, we learned that $Δ V_{R} = ε$ . That is, the potential difference across the resistor is exactly the emf supplied by the battery. But then Equations 28.8 and 28.10, for $P_{bat}$ and $P_{R}$ , are numerically equal, and we find that

P_{R} = P_{bat}

(28.11)

The answer to the question “What happens to the energy supplied by the battery?” is “The battery’s chemical energy is transformed into the thermal energy of the resistor.” The rate at which the battery supplies energy is exactly equal to the rate at which the resistor dissipates energy. This is, of course, exactly what we would have expected from energy conservation.

EXAMPLE 28.2 The power of light

How much current is “drawn” by a 100 W incandescent lightbulb connected to a 120 V outlet?

MODEL Most household appliances, such as a 100 W lightbulb or a 1500 W hair dryer, have a power rating. The rating does not mean that these appliances always dissipate that much power. These appliances are intended for use at a standard household voltage of 120 V, and their rating is the power they will dissipate if operated with a potential difference of 120 V. Their power consumption will differ from the rating if they are operated at any other potential difference.

SOLVE Because the lightbulb is operating as intended, it will dissipate 100 W of power. Thus

I = \frac{P_{R}}{Δ V_{R}} = \frac{100 W}{120 V} = 0.833 A

REVIEW A current of 0.833 A in this lightbulb transfers $100 J / s$ to the thermal energy of the filament, which, in turn, dissipates $100 J / s$ as heat and light to its surroundings.

A resistor obeys Ohm’s law, $Δ V_{R} = I R$ . (Remember that Ohm’s law gives only the magnitude of $Δ V_{R}$ .) This gives us two alternative ways of writing the power dissipated by a resistor. We can either substitute IR for $Δ V_{R}$ or substitute $Δ V_{R} / R$ for I. Thus

P_{R} = I Δ V_{R} = I^{2} R = \frac{(Δ V_{R})^{2}}{R} (power dissipated by a resistor)

(28.12)

If the same current $I$ passes through several resistors in series, then $P_{R} = I^{2} R$ tells us that most of the power will be dissipated by the largest resistance. This is why a lightbulb filament glows but the connecting wires do not. Essentially all of the power supplied by the battery is dissipated by the high-resistance lightbulb filament and essentially no power is dissipated by the low-resistance wires. The filament gets very hot, but the wires do not.

EXAMPLE 28.3 The power of sound

Most loudspeakers are designed to have a resistance of $8 Ω$ . If an $8 Ω$ loudspeaker is connected to a stereo amplifier with a rating of 100 W, what is the maximum possible current to the loudspeaker?

MODEL The rating of an amplifier is the maximum power it can deliver. Most of the time it delivers far less, but the maximum might be reached for brief, intense sounds like cymbal crashes.

SOLVE The loudspeaker is a resistive load. The maximum current to the loudspeaker occurs when the amplifier delivers maximum power $P_{max} = (I_{max})^{2} R$ . Thus

I_{max} = \sqrt{\frac{P_{max}}{R}} = \sqrt{\frac{100 W}{8 Ω}} = 3.5 A