The circuit of FIGURE 28.11 has two identical lightbulbs, A and B. Which is brighter? Or are they equally bright? Think about this before going on.

FIGURE 28.11 Which lightbulb is brighter?

You might have been tempted to say that A is brighter. After all, the current gets to A first, so A might “use up” some of the current and leave less for B. But this would violate the law of conservation of charge. There are no junctions between A and B, so the current through the two bulbs must be the same. Hence the bulbs are equally bright.

What a bulb uses is not current but energy. The battery’s charge escalator is an energy-transfer process, transferring chemical energy $E_{\text{chem}}$ stored in the battery to the potential energy $U$ of the charges. That energy is, in turn, transformed into the thermal energy of the resistors, increasing their temperature until—in the case of lightbulb filaments—they glow.

A charge gains potential energy $\text{Δ}U=q\text{Δ}{V}_{\text{bat}}$ as it moves up the charge escalator in the battery. For an ideal battery, with $\text{Δ}{V}_{\text{bat}}=\epsilon$, the battery supplies energy $\text{Δ}U=q\epsilon$ as it lifts charge $q$ from the negative to the positive terminal.

It is useful to know the rate at which the battery supplies energy to the charges. Recall from Chapter 9 that the rate at which energy is transferred is power, measured in joules per second or watts. If energy $\text{Δ}U=q\epsilon$ is transferred to charge $q$, then the rate at which energy is transferred from the battery to the moving charges is

But $dq\text{/}dt$, the rate at which charge moves through the battery, is the current I. Hence the power supplied by a battery, or the rate at which the battery (or any other source of emf) transfers energy to the charges passing through it, is

$I\epsilon$ has units of $\text{J}\text{/}\text{s}$, or W. For example, a 120 V battery that generates 2 A of current is delivering 240 W of power to the circuit.