Consider the three lightbulbs in FIGURE 28.13. The batteries are identical and the bulbs are identical. You learned in the previous section that B and C are equally bright, because the current is the same through both, but how does the brightness of B compare to that of A? Think about this before going on.

FIGURE 28.13 How does the brightness of bulb B compare to that of A?

FIGURE 28.14a shows two resistors placed end to end between points a and b. Resistors that are aligned end to end, with no junctions between them, are called series resistors or, sometimes, resistors “in series.” Because there are no junctions, the current $I$ must be the same through each of these resistors. That is, the current out of the last resistor in a series is equal to the current into the first resistor.

FIGURE 28.14 Replacing two series resistors with an equivalent resistor.

The potential differences across the two resistors are $\text{Δ}{V}_1=I{R}_1$ and $\text{Δ}{V}_2=I{R}_2$. The total potential difference $\text{Δ}{V}_{\text{ab}}$ between points a and b is the sum of the individual potential differences:

Suppose, as in FIGURE 28.14b, we replaced the two resistors with a single resistor having current $I$ and potential difference $\text{Δ}{V}_{\text{ab}}=\text{Δ}{V}_1+\text{Δ}{V}_2$. We can then use Ohm’s law to find that the resistance $R_{\text{ab}}$ between points a and b is

Because the battery has to establish the same potential difference across the load and provide the same current in both cases, the two resistors $R_1$ and $R_2$ act exactly the same as a single resistor of value $R_1+R_2$. We can say that the single resistor $R_{\text{ab}}$ is equivalent to the two resistors in series.

There was nothing special about having only two resistors. If we have $N$ resistors in series, their equivalent resistance is

The current and the power output of the battery will be unchanged if the $N$ series resistors are replaced by the single resistor $R_{\text{eq}}$. The key idea in this analysis is that resistors in series all have the same current.

Now we can answer the lightbulb question posed at the beginning of this section. Suppose the resistance of each lightbulb is R. The battery drives current $I_{\text{A}}=\epsilon \text{/}R$ through bulb A. Bulbs B and C are in series, with an equivalent resistance $R_{\text{eq}}=2R$, but the battery has the same emf $\epsilon$. Thus the current through bulbs B and C is $I_{\text{B}+\text{C}}=\epsilon \text{/}{R}_{\text{eq}}=\epsilon \text{/}2R={\displaystyle \frac{1}{2}}{I}_{\text{A}}$. Bulb B has only half the current of bulb A, so B is dimmer.

Many people predict that A and B should be equally bright. It’s the same battery, so shouldn’t it provide the same current to both circuits? No! A battery is a source of emf, not a source of current. In other words, the battery’s emf is the same no matter how the battery is used. When you buy a 1.5 V battery you’re buying a device that provides a specified amount of potential difference, not a specified amount of current. The battery does provide the current to the circuit, but the amount of current depends on the resistance of the load. Your 1.5 V battery causes 1 A to pass through a $1.5\text{Ω}$ load but only 0.1 A to pass through a $15\text{Ω}$ load. As an analogy, think about a water faucet. The pressure in the water main underneath the street is a fixed and unvarying quantity set by the water company, but the amount of water coming out of a faucet depends on how far you open it. A faucet opened slightly has a “high resistance,” so only a little water flows. A wide-open faucet has a “low resistance,” and the water flow is large.

In summary, a battery provides a fixed and unvarying emf (potential difference). It does not provide a fixed and unvarying current. The amount of current depends jointly on the battery’s emf and the resistance of the circuit attached to the battery.

EXAMPLE 28.4 A series resistor circuit

a. What is the current in the circuit of FIGURE 28.15a?

b. Draw a graph of potential versus position in the circuit, going cw from $V=0\text{V}$ at the battery’s negative terminal.

FIGURE 28.15 Analyzing a circuit with series resistors.

FIGURE 28.15 Full Alternative Text

MODEL The three resistors are end to end, with no junctions between them, and thus are in series. Assume ideal connecting wires and an ideal battery.

SOLVE

a. The battery “acts” the same—it provides the same current at the same potential difference—if we replace the three series resistors by their equivalent resistance

$$R_{\text{eq}}=15\text{Ω}+4\text{Ω}+8\text{Ω}=27\text{Ω}$$

This is shown as an equivalent circuit in FIGURE 28.15b. Now we have a circuit with a single battery and a single resistor, for which we know the current to be

$$I={\displaystyle \frac{\epsilon }{R_{\text{eq}}}}={\displaystyle \frac{9\text{V}}{27\text{Ω}}}=0.333\text{A}$$

b. $I=0.333\text{A}$ is the current in each of the three resistors in the original circuit. Thus the potential differences across the resistors are $\text{Δ}{V}_{\text{res}1}=-I{R}_1=-5.0\text{V,}\text{Δ}{V}_{\text{res}2}=-I{R}_2=-1.3\text{V,}$ and $\text{Δ}{V}_{\text{res}3}=-I{R}_3=-2.7\text{V}$ for the $15\text{Ω}$, the $4\text{Ω}$, and the $8\text{Ω}$ resistors. FIGURE 28.15c shows that the potential increases by 9 V due to the battery, then decreases by 9 V in three steps.