FIGURE 28.20 is another lightbulb puzzle. Initially the switch is open. The current is the same through bulbs A and B and they are equally bright. Bulb C is not glowing. What happens to the brightness of A and B when the switch is closed? And how does the brightness of C then compare to that of A and B? Think about this before going on.

FIGURE 28.20 What happens to the brightness of the bulbs when the switch is closed?

FIGURE 28.21a shows two resistors aligned side by side with their ends connected at c and d. Resistors connected at both ends are called parallel resistors or, sometimes, resistors “in parallel.” The left ends of both resistors are at the same potential $V_{\text{c}}$. Likewise, the right ends are at the same potential $V_{\text{d}}$. Thus the potential differences $\text{Δ}{V}_1$ and $\text{Δ}{V}_2$ are the same and are simply $\text{Δ}{V}_{\text{cd}}$.

FIGURE 28.21 Replacing two parallel resistors with an equivalent resistor.

Kirchhoff’s junction law applies at the junctions. The input current $I$ splits into currents $I_1$ and $I_2$ at the left junction. On the right, the two currents are recombined into current I. According to the junction law,

We can apply Ohm’s law to each resistor, along with $\text{Δ}{V}_1=\text{Δ}{V}_2=\text{Δ}{V}_{\text{cd}}$, to find that the current is

Suppose, as in FIGURE 28.21b, we replaced the two resistors with a single resistor having current $I$ and potential difference $\text{Δ}{V}_{\text{cd}}$. This resistor is equivalent to the original two because the battery has to establish the same potential difference and provide the same current in either case. A second application of Ohm’s law shows that the resistance between points c and d is

The single resistor $R_{\text{cd}}$ draws the same current as resistors $R_1$ and $R_2$, so, as far as the battery is concerned, resistor $R_{\text{cd}}$ is equivalent to the two resistors in parallel.

There is nothing special about having chosen two resistors to be in parallel. If we have $N$ resistors in parallel, the equivalent resistance is

The behavior of the circuit will be unchanged if the $N$ parallel resistors are replaced by the single resistor $R_{\text{eq}}$. The key idea of this analysis is that resistors in parallel all have the same potential difference.

EXAMPLE 28.7 A parallel resistor circuit

The three resistors of FIGURE 28.22 are connected to a 9 V battery. Find the potential difference across and the current through each resistor.

FIGURE 28.22 Parallel resistor circuit of Example 28.7.

FIGURE 28.22 Full Alternative Text

MODEL The resistors are in parallel. Assume an ideal battery and ideal connecting wires.

SOLVE The three parallel resistors can be replaced by a single equivalent resistor

$$R_{\text{eq}}={({\displaystyle \frac{1}{15\text{Ω}}}+{\displaystyle \frac{1}{4\text{Ω}}}+{\displaystyle \frac{1}{8\text{Ω}}})}^{-1}={\left(0.4417{\text{Ω}}^{-1}\right)}^{-1}=2.26\text{Ω}$$

The equivalent circuit is shown in FIGURE 28.23a, from which we find the current to be

$$I={\displaystyle \frac{\epsilon }{R_{\text{eq}}}}={\displaystyle \frac{9\text{V}}{2.26\text{Ω}}}=3.98\text{A}$$

The potential difference across $R_{\text{eq}}$ is $\text{Δ}{V}_{\text{eq}}=\epsilon =9.0\text{V}$. Now we have to be careful. Current $I$ divides at the junction into the smaller currents $I_1,I_2$, and $I_3$ shown in FIGURE 28.23b. However, the division is not into three equal currents. According to Ohm’s law, resistor $i$ has current $I_i=\text{Δ}{V}_i\text{/}{R}_i$. Because the three resistors are each connected to the battery by ideal wires, as is the equivalent resistor, their potential differences are equal:

$$\text{Δ}{V}_1=\text{Δ}{V}_2=\text{Δ}{V}_3=\text{Δ}{V}_{\text{eq}}=9.0\text{V}$$

Thus the currents are

$$\begin{array}{ccc}{I}_1=\frac{9\text{ V}}{15\text{ Ω}}=0.60\text{ A}& I_2=\frac{9\text{ V}}{4\text{ Ω}}=2.25\text{ A}& I_3=\frac{9\text{ V}}{8\text{ Ω}}=1.13\text{ A}\end{array}$$

FIGURE 28.23 The parallel resistors can be replaced by a single equivalent resistor.

FIGURE 28.23 Full Alternative Text

REVIEW The sum of the three currents is 3.98 A, as required by Kirchhoff’s junction law.

The result of Example 28.7 seems surprising. The equivalent of a parallel combination of $15\text{Ω},4\text{Ω}$, and $8\text{Ω}$ was found to be $2.26\text{Ω}$. How can the equivalent of a group of resistors be less than any single resistance in the group? Shouldn’t more resistors imply more resistance? The answer is yes for resistors in series but not for resistors in parallel. Even though a resistor is an obstacle to the flow of charge, parallel resistors provide more pathways for charge to get through. Consequently, the equivalent of several resistors in parallel is always less than any single resistor in the group.

Complex combinations of resistors can often be reduced to a single equivalent resistance through a step-by-step application of the series and parallel rules. The final example in this section illustrates this idea.

EXAMPLE 28.8 A combination of resistors

What is the equivalent resistance of the group of resistors shown in FIGURE 28.24?

FIGURE 28.24 A combination of resistors.

FIGURE 28.24 Full Alternative Text

MODEL This circuit contains both series and parallel resistors.

SOLVE Reduction to a single equivalent resistance is best done in a series of steps, with the circuit being redrawn after each step. The procedure is shown in FIGURE 28.25. Note that the $10\text{Ω}$ and $25\text{Ω}$ resistors are not in parallel. They are connected at their top ends but not at their bottom ends. Resistors must be connected to each other at both ends to be in parallel. Similarly, the $10\text{Ω}$ and $45\text{Ω}$ resistors are not in series because of the junction between them. If the original group of four resistors occurred within a larger circuit, they could be replaced with a single $15.4\text{Ω}$ resistor without having any effect on the rest of the circuit.

FIGURE 28.25 The combination is reduced to a single equivalent resistor.

FIGURE 28.25 Full Alternative Text

To return to the lightbulb question that began this section, FIGURE 28.26 has redrawn the circuit with each bulb shown as a resistance $R$. Initially, before the switch is closed, bulbs A and B are in series with equivalent resistance 2R. The current from the battery is

FIGURE 28.26 The lightbulbs of Figure 28.20 with the switch open and closed.

This is the current in both bulbs.

Closing the switch places bulbs B and C in parallel. The equivalent resistance of two identical resistors in parallel is $R_{\text{eq}}=\frac{1}{2}R$. This equivalent resistance of B and C is in series with bulb A; hence the total resistance of the circuit is $\frac{3}{2}R$ and the current leaving the battery is

Closing the switch decreases the circuit resistance and thus increases the current leaving the battery.

All the charge flows through A, so A increases in brightness when the switch is closed. The current $I_{\text{after}}$ then splits at the junction. Bulbs B and C have equal resistance, so the current splits equally. The current in B is $\frac{1}{3}\left(\epsilon \text{/}R\right)$, which is less than $I_{\text{before}}$. Thus B decreases in brightness when the switch is closed. Bulb C has the same brightness as bulb B.