Contents

Preface
1What is Physics?
2Motion in One Dimension
3Acceleration
4Forces and Newton’s Laws of Motion
5Motion in Two Dimensions
6Circular and Rotational Motion
7Newton's Law of Gravitation
8Momentum
9Work, Energy, and Simple Machines
10Special Relativity
11Thermal Energy, Heat, and Work
12Thermodynamics
13Waves and Their Properties
14Sound
15Light
16Mirrors and Lenses
- Introduction
- 16.1Reflection
- 16.2Refraction
- 16.3Lenses
- Key Terms
- Section Summary
- Key Equations
- Chapter Review
17Diffraction and Interference
18Static Electricity
19Electrical Circuits
20Magnetism
21The Quantum Nature of Light
22The Atom
23Particle Physics
- Introduction
- 23.1The Four Fundamental Forces
- 23.2Quarks
- 23.3The Unification of Forces
- Key Terms
- Section Summary
- Chapter Review
AReference Tables
Index

Parallel Circuits

Section Learning Objectives

By the end of this section, you will be able to do the following:

Interpret circuit diagrams with parallel resistors
Calculate equivalent resistance of resistor combinations containing series and parallel resistors

Section Key Terms

in parallel

Resistors in Parallel

In the previous section, we learned that resistors in series are resistors that are connected one after the other. If we instead combine resistors by connecting them next to each other, as shown in Figure 19.16, then the resistors are said to be connected in parallel. Resistors are in parallel when both ends of each resistor are connected directly together.

Note that the tops of the resistors are all connected to the same wire, so the voltage at the top of the each resistor is the same. Likewise, the bottoms of the resistors are all connected to the same wire, so the voltage at the bottom of each resistor is the same. This means that the voltage drop across each resistor is the same. In this case, the voltage drop is the voltage rating V of the battery, because the top and bottom wires connect to the positive and negative terminals of the battery, respectively.

Although the voltage drop across each resistor is the same, we cannot say the same for the current running through each resistor. Thus, $I_{1}, I_{2}, and I_{3} I_{1}, I_{2}, and I_{3}$ are not necessarily the same, because the resistors $R_{1}, R_{2}, and R_{3} R_{1}, R_{2}, and R_{3}$ do not necessarily have the same resistance.

Note that the three resistors in Figure 19.16 provide three different paths through which the current can flow. This means that the equivalent resistance for these three resistors must be less than the smallest of the three resistors. To understand this, imagine that the smallest resistor is the only path through which the current can flow. Now add on the alternate paths by connecting other resistors in parallel. Because the current has more paths to go through, the overall resistance (i.e., the equivalent resistance) will decrease. Therefore, the equivalent resistance must be less than the smallest resistance of the parallel resistors.

On the left is a circuit diagram with three resistors connected in parallel. On the right is a circuit diagram with only one resistor that has equivalent resistance to the three resistors shown on the left.

Figure 19.16 The left circuit diagram shows three resistors in parallel. The voltage V of the battery is applied across all three resistors. The currents that flow through each branch are not necessarily equal. The right circuit diagram shows an equivalent resistance that replaces the three parallel resistors.

To find the equivalent resistance $R_{equiv} R_{equiv}$ of the three resistors $R_{1}, R_{2}, and R_{3} R_{1}, R_{2}, and R_{3}$ , we apply Ohm’s law to each resistor. Because the voltage drop across each resistor is V, we obtain

V = I_{1} R_{1}, V = I_{2} R_{2}, V = I_{3} R_{3} V = I_{1} R_{1}, V = I_{2} R_{2}, V = I_{3} R_{3}

19.21

I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}, I_{3} = \frac{V}{R_{3} .} I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}, I_{3} = \frac{V}{R_{3} .}

19.22

We also know from conservation of charge that the three currents $I_{1}, I_{2}, and I_{3} I_{1}, I_{2}, and I_{3}$ must add up to give the current I that goes through the battery. If this were not true, current would have to be mysteriously created or destroyed somewhere in the circuit, which is physically impossible. Thus, we have

I = I_{1} + I_{2} + I_{3} . I = I_{1} + I_{2} + I_{3} .

19.23

Inserting the expressions for $I_{1}, I_{2}, and I_{3} I_{1}, I_{2}, and I_{3}$ into this equation gives

I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}} = V (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}) I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}} = V (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}})

19.24

V = I (\frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}}) . V = I (\frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}}) .

19.25

This formula is just Ohm’s law, with the factor in parentheses being the equivalent resistance.

V = I (\frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}}) = I R_{equiv.} V = I (\frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}}) = I R_{equiv.}

19.26

Thus, the equivalent resistance for three resistors in parallel is

R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} . R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} .

19.27

The same logic works for any number of resistors in parallel, so the general form of the equation that gives the equivalent resistance of N resistors connected in parallel is

R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + \dots + 1 / R_{N}} . R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + \dots + 1 / R_{N}} .

19.28

Worked Example

Find the Current through Parallel Resistors

The three circuits below are equivalent. If the voltage rating of the battery is $V_{battery} = 3 V V_{battery} = 3$ , what is the equivalent resistance of the circuit and what current runs through the circuit?

Three equivalent circuit diagrams are shown, each with three resistors connected in parallel.

Strategy

The three resistors are connected in parallel and the voltage drop across them is V_battery. Thus, we can apply the equation for the equivalent resistance of resistors in parallel, which takes the form

R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} . R_{equiv} = \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} .

19.29

The circuit with the equivalent resistance is shown below. Once we know the equivalent resistance, we can use Ohm’s law to find the current in the circuit.

A circuit diagram showing only one resistor that is equivalent to the three resistors shown in each of the three diagrams shown above.

Show solution

Solution

Inserting the given values for the resistance into the equation for equivalent resistance gives

\begin{matrix} R_{equiv} & = & \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} \\ = & \frac{1}{1 / 10 Ω + 1 / 25 Ω + 1 / 15 Ω} \\ = & 4.84 Ω. \end{matrix} \begin{matrix} R_{equiv} & = & \frac{1}{1 / R_{1} + 1 / R_{2} + 1 / R_{3}} \\ = & \frac{1}{1 / 10 + 1 / 25 + 1 / 15} \\ = & 4.84 \end{matrix}

19.30

The current through the circuit is thus

\begin{matrix} V & = & I R \\ I & = & \frac{V}{R} \\ = & \frac{3 V}{4.84 Ω} \\ = & 0.62 A. \end{matrix} \begin{matrix} V & = & I R \\ I & = & \frac{V}{R} \\ = & \frac{3}{4.84} \\ = & 0.62 \end{matrix}

19.31

Discussion

Although 0.62 A flows through the entire circuit, note that this current does not flow through each resistor. However, because electric charge must be conserved in a circuit, the sum of the currents going through each branch of the circuit must add up to the current going through the battery. In other words, we cannot magically create charge somewhere in the circuit and add this new charge to the current. Let’s check this reasoning by using Ohm’s law to find the current through each resistor.

\begin{array}{l} I_{1} = \frac{V}{R_{1}} = \frac{3 V}{10 Ω} = 0.30 A \\ I_{2} = \frac{V}{R_{2}} = \frac{3 V}{25 Ω} = 0.12 A \\ I_{3} = \frac{V}{R_{3}} = \frac{3 V}{15 Ω} = 0.20 A \end{array} \begin{array}{l} I_{1} = \frac{V}{R_{1}} = \frac{3}{10} = 0.30 \\ I_{2} = \frac{V}{R_{2}} = \frac{3}{25} = 0.12 \\ I_{3} = \frac{V}{R_{3}} = \frac{3}{15} = 0.20 \end{array}

19.32

As expected, these currents add up to give 0.62 A, which is the total current found going through the equivalent resistor. Also, note that the smallest resistor has the largest current flowing through it, and vice versa.

Worked Example

Reasoning with Parallel Resistors

Without doing any calculation, what is the equivalent resistance of three identical resistors R in parallel?

Strategy

Three identical resistors R in parallel make three identical paths through which the current can flow. Thus, it is three times easier for the current to flow through these resistors than to flow through a single one of them.

Show solution

Solution

If it is three times easier to flow through three identical resistors R than to flow through a single one of them, the equivalent resistance must be three times less: R/3.

Discussion

Let’s check our reasoning by calculating the equivalent resistance of three identical resistors R in parallel. The equation for the equivalent resistance of resistors in parallel gives

\begin{matrix} R_{equiv} & = & \frac{1}{1 / R + 1 / R + 1 / R} \\ = & \frac{1}{3 / R} \\ = & \frac{R}{3} . \end{matrix} \begin{matrix} R_{equiv} & = & \frac{1}{1 / R + 1 / R + 1 / R} \\ = & \frac{1}{3 / R} \\ = & \frac{R}{3} . \end{matrix}

19.33

Thus, our reasoning was correct. In general, when more paths are available through which the current can flow, the equivalent resistance decreases. For example, if we have identical resistors R in parallel, the equivalent resistance would be R/10.

Practice Problems

10.

Three resistors, 10, 20, and 30 Ω, are connected in parallel. What is the equivalent resistance?

The equivalent resistance is 5.5 Ω
The equivalent resistance is 60 Ω
The equivalent resistance is 6 × 103 Ω
The equivalent resistance is 6 × 104 Ω

11 .

Watch Physics: Resistors in Parallel. This video introduces and explains how resistors work when parallel.

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Grasp Check

True or false—In a circuit diagram, we can assume that the voltage is the same at every point in a given wire.

false
true

Watch Physics

Resistors in Series and in Parallel

This video shows how to calculate the equivalent resistance of a circuit containing resistors in parallel and in series. The lecturer uses the same approach as outlined above for finding the equivalent resistance.

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Grasp Check

Imagine connected N identical resistors in parallel. Each resistor has a resistance of R. What is the equivalent resistance for this group of parallel resistors?

The equivalent resistance is (R)^N.
The equivalent resistance is NR.
The equivalent resistance is $\frac{R}{N} . \frac{R}{N} .$
The equivalent resistance is $\frac{N}{R} . \frac{N}{R} .$

Worked Example

Find the Current through a Complex Resistor Circuit

The battery in the circuit below has a voltage rating of 10 V. What current flows through the circuit and in what direction?

A circuit diagram showing six resistors, some in series and others parallel.

Strategy

Apply the Strategy for finding equivalent resistance to replace all the resistors with a single equivalent resistance, then use Ohm’s law to find the current through the equivalent resistor.

Show solution

Solution

The resistor combination $R_{4} R_{4}$ and $R_{5} R_{5}$ can be reduced to an equivalent resistance of

R_{P1} = \frac{1}{1 / R_{4} + 1 / R_{5}} = \frac{1}{1 / 45 Ω + 1 / 60 Ω} = 25.71 Ω R. R_{P1} = \frac{1}{1 / R_{4} + 1 / R_{5}} = \frac{1}{1 / 45 + 1 / 60} = 25.71

19.38

Replacing $R_{4} R_{4}$ and $R_{5} R_{5}$ with this equivalent resistance gives the circuit below.

The same circuit diagram from above but with the two parallel resistors combined.

We now replace the two upper resistors $R_{2} R_{2}$ and $R_{3} R_{3}$ by the equivalent resistor $R_{S 1} R_{S 1}$ and the two lower resistors $R_{P 1} R_{P 1}$ and $R_{6} R_{6}$ by their equivalent resistor $R_{S2} R_{S2}$ . These resistors are in series, so we add them together to find the equivalent resistance.

\begin{matrix} R_{S1} & = & R_{2} + R_{3} & = & 50 Ω + 30 Ω & = & 80 Ω \\ R_{S2} & = & R_{P1} + R_{6} & = & 25.71 Ω + 20 Ω & = & 45.71 Ω \end{matrix} \begin{matrix} R_{S1} & = & R_{2} + R_{3} & = & 50 & = & 80 \\ R_{S2} & = & R_{P1} + R_{6} & = & 25.71 & = & 45.71 \end{matrix}

19.39

Replacing the relevant resistors with their equivalent resistor gives the circuit below.

The same circuit diagram as above, but with two pairs of resistors in series combined.

Now replace the two resistors $R_{S1} and R_{S2} R_{S1} and R_{S2}$ , which are in parallel, with their equivalent resistor $R_{P2} R_{P2}$ . The resistance of $R_{P2} R_{P2}$ is

R_{P2} = \frac{1}{1 / R_{S1} + 1 / R_{S2}} = \frac{1}{1 / 80 Ω + 1 / 45.71 Ω} = 29.09 Ω. R_{P2} = \frac{1}{1 / R_{S1} + 1 / R_{S2}} = \frac{1}{1 / 80} = 29.09

19.40

Updating the circuit diagram by replacing $R_{S1} and R_{S2} R_{S1} and R_{S2}$ with this equivalent resistance gives the circuit below.

The same circuit diagram shown above but with the two resistors in series combined.

Finally, we combine resistors $R_{1} and R_{P2} R_{1} and R_{P2}$ , which are in series. The equivalent resistance is $R_{S3} = R_{1} + R_{P2} = 75 Ω + 29.09 Ω = 104.09 Ω . R_{S3} = R_{1} + R_{P2} = 75$ The final circuit is shown below.

The final circuit diagram showing the final two resistors combined.

We now use Ohm’s law to find the current through the circuit.

\begin{matrix} V & = & I R_{S3} \\ I & = & \frac{V}{R_{S3}} & = & \frac{10 V}{104.09 Ω} & = & 0.096 A \end{matrix} \begin{matrix} V & = & I R_{S3} \\ I & = & \frac{V}{R_{S3}} & = & \frac{10}{104.09} & = & 0.096 \end{matrix}

19.41

The current goes from the positive terminal of the battery to the negative terminal of the battery, so it flows clockwise in this circuit.

Discussion

This calculation may seem rather long, but with a little practice, you can combine some steps. Note also that extra significant digits were carried through the calculation. Only at the end was the final result rounded to two significant digits.

Worked Example

Strange-Looking Circuit Diagrams

Occasionally, you may encounter circuit diagrams that are not drawn very neatly, such as the diagram shown below. This circuit diagram looks more like how a real circuit might appear on the lab bench. What is the equivalent resistance for the resistors in this diagram, assuming each resistor is 10 $Ω Ω$ and the voltage rating of the battery is 12 V.

A circuit diagram with curved lines between all components instead of straight lines.

Strategy

Let’s redraw this circuit diagram to make it clearer. Then we’ll apply the Strategy outlined above to calculate the equivalent resistance.

Show solution

Solution

To redraw the diagram, consider the figure below. In the upper circuit, the blue resistors constitute a path from the positive terminal of the battery to the negative terminal. In parallel with this circuit are the red resistors, which constitute another path from the positive to negative terminal of the battery. The blue and red paths are shown more cleanly drawn in the lower circuit diagram. Note that, in both the upper and lower circuit diagrams, the blue and red paths connect the positive terminal of the battery to the negative terminal of the battery.

The circuit diagram above redrawn with straight lines. In the upper circuit, the blue resistors constitute a path from the positive terminal of the battery to the negative terminal. In parallel with this circuit are the red resistors, which constitute another path from the positive to negative terminal of the battery.

Now it is easier to see that $R_{1} and R_{2} R_{1} and R_{2}$ are in parallel, and the parallel combination is in series with $R_{4} R_{4}$ . This combination in turn is in parallel with the series combination of $R_{3} and R_{5} R_{3} and R_{5}$ . First, we calculate the blue branch, which contains $R_{1}, R_{2}, and R_{4} R_{1}, R_{2}, and R_{4}$ . The equivalent resistance is

R_{blue} = \frac{1}{1 / R_{1} + 1 / R_{2}} + R_{4} = \frac{1}{1 / 10 Ω + 1 / 10 Ω} + 10 Ω = 15 Ω. R_{blue} = \frac{1}{1 / R_{1} + 1 / R_{2}} + R_{4} = \frac{1}{1 / 10 Ω + 1 / 10} + 10

19.42

where we show the contribution from the parallel combination of resistors and from the series combination of resistors. We now calculate the equivalent resistance of the red branch, which is

R_{red} = R_{3} + R_{5} = 10 Ω + 10 Ω = 20 Ω . R_{red} = R_{3} + R_{5} = 10

19.43

Inserting these equivalent resistors into the circuit gives the circuit below.

The same circuit diagram but with the red resistors combined and the blue resistors combined.

These two resistors are in parallel, so they can be replaced by a single equivalent resistor with a resistance of

R_{equiv} = \frac{1}{1 / R_{blue} + 1 / R_{red}} = \frac{1}{1 / 15 Ω + 1 / 20 Ω} = 8.6 Ω. R_{equiv} = \frac{1}{1 / R_{blue} + 1 / R_{red}} = \frac{1}{1 / 15 + 1 / 20} = 8.6

19.44

The final equivalent circuit is show below.

The simplified circuit diagram with one resistor equivalent to the original five resistors.

Discussion

Finding the equivalent resistance was easier with a clear circuit diagram. This is why we try to make clear circuit diagrams, where the resistors in parallel are lined up parallel to each other and at the same horizontal position on the diagram.

We can now use Ohm’s law to find the current going through each branch to this circuit. Consider the circuit diagram with $R_{blue} R_{blue}$ and $R_{red} R_{red}$ . The voltage across each of these branches is 12 V (i.e., the voltage rating of the battery). The current in the blue branch is

I_{blue} = \frac{V}{R_{blue}} = \frac{12 V}{15 Ω} = 0.80 A. I_{blue} = \frac{V}{R_{blue}} = \frac{12}{15} = 0.80

19.45

The current across the red branch is

I_{red} = \frac{V}{R_{red}} = \frac{12 V}{20 Ω} = 0.60 A. I_{red} = \frac{V}{R_{red}} = \frac{12}{20} = 0.60

19.46

The current going through the battery must be the sum of these two currents (can you see why?), or 1.4 A.

Practice Problems

12.

What is the formula for the equivalent resistance of two parallel resistors with resistance R₁ and R₂?

Equivalent resistance of two parallel resistors $R_{eqv} = R_{1} + R_{2} R_{eqv} = R_{1} + R_{2}$
Equivalent resistance of two parallel resistors $R_{eqv} = R_{1} \times R_{2} R_{eqv} = R_{1}$
Equivalent resistance of two parallel resistors $R_{eqv} = R_{1} - R_{2} R_{eqv} = R_{1} - R_{2}$
Equivalent resistance of two parallel resistors $R_{eqv} = \frac{1}{1 / R_{1} + 1 / R_{2}} R_{eqv} = \frac{1}{1 / R_{1} + 1 / R_{2}}$

13.

This shows a vertical parallel circuit. In the middle of either side there are resistors. The left resistor is labeled R1 = 35Ω. The right resistor is labeled R1 = 55Ω

Figure 19.19

What is the equivalent resistance for the two resistors shown?

The equivalent resistance is 20 Ω
The equivalent resistance is 21 Ω
The equivalent resistance is 90 Ω
The equivalent resistance is 1,925 Ω

Check Your Understanding

14.

The voltage drop across parallel resistors is ________.

the same for all resistors
greater for the larger resistors
less for the larger resistors
greater for the smaller resistors

15.

Consider a circuit of parallel resistors. The smallest resistor is 25 Ω . What is the upper limit of the equivalent resistance?

The upper limit of the equivalent resistance is 2.5 Ω.
The upper limit of the equivalent resistance is 25 Ω.
The upper limit of the equivalent resistance is 100 Ω.
There is no upper limit.