Circuits and DC Instruments

27 Resistors in Series and Parallel

Learning Objectives

  • Draw a circuit with resistors in parallel and in series.
  • Calculate the voltage drop of a current across a resistor using Ohm’s law.
  • Contrast the way total resistance is calculated for resistors in series and in parallel.
  • Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit.
  • Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel.

Most circuits have more than one component. One common component is a resistor, which limits the flow of charge in the circuit. A measure of this opposition to charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 27.1. The total resistance of a combination of resistors depends on both their individual values and how they are connected.

In part a of the figure, resistors labeled R sub 1, R sub 2, R sub 3, and R sub 4 are connected in series along one path of a circuit. In part b of the figure, the same resistors are connected along parallel paths of a circuit.
Figure 27.1: (a) A series connection of resistors. (b) A parallel connection of resistors.

Resistors in Series

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must pass through each device sequentially, one after another. In this configuration, the same current flows through every component in the series path.

For example, if current flows through a person holding a screwdriver and into the Earth, then [latex]{R}_{1}[/latex] in Figure 27.1(a) could represent the resistance of the screwdriver’s shaft, [latex]{R}_{2}[/latex] the resistance of its handle, [latex]{R}_{3}[/latex] the person’s body resistance, and [latex]{R}_{4}[/latex] the resistance of their shoes. This example highlights how resistances in series can arise naturally in real-world situations, including those involving electrical safety.

Figure 27.2 shows resistors in series connected to a voltage source. It is reasonable to expect that the total resistance is the sum of the individual resistances, since the current must pass through each resistor in sequence.

This idea has practical implications. For example, a person wishing to reduce the risk of electrical shock could increase the total resistance in the path by wearing high-resistance rubber-soled shoes. On the other hand, a faulty appliance cord with unexpectedly high resistance could reduce the operating current and impair device performance.

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Figure 27.2: Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical potential, called a voltage drop, across each resistor in Figure 27.2.

According to Ohm’s law, the voltage drop [latex]V[/latex] across a resistor when a current flows through it is given by

[latex]V = IR[/latex]

where [latex]I[/latex] is the current in amperes (A) and [latex]R[/latex] is the resistance in ohms [latex](\Omega)[/latex]. Another way to interpret this relationship is that [latex]V[/latex] represents the voltage required to drive a current [latex]I[/latex] through a resistance [latex]R[/latex].

Thus, the voltage drop across [latex]{R}_{1}[/latex] is [latex]{V}_{1} = I{R}_{1}[/latex], across [latex]{R}_{2}[/latex] is [latex]{V}_{2} = I{R}_{2}[/latex], and across [latex]{R}_{3}[/latex] is [latex]{V}_{3} = I{R}_{3}[/latex]. The sum of these voltage drops equals the total voltage supplied by the source:

[latex]V = V_{1} + V_{2} + V_{3}[/latex]

This result follows directly from the principles of conservation of energy and conservation of charge. Electrical potential energy is given by

[latex]\text{PE} = qV[/latex]

where [latex]q[/latex] is the electric charge and [latex]V[/latex] is the voltage. The energy supplied by the source is therefore [latex]qV[/latex], while the energy dissipated by the resistors is

[latex]qV_{1} + qV_{2} + qV_{3}[/latex]

Since energy is conserved, the total energy supplied must equal the total energy dissipated. This confirms that the total voltage is distributed across the resistors and that their individual voltage drops add to the source voltage.

Conceptual Check

If one resistor in a series circuit increases in value, what happens to the total resistance and the current in the circuit? Explain your reasoning using Ohm’s law.

Connections: Conservation Laws

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are fundamental to all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electric circuits.

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus,

[latex]qV = qV_{1} + qV_{2} + qV_{3}[/latex]

The charge [latex]q[/latex] cancels from both sides, yielding

[latex]V = V_{1} + V_{2} + V_{3}[/latex]

as stated. Note that the same amount of charge passes through the battery and each resistor in a given time interval. Since there is no capacitance to store charge and no place for charge to accumulate or leak, charge is conserved throughout the circuit.

Now substituting the expressions for the individual voltage drops gives

[latex]V = IR_{1} + IR_{2} + IR_{3} = I(R_{1} + R_{2} + R_{3})[/latex]

For the equivalent single series resistance [latex]R_{\text{s}}[/latex], we also have

[latex]V = IR_{\text{s}}[/latex]

Comparing these two expressions shows that the total, or equivalent, series resistance [latex]R_{\text{s}}[/latex] of three resistors is

[latex]R_{\text{s}} = R_{1} + R_{2} + R_{3}[/latex]

This reasoning is valid in general for any number of resistors connected in series. Thus, the total resistance [latex]R_{\text{s}}[/latex] of a series connection is

[latex]R_{\text{s}} = R_{1} + R_{2} + R_{3} + \cdots[/latex]

Since all of the current must pass through each resistor, it experiences the resistance of each component. Therefore, resistances in series simply add together.

Conceptual Check

Why does adding more resistors in series always increase the total resistance of a circuit? How does this affect the current for a fixed voltage source?

Example 27.1: Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the battery in Figure 27.2 is [latex]12.0\ \text{V}[/latex], and the resistances are [latex]R_{1} = 1.00\ \Omega[/latex], [latex]R_{2} = 6.00\ \Omega[/latex], and [latex]R_{3} = 13.0\ \Omega[/latex].
(a) What is the total resistance?
(b) Find the current.
(c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source.
(d) Calculate the power dissipated by each resistor.
(e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances:

[latex]R_{\text{s}} = R_{1} + R_{2} + R_{3}[/latex]
[latex]= 1.00\ \Omega + 6.00\ \Omega + 13.0\ \Omega[/latex]
[latex]= 20.0\ \Omega[/latex]

Strategy and Solution for (b)

The current is found using Ohm’s law, [latex]V = IR[/latex]. Substituting the applied voltage and total resistance gives:

[latex]I = \frac{V}{R_{\text{s}}} = \frac{12.0\ \text{V}}{20.0\ \Omega} = 0.600\ \text{A}[/latex]

Strategy and Solution for (c)

The voltage drop (or [latex]IR[/latex] drop) across each resistor is given by Ohm’s law.

[latex]V_{1} = IR_{1} = (0.600\ \text{A})(1.00\ \Omega) = 0.600\ \text{V}[/latex]

Similarly,

[latex]V_{2} = IR_{2} = (0.600\ \text{A})(6.00\ \Omega) = 3.60\ \text{V}[/latex]

and

[latex]V_{3} = IR_{3} = (0.600\ \text{A})(13.0\ \Omega) = 7.80\ \text{V}[/latex]

Discussion for (c)

The three voltage drops add to the source voltage, as expected:

[latex]V_{1} + V_{2} + V_{3} = 0.600 + 3.60 + 7.80 = 12.0\ \text{V}[/latex]

Strategy and Solution for (d)

The power dissipated by each resistor can be calculated using Joule’s law, [latex]P = IV[/latex], or more conveniently for resistors, [latex]P = I^{2}R[/latex].

[latex]P_{1} = I^{2}R_{1} = (0.600\ \text{A})^{2}(1.00\ \Omega) = 0.360\ \text{W}[/latex]

Similarly,

[latex]P_{2} = I^{2}R_{2} = (0.600\ \text{A})^{2}(6.00\ \Omega) = 2.16\ \text{W}[/latex]

and

[latex]P_{3} = I^{2}R_{3} = (0.600\ \text{A})^{2}(13.0\ \Omega) = 4.68\ \text{W}[/latex]

Discussion for (d)

Power can also be calculated using [latex]P = IV[/latex] or [latex]P = \frac{V^{2}}{R}[/latex], where [latex]V[/latex] is the voltage drop across each resistor. All methods yield the same results.

Strategy and Solution for (e)

The power output of the source is found using [latex]P = IV[/latex], where [latex]V[/latex] is the source voltage:

[latex]P = (0.600\ \text{A})(12.0\ \text{V}) = 7.20\ \text{W}[/latex]

Discussion for (e)

The total power dissipated by the resistors is:

[latex]P_{1} + P_{2} + P_{3} = 0.360 + 2.16 + 4.68 = 7.20\ \text{W}[/latex]

This equals the power supplied by the source. Since power is energy per unit time, conservation of energy requires that the power output of the source equals the total power dissipated by the resistors.

Quick Check

If one of the resistors were doubled, how would that affect the total resistance and the current in the circuit? Would the power dissipated by the circuit increase or decrease?

Major Features of Resistors in Series

  1. Series resistances add:
    [latex]R_s = R_1 + R_2 + R_3 + \cdots[/latex]
  2. The same current flows through each resistor in series.Because there is only one path for charge flow, the current is identical at every point in the circuit.
  3. Voltage is divided among the resistors.Each resistor receives only a portion of the total source voltage, with larger resistances producing larger voltage drops.

Resistors in Parallel

Figure 27.3 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by conducting wires with negligible resistance. Each resistor therefore has the full voltage of the source applied to it.

Each resistor draws the same current it would if it alone were connected to the voltage source (provided the source is not overloaded). For example, an automobile’s headlights, radio, and other components are wired in parallel so that they each receive the full voltage and can operate independently. The same is true in homes and buildings, where appliances are connected in parallel. (See Figure 27.3(b).)

Parallel circuit and household wiring
Figure 27.3: (a) Three resistors connected in parallel to a battery and the equivalent single (parallel) resistance. (b) Electrical power distribution in a house. (credit: Dmitry G, Wikimedia Commons)

To find an expression for the equivalent parallel resistance [latex]R_{\text{p}}[/latex], we consider the currents that flow and how they are related to resistance. Since each resistor has the full voltage across it, the currents through the individual resistors are:

[latex]I_{1} = \frac{V}{R_{1}}, \quad I_{2} = \frac{V}{R_{2}}, \quad I_{3} = \frac{V}{R_{3}}[/latex]

Conservation of charge implies that the total current [latex]I[/latex] supplied by the source is the sum of the individual currents:

[latex]I = I_{1} + I_{2} + I_{3}[/latex]

Substituting the expressions for the individual currents gives:

[latex]I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}} = V\left(\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\right)[/latex]

From Ohm’s law for the equivalent resistance, we also have:

[latex]I = \frac{V}{R_{\text{p}}} = V\left(\frac{1}{R_{\text{p}}}\right)[/latex]

The expressions in parentheses must therefore be equal. Generalizing to any number of resistors, the total resistance [latex]R_{\text{p}}[/latex] of a parallel combination is given by:

[latex]\frac{1}{R_{\text{p}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \cdots[/latex]

This relationship results in a total resistance [latex]R_{\text{p}}[/latex] that is always less than the smallest individual resistance. When resistors are connected in parallel, more current flows from the source than would flow through any single resistor alone, and so the overall resistance of the circuit is reduced.

Conceptual Insight

Why does adding more resistors in parallel decrease the total resistance? Think of each additional branch as providing another path for charge to flow. More paths mean less opposition to current, which lowers the overall resistance.

Example 27.2: Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

Let the voltage output of the battery and resistances in the parallel connection in Figure 27.3 be the same as the previously considered series connection:
[latex]V = 12.0\ \text{V}[/latex],
[latex]R_{1} = 1.00\ \Omega[/latex],
[latex]R_{2} = 6.00\ \Omega[/latex], and
[latex]R_{3} = 13.0\ \Omega[/latex].
(a) What is the total resistance?
(b) Find the total current.
(c) Calculate the currents in each resistor, and show these add to equal the total current output of the source.
(d) Calculate the power dissipated by each resistor.
(e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance for a parallel combination is found using:

[latex]\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}[/latex]
[latex]= \frac{1}{1.00\ \Omega} + \frac{1}{6.00\ \Omega} + \frac{1}{13.0\ \Omega}[/latex]
[latex]= \frac{1.00}{\Omega} + \frac{0.1667}{\Omega} + \frac{0.07692}{\Omega} = \frac{1.2436}{\Omega}[/latex]

(Note that intermediate values are shown with extra digits for accuracy.)

We invert to find the equivalent resistance:

[latex]R_{p} = \frac{1}{1.2436}\ \Omega = 0.8041\ \Omega[/latex]

With proper significant figures:

[latex]R_{p} = 0.804\ \Omega[/latex]

Discussion for (a)

As expected, the equivalent resistance is less than the smallest individual resistance.

Strategy and Solution for (b)

Using Ohm’s law with the equivalent resistance:

[latex]I = \frac{V}{R_{p}} = \frac{12.0\ \text{V}}{0.8041\ \Omega} = 14.92\ \text{A}[/latex]

Discussion for (b)

The total current is much larger than in the series case. Parallel circuits reduce total resistance, allowing more current to flow.

Strategy and Solution for (c)

Each resistor experiences the full voltage, so currents are:

[latex]I_{1} = \frac{V}{R_{1}} = \frac{12.0\ \text{V}}{1.00\ \Omega} = 12.0\ \text{A}[/latex]

Similarly,

[latex]I_{2} = \frac{12.0\ \text{V}}{6.00\ \Omega} = 2.00\ \text{A}[/latex]

and

[latex]I_{3} = \frac{12.0\ \text{V}}{13.0\ \Omega} = 0.92\ \text{A}[/latex]

Discussion for (c)

The total current is:

[latex]I_{1} + I_{2} + I_{3} = 14.92\ \text{A}[/latex]

This agrees with the total current and confirms conservation of charge.

Strategy and Solution for (d)

Using [latex]P = \frac{V^{2}}{R}[/latex], since each resistor has the full voltage:

[latex]P_{1} = \frac{(12.0\ \text{V})^{2}}{1.00\ \Omega} = 144\ \text{W}[/latex]

Similarly,

[latex]P_{2} = \frac{(12.0\ \text{V})^{2}}{6.00\ \Omega} = 24.0\ \text{W}[/latex]

and

[latex]P_{3} = \frac{(12.0\ \text{V})^{2}}{13.0\ \Omega} = 11.1\ \text{W}[/latex]

Discussion for (d)

Each resistor dissipates significantly more power in parallel than in series because each receives the full voltage.

Strategy and Solution for (e)

Using total current:

[latex]P = IV = (14.92\ \text{A})(12.0\ \text{V}) = 179\ \text{W}[/latex]

Discussion for (e)

Total power dissipated:

[latex]P_{1} + P_{2} + P_{3} = 144 + 24.0 + 11.1 = 179\ \text{W}[/latex]

This matches the source power, confirming conservation of energy.

Overall Discussion

Both total current and power are greater in parallel circuits than in series circuits with the same components and voltage source.

Conceptual Takeaway

In parallel circuits, each component receives the full voltage. This leads to larger currents and greater total power consumption compared to series circuits. This is why household electrical systems must be carefully protected with circuit breakers.

Major Features of Resistors in Parallel

  1. Parallel resistance is found using:
    [latex]\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \cdots[/latex]

    The equivalent resistance is always less than the smallest individual resistance in the combination.

  2. Each resistor has the same voltage across it.Every branch in a parallel circuit is directly connected to the source, so each resistor experiences the full source voltage. This is why power distribution systems (such as those in homes and hospitals) use parallel connections—devices receive the same voltage and can operate independently.
  3. Current is divided among the branches.The total current supplied by the source splits between the parallel paths. The sum of the branch currents equals the total current:
    [latex]I = I_{1} + I_{2} + I_{3} + \cdots[/latex]

Combinations of Series and Parallel

More complex connections of resistors are often combinations of both series and parallel arrangements. These configurations are commonly encountered in real circuits, especially when the resistance of connecting wires is taken into account. In such cases, the resistance of the wires is typically in series with other components that may themselves be arranged in parallel.

Combinations of series and parallel resistors can be reduced to a single equivalent resistance using a systematic step-by-step technique, illustrated in Figure 27.4. In this process, groups of resistors are identified as either series or parallel, replaced by their equivalent resistance, and then further simplified until only one equivalent resistance remains. Although the procedure may involve several steps, it is conceptually straightforward.

Reduction of a complex resistor network
Figure 27.4: This combination of seven resistors contains both series and parallel components. Each group is identified and replaced with its equivalent resistance. The process continues step by step until a single equivalent resistance is obtained.

The simplest combination of series and parallel resistors, shown in Figure 27.5, is also one of the most useful, because it appears in many practical applications. For example, [latex]R_{1}[/latex] could represent the resistance of wires connecting a car battery to electrical devices, which are connected in parallel. The resistors [latex]R_{2}[/latex] and [latex]R_{3}[/latex] could represent devices such as a starter motor and an interior light.

In earlier discussions, we often assumed that wire resistance is negligible. However, when wire resistance is significant, it can have important effects on circuit behavior, as the next example demonstrates.

Problem-Solving Strategy: Reducing Complex Circuits

To analyze circuits with both series and parallel components:

  1. Identify groups of resistors that are clearly in series or parallel.
  2. Replace each group with its equivalent resistance.
  3. Redraw the simplified circuit.
  4. Repeat the process until only one equivalent resistance remains.
  5. Use Ohm’s law to find current and voltage as needed.

Working step-by-step and redrawing the circuit at each stage helps avoid mistakes and makes complex problems manageable.

Example 27.3: Calculating Resistance, [latex]\text{IR}[/latex] Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits

Figure 27.5 shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider [latex]{R}_{1}[/latex] to be the resistance of wires leading to [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex]. (a) Find the total resistance. (b) What is the [latex]\text{IR}[/latex] drop in [latex]{R}_{1}[/latex]? (c) Find the current [latex]{I}_{2}[/latex] through [latex]{R}_{2}[/latex]. (d) What power is dissipated by [latex]{R}_{2}[/latex]?

Circuit diagram in which a battery of twelve point zero volts is connected to a combination of three resistors. Resistors R sub two and R sub three are connected in parallel to each other, and their combination is connected in series to resistor R sub one. R sub one has a resistance of one point zero zero ohms, R sub two has a resistance of six point zero zero ohms, and R sub three has a resistance of thirteen point zero ohms.
Figure 27.5: These three resistors are connected to a voltage source so that [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex] are in parallel with one another and that combination is in series with [latex]{R}_{1}[/latex].

Strategy and Solution for (a)

To find the total resistance, we note that [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex] are in parallel and their combination [latex]{R}_{\text{p}}[/latex] is in series with [latex]{R}_{1}[/latex]. Thus the total (equivalent) resistance of this combination is

[latex]{R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}.[/latex]

First, we find [latex]{R}_{\text{p}}[/latex] using the equation for resistors in parallel and entering known values:

[latex]\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}=\frac{1}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }+\frac{1}{\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega }=\frac{0\text{.}\text{2436}}{\Omega }.[/latex]

Inverting gives

[latex]{R}_{\text{p}}=\frac{1}{0\text{.}\text{2436}}\Omega =4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .[/latex]

So the total resistance is

[latex]{R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega +4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega =5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .[/latex]

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values ([latex]20.0 \Omega[/latex] and [latex]0.804 \Omega[/latex], respectively) found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find the [latex]\text{IR}[/latex] drop in [latex]{R}_{1}[/latex], we note that the full current [latex]I[/latex] flows through [latex]{R}_{1}[/latex]. Thus its [latex]\text{IR}[/latex] drop is

[latex]{V}_{1}={\text{IR}}_{1}.[/latex]

We must find [latex]I[/latex] before we can calculate [latex]{V}_{1}[/latex]. The total current [latex]I[/latex] is found using Ohm’s law for the circuit. That is,

[latex]I=\frac{V}{{R}_{\text{tot}}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega }=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A}.[/latex]

Entering this into the expression above, we get

[latex]{V}_{1}={\text{IR}}_{1}=\left(2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}.[/latex]

Discussion for (b)

The voltage applied to [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex] is less than the total voltage by an amount [latex]{V}_{1}[/latex]. When wire resistance is large, it can significantly affect the operation of the devices represented by [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex].

Strategy and Solution for (c)

To find the current through [latex]{R}_{2}[/latex], we must first find the voltage applied to it. We call this voltage [latex]{V}_{\text{p}}[/latex], because it is applied to a parallel combination of resistors. The voltage applied to both [latex]{R}_{2}[/latex] and [latex]{R}_{3}[/latex] is reduced by the amount [latex]{V}_{1}[/latex], and so it is

[latex]{V}_{\text{p}}=V-{V}_{1}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}-2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}=9\text{.}\text{65}\phantom{\rule{0.25em}{0ex}}\text{V}.[/latex]

Now the current [latex]{I}_{2}[/latex] through resistance [latex]{R}_{2}[/latex] is found using Ohm’s law:

[latex]{I}_{2}=\frac{{V}_{\text{p}}}{{R}_{2}}=\frac{9\text{.}\text{65 V}}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}.[/latex]

Discussion for (c)

The current is less than the 2.00 A that flowed through [latex]{R}_{2}[/latex] when it was connected in parallel to the battery in the previous parallel circuit example.

Strategy and Solution for (d)

The power dissipated by [latex]{R}_{2}[/latex] is given by

[latex]{P}_{2}=({I}_{2}{)}^{2}{R}_{2}=(1\text{.}\text{61}\phantom{\rule{0.25em}{0ex}}\text{A}{)}^{2}(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega )=\text{15}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{W}.[/latex]

Discussion for (d)

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

Practical Implications

One important implication of the previous example is that resistance in wires reduces the current and power delivered to a device. If wire resistance is relatively large, as in a worn or very long extension cord, then this loss can become significant. When a large current flows, the resulting [latex]IR[/latex] drop in the wires can also be substantial.

For example, when you are rummaging in the refrigerator and the motor turns on, the interior light often dims momentarily. Similarly, you may notice the passenger compartment light dim when you start the engine of your car (although part of this effect may also be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in Figure 27.6. The device represented by [latex]R_{3}[/latex] has a very low resistance, and when it is switched on, a large current flows. This increased current produces a larger [latex]IR[/latex] drop in the wires, represented by [latex]R_{1}[/latex]. As a result, the voltage available across the light bulb (represented by [latex]R_{2}[/latex]) is reduced, causing the light to dim noticeably.

Circuit showing voltage drop causing dimming lights
Figure 27.6: Why do lights dim when a large appliance is switched on? The large current drawn by the appliance motor causes a significant [latex]IR[/latex] drop in the wires, reducing the voltage across the light.

Real-World Insight

This effect is especially important in electrical safety and system design. Excessive current in wires can lead not only to reduced performance (such as dimming lights), but also to overheating. This is why electrical systems use appropriately sized wires and include circuit breakers or fuses to prevent dangerous current levels.

Check Your Understanding

Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel.

No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in Kirchhoff’s Rules, will allow you to analyze the circuit.

Problem-Solving Strategies for Series and Parallel Resistors

  1. Draw a clear circuit diagram.Label all resistors and voltage sources. This step should also include listing all known quantities directly on the diagram, since they are easier to track when visually organized.
  2. Identify the unknowns.Determine exactly what needs to be calculated. Writing a list of unknown quantities helps keep the solution focused and organized.
  3. Classify the circuit elements.Determine whether resistors are in series, in parallel, or part of a combination. Examine the diagram carefully. Resistors are in series if the same current must pass sequentially through them; they are in parallel if they share the same two connection points and therefore have the same voltage across them.
  4. Apply the appropriate relationships.Use the major features of series or parallel circuits to solve for the unknowns. If the circuit contains both types, reduce it step-by-step by simplifying groups of series or parallel resistors, as demonstrated in the examples in this chapter.Special note: When calculating [latex]R_{\text{p}}[/latex], be careful to correctly take the reciprocal after summing the inverse resistances.
  5. Check your results.Verify that answers are reasonable and consistent. Units must be correct, and numerical values should make physical sense. For example, total resistance in series should be larger than any individual resistance, while total resistance in parallel should be smaller. Power should also be greater for the same devices in parallel compared with series.

Section Summary

  • The total resistance of an electrical circuit with resistors wired in series is the sum of the individual resistances:
    [latex]R_{\text{s}} = R_1 + R_2 + R_3 + \cdots[/latex]
  • Each resistor in a series circuit has the same current flowing through it.
  • The voltage drop (and corresponding power dissipation) across each individual resistor in series may differ, but their combined total equals the voltage supplied by the source.
  • The total resistance of an electrical circuit with resistors wired in parallel is less than the smallest individual resistance and is determined by:
    [latex]\frac{1}{R_{\text{p}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots[/latex]
  • Each resistor in a parallel circuit has the same full voltage of the source applied to it.
  • The current through each resistor in a parallel circuit can differ and depends on the resistance of each branch.
  • More complex circuits that combine series and parallel elements can be reduced to a single equivalent resistance by identifying series and parallel sections, simplifying them step-by-step, and continuing until only one equivalent resistance remains.

Conceptual Questions

  1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 27.7 has on current when open and when closed.
    The diagram shows a circuit with a voltage source and internal resistance small r connected in series with a resistance R and a switch.
    Figure 27.7: A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.)
  2. What is the voltage across the open switch in Figure 27.7?
  3. There is a voltage across an open switch, such as in Figure 27.7. Why, then, is the power dissipated by the open switch small?
  4. Why is the power dissipated by a closed switch, such as in Figure 27.7, small?
  5. A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 27.8. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Do not try this—it is hard on the battery!)
    This diagram shows a circuit with a voltage source and internal resistance small r. A resistance R and an open switch are connected in parallel to it.
    Figure 27.8: A wiring mistake put this switch in parallel with the device represented by [latex]R[/latex]. (Note that in this diagram, the script E represents the voltage (or electromotive force) of the battery.)
  6. Knowing that the severity of a shock depends on the magnitude of the current through your body, would you prefer to be in series or parallel with a resistance, such as the heating element of a toaster, if shocked by it? Explain.
  7. Would your headlights dim when you start your car’s engine if the wires in your automobile were superconductors? (Do not neglect the battery’s internal resistance.) Explain.
  8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each?
  9. If two household lightbulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? Explain.
  10. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for?
  11. Before World War II, some radios got power through a “resistance cord” that had a significant resistance. Such a resistance cord reduces the voltage to a desired level for the radio’s tubes and the like, and it saves the expense of a transformer. Explain why resistance cords become warm and waste energy when the radio is on.
  12. Some light bulbs have three power settings (not including zero), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings?

Problem Exercises

Note: Data taken from figures can be assumed to be accurate to three significant digits.

  1. (a) What is the resistance of ten [latex]\text{275-Ω}[/latex] resistors connected in series? (b) In parallel?
  2. (a) What is the resistance of a [latex]\text{1.00}×{10}^{2}-\Omega[/latex], a [latex]2\text{.}\text{50-kΩ}[/latex], and a [latex]4\text{.}\text{00-k}\Omega[/latex] resistor connected in series? (b) In parallel?
  3. What are the largest and smallest resistances you can obtain by connecting a [latex]\text{36}\text{.}0-\Omega[/latex], a [latex]\text{50}\text{.}0-\Omega[/latex], and a [latex]\text{700-Ω}[/latex] resistor together?
  4. An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse?
  5. Your car’s 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.)
  6. (a) Given a 48.0-V battery and [latex]\text{24}\text{.}0-\Omega[/latex] and [latex]\text{96}\text{.}0-\Omega[/latex] resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.
  7. Referring to the example combining series and parallel circuits and Figure 27.5, calculate [latex]{I}_{3}[/latex] in the following two different ways: (a) from the known values of [latex]I[/latex] and [latex]{I}_{2}[/latex]; (b) using Ohm’s law for [latex]{R}_{3}[/latex]. In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.
  8. Referring to Figure 27.5: (a) Calculate [latex]{P}_{3}[/latex] and note how it compares with [latex]{P}_{3}[/latex] found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors.
  9. Refer to Figure 27.6 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is [latex]0\text{.}\text{400}\phantom{\rule{0.25em}{0ex}}\Omega[/latex], and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor?
  10. A 240-kV power transmission line carrying [latex]5.00×{10}^{2}\phantom{\rule{0.25em}{0ex}}\text{A}[/latex] is hung from grounded metal towers by ceramic insulators, each having a [latex]1\text{.}\text{00}×{\text{10}}^{9}-\Omega[/latex] resistance. Figure 27.9. (a) What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.
    The diagram shows a grounded metal transmission tower. Two ground conductors on top of the tower point out like antennas. Hanging from the tower are a set of three bundled conductors, one on either end and one in the middle.
    Figure 27.9: High-voltage (240-kV) transmission line carrying [latex]5.00×{10}^{2}\phantom{\rule{0.25em}{0ex}}\text{A}[/latex] is hung from a grounded metal transmission tower. The row of ceramic insulators provide [latex]1.00×{\text{10}}^{9}\Omega [/latex] of resistance each.
  11. Show that if two resistors [latex]{R}_{1}[/latex] and [latex]{R}_{2}[/latex] are combined and one is much greater than the other ([latex]{R}_{1}\text{>>}{R}_{2}[/latex]): (a) Their series resistance is very nearly equal to the greater resistance [latex]{R}_{1}[/latex]. (b) Their parallel resistance is very nearly equal to smaller resistance [latex]{R}_{2}[/latex].
  12. Unreasonable Results Two resistors, one having a resistance of [latex]1\text{45}\phantom{\rule{0.25em}{0ex}}\Omega[/latex], are connected in parallel to produce a total resistance of [latex]150\phantom{\rule{0.25em}{0ex}}\Omega[/latex]. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
  13. Unreasonable Results Two resistors, one having a resistance of [latex]9\text{00 kΩ}[/latex], are connected in series to produce a total resistance of [latex]0\text{.}\text{500 MΩ}[/latex]. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Glossary

series
a sequence of resistors or other components wired into a circuit one after the other
resistor
a component that provides resistance to the current flowing through an electrical circuit
resistance
causing a loss of electrical power in a circuit
Ohm’s law
the relationship between current, voltage, and resistance within an electrical circuit: [latex]V=\text{IR}[/latex]
voltage
the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery
voltage drop
the loss of electrical power as a current travels through a resistor, wire or other component
current
the flow of charge through an electric circuit past a given point of measurement
Joule’s law
the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: [latex]{P}_{e}=\text{IV}[/latex]
parallel
the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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