Resistors in Series and Parallel

We have already covered combinations of capacitors in series and parallel. We can make the same two basic types of arrangement with resistors. Just as with capacitors, we can simplify the analysis of a resistor circuit step-by-step, by imagining replacing a combination of resistors by a single ("effective") resistor. We can do this because the internal details of a resistor are irrelevant to its function in a circuit. All we really care about is the value of the resistance that needs to be plugged into Ohm's Law, as applied to the current and voltage found at the two terminals leading into and out of the resistor.

You could say the idea behind analyzing circuits is to get as far as possible by treating parts of the circuit as a black box. And we can indeed treat complicated arrangements of resistors as a black box, so long as there are just two terminals connecting the black box to the outside (because resistors must have two terminals).

It does not matter whether several individual resistors exist between the two terminals in question, or just a single wire, as long as we know the resistance of our black box. Read this text, which implements this strategy, with the aid of two basic rules: one for series and one for parallel resistors.

Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 21.2. The total resistance of a combination of resistors depends on both their individual values and how they are connected.

Two schematics, showing a series of connection resistors, plus a parallel connection of resistors.

Figure 21.2 (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 flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then $R_{1}$ in Figure 21.2(a) could be the resistance of the screwdriver’s shaft, $R_{2}$ the resistance of its handle, $R_{3}$ the person’s body resistance, and $R_{4}$ the resistance of her shoes.

Figure 21.3 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Schematic of three resistors connected in series to a battery

Figure 21.3 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 power, called a voltage drop, in each resistor in Figure 21.3.

According to Ohm’s law, the voltage drop, $V$ , across a resistor when a current flows through it is calculated using the equation $V=IR$ , where $I$ equals the current in amps (A) and $R$ is the resistance in ohms $\Omega$ . Another way to think of this is that $V$ is the voltage necessary to make a current $I$ flow through a resistance $R$ .

So the voltage drop across $R_{1}$ is $V_{1}=IR_{1}$ , that across $R_{2}$ is $V_{2}=IR_{2}$ , and that across $R_{3}$ is $V_{3}=IR_{3}$ . The sum of these voltages equals the voltage output of the source; that is,

$V=V_{1}+V_{2}+V_{3}$ [Equation 21.1]

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation $PE=qV$ , where $q$ is the electric charge and $V$ is the voltage. Thus the energy supplied by the source is $aV$ , while that dissipated by the resistors is

$qV_{1}+qV_{2}+qV_{3}$ [Equation 21.2]

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 directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, $qV=qV_{1}+qV_{2}+qV_{3}$ . The charge $q$ cancels, yielding $V=V_{1}+V_{2}+V_{3}$ , as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

$V=IR_{1}+IR_{2}+IR_{3}=I(R_{1}+R_{2}+R_{3})$ [Equation 21.3]

Note that for the equivalent single series resistance $R_{s}$ , we have

$V=IR_{s}$ [Equation 21.4]

This implies that the total or equivalent series resistance $R_{s}$ of three resistors is $R_{s}=R_{1}+R_{2}+R_{3}$ .

This logic is valid in general for any number of resistors in series; thus, the total resistance $R_{s}$ of a series connection is

$R_{s}=R_{1}+R_{2}+R_{3}+ \cdots$ [Equation 21.5]

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

Example 21.1 Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the battery in Figure 21.3 is $12.0\: V$ , and the resistances are $R_{1}=1.00\: \Omega$ , $R_{2}=6.00\: \Omega$ , and $R_{3}=13.00\: \Omega$ . (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, as given by this equation:

$R_{s}=R_{1}+R_{2}+R_{3}$ [Equation 21.6]

$=1.00\: \Omega+6.00\: \Omega+13.00\: \Omega$

$20\:\Omega$

Strategy and Solution for (b)

The current is found using Ohm’s law, $V=1R$ . Entering the value of the applied voltage and the total resistance yields the current for the circuit:

$I=\frac{V}{R_{s}}=\frac{12.0\: V}{20.0\:\Omega}=0.600\: A$ [Equation 21.7]

Strategy and Solution for (c)

The voltage – or $IR$ drop – in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

$V_{1}=IR_{1}=(0.600\:A)(1.0\:\Omega)=0.600\:A$ [Equation 21.8]

and

$V_{3}=IR_{3}=(0.600\:A)(13.0\:\Omega)=7.80\:A$ [Equation 21.10]

Discussion for (c)

The three $IR$ drops add to $12.0\:V$ , as predicted:

$V_{1}+V_{2}+V_{3}=(0.600+3.60+7.80)V=12.0\:V$ [Equation 21.11]

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law, $P=IV$ , where $P$ is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law $V=IR$ into Joule’s law, we get the power dissipated by the first resistor as

$P_{1}=I^{2}R_{1}=(0.600\: A)^{2}(1.00\:\Omega )=0.360\:W$ [Equation 21.12]

and

$P_{3}=I^{2}R_{3}=(0.600\: A)^{2}(13.0\:\Omega )=4.68\:W$ [Equation 21.14]

Discussion for (d)

Power can also be calculated using either $P=IV$ or $P=\frac{V^{2}}{R}$ , where $V$ is the voltage drop across the resistor (not the full voltage of the source). The same values will be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use $P=IV$ , where $V$ is the source voltage. This gives

$P=(0.600\:A)(12.0\:V)=7.20\:W$ [Equation 21.15]

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is,

$P_{1}+P_{2}+P_{3}=(0.360+2.16+4.68)W=7.20\:W$ [Equation 21.16]

Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors.

Major Features of Resistors in Series

Series resistances add: $R_{s}=R_{1}+R_{2}+R_{3}+ \cdots$
The same current flows through each resistor in series.
Individual resistors in series do not get the total source voltage, but divide it.

Resistors in Parallel

Figure 21.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus 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 voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure 21.4(b).)

Two images: a schematic of three resistors connected in parallel to a battery, an electric power setup in a house.

Figure 21.4 (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G, Wikimedia Commons)

To find an expression for the equivalent parallel resistance $R_{p}$ , let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are $I_{1}=\frac{V}{R_{1}}$ , $I_{2}=\frac{V}{R_{2}}$ , and $I_{3}=\frac{V}{R_{3}}$ . Conservation of charge implies that the total current $I$ produced by the source is the sum of these currents:

$I=I_{1}+I_{2}+I_{3}$ [Equation 21.17]

Substituting the expressions for the individual currents gives

$I=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}=V(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}})$ [Equation 21.18]

Note that Ohm’s law for the equivalent single resistance gives

$I=\frac{V}{R_{p}}=V(\frac{1}{R_{p}})$ [Equation 21.19]

The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance $R_{p}$ of a parallel connection is related to the individual resistances by

$\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\cdots$ [Equation 21.20]

This relationship results in a total resistance $R_{p}$ that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower.

Example 21.2 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 21.4 be the same as the previously considered series connection: $R_{1}=1.00\: \Omega$ , $R_{2}=6.00\: \Omega$ , and $R_{3}=13.00\: \Omega$ . (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.