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 this figure (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.

This figure 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.)

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.

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 this figure.

According to Ohm’s law, the voltage drop, \(V\), across a resistor when a current flows through it is calculated using the equation \(V=\text{IR}\), 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 \({V}_{1}={\mathit{IR}}_{1}\), that across \({R}_{2}\) is \({V}_{2}={\mathit{IR}}_{2}\), and that across \({R}_{3}\) is \({V}_{3}={\mathit{IR}}_{3}\). The sum of these voltages equals the voltage output of the source; that is,

\(V={V}_{1}+{V}_{2}+{V}_{3}.\)

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

\({\text{qV}}_{1}+{\text{qV}}_{2}+{\text{qV}}_{3}.\)

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, \(\mathit{qV}={\mathit{qV}}_{1}+{\mathit{qV}}_{2}+{\mathit{qV}}_{3}\). 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

\(V={\text{IR}}_{1}+{\text{IR}}_{2}+{\text{IR}}_{3}=I({R}_{1}+{R}_{2}+{R}_{3}).\)

Note that for the equivalent single series resistance \({R}_{\text{s}}\), we have

\(V={\text{IR}}_{\text{s}}.\)

This implies that the total or equivalent series resistance \({R}_{\text{s}}\) of three resistors is \({R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}\).

This logic is valid in general for any number of resistors in series; thus, the total resistance \({R}_{\text{s}}\) of a series connection is

\({R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}+\text{.}\text{.}\text{.},\)

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

Suppose the voltage output of the battery in this figure is \(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\), and the resistances are \({R}_{1}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \), \({R}_{2}=6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \), and \({R}_{3}=\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \). (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:

\(\begin{array}{lll}{R}_{\text{s}}& =& {R}_{1}+{R}_{2}+{R}_{3}\\ & =& 1.00 \Omega +6.00 \Omega +\text{13.0 Ω}\\ & =& \text{20.0 Ω.}\end{array}\)

Strategy and Solution for (b)

The current is found using Ohm’s law, \(V=\text{IR}\). Entering the value of the applied voltage and the total resistance yields the current for the circuit:

\(I=\cfrac{V}{{R}_{\text{s}}}=\cfrac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{20}\text{.}\text{0}\phantom{\rule{0.25em}{0ex}}\Omega }=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}.\)

Strategy and Solution for (c)

The voltage—or \(\text{IR}\) drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

\({V}_{1}={\mathit{IR}}_{1}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A})(1\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega )=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{V}.\)

Similarly,

\({V}_{2}={\mathit{IR}}_{2}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A})(6\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega )=3\text{.}\text{60}\phantom{\rule{0.25em}{0ex}}\text{V}\)

and

\({V}_{3}={\mathit{IR}}_{3}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A})(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega )=7\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}\text{V}.\)

Discussion for (c)

The three \(\text{IR}\) drops add to \(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\), as predicted:

\({V}_{1}+{V}_{2}+{V}_{3}=(0\text{.}\text{600}+3\text{.}\text{60}+7\text{.}\text{80})\phantom{\rule{0.25em}{0ex}}\text{V}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}.\)

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, \(P=\text{IV}\), where \(P\) is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law \(V=\text{IR}\) into Joule’s law, we get the power dissipated by the first resistor as

\({P}_{1}={I}^{2}{R}_{1}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{)}^{2}(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega )=0\text{.}\text{360}\phantom{\rule{0.25em}{0ex}}\text{W}.\)

Similarly,

\({P}_{2}={I}^{2}{R}_{2}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{)}^{2}(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega )=2\text{.}\text{16}\phantom{\rule{0.25em}{0ex}}\text{W}\)

and

\({P}_{3}={I}^{2}{R}_{3}=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{)}^{2}(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega )=4\text{.}\text{68}\phantom{\rule{0.25em}{0ex}}\text{W}.\)

Discussion for (d)

Power can also be calculated using either \(P=\text{IV}\) or \(P=\cfrac{{V}^{2}}{R}\), 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 \(P=\text{IV}\), where \(V\) is the source voltage. This gives

\(P=(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A})(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V})=7\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{W}.\)

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,

\({P}_{1}+{P}_{2}+{P}_{3}=(0\text{.}\text{360}+2\text{.}\text{16}+4\text{.}\text{68})\phantom{\rule{0.25em}{0ex}}\text{W}=7\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{W}.\)

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: \({R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}+\text{.}\text{.}\text{.}\text{.}\)
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 Series