Basic Electrical Wiring and Installation

Lessons

Senior 3 · Physics

Basic Electrical Wiring and Installation

5 lessons

Lesson 2 of 5

Electromotive Force, Internal Resistance and Cells in Series and Parallel

25 min

Why do the headlights dim?

Switch on a car's headlights and then start the engine: while the starter motor turns, the lights go dim, then brighten again as soon as the engine runs. The battery has not suddenly lost its charge. The starter motor draws a very large current, and part of the battery's voltage is used up inside the battery itself, leaving less for the lights. Every real cell and battery behaves like this, in a torch, a phone or a solar home system, and this lesson shows how to calculate it.

Emf and terminal potential difference

A cell does work on the charges that pass through it, giving each coulomb a certain amount of energy. That energy per coulomb is the cell's electromotive force (emf), $E$ . But the chemicals inside the cell also resist the current: the cell has an internal resistance $r$ . When a current $I$ flows, some of the energy is used driving it through $r$ , and the voltage available at the terminals, the terminal potential difference $V$ , is less than $E$ :

E = V + Ir \qquad\text{and, with an external resistance } R,\qquad E = I(R + r)

The part $I r$ is called the lost volts. On open circuit (no current) $V = E$ , so a high-resistance voltmeter connected across a cell with nothing else attached reads its emf.

Definition

Electromotive force (emf, $E$ ): the energy converted into electrical energy by a source for each coulomb of charge that passes through it. Unit: the volt ( $1\ \text{V} = 1\ \mathrm{J\,C^{-1}}$ ).

Internal resistance ( $r$ ): the resistance of the materials inside a cell or battery. Terminal potential difference ( $V$ ): the potential difference across the terminals when the source supplies a current, $V = E - I r$ .

Worked example 1: a cell and a resistor

A cell of emf 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current, the terminal pd and the lost volts.

I = \frac{E}{R + r} = \frac{1.5}{2.5 + 0.50} = 0.50\ \text{A}

V = IR = 0.50 \times 2.5 = 1.25\ \text{V} \qquad \text{lost volts} = Ir = 0.50 \times 0.50 = 0.25\ \text{V}

Check: $1.25 + 0.25 = 1.5\ \text{V} = E$ .

Worked example 2: the car battery

A car battery has emf 12.6 V and internal resistance 0.020 Ω. The starter motor draws 150 A. Find the terminal pd while the engine is being started.

V = E - Ir = 12.6 - 150 \times 0.020 = 12.6 - 3.0 = 9.6\ \text{V}

The lights get only 9.6 V instead of about 12.6 V, so they dim; when the starter stops, the current falls and the terminal pd rises again.

Experiment: measuring $E$ and $r$

Set up the circuit. Start with the variable resistor at its largest value.
Record the ammeter reading $I$ and the voltmeter reading $V$ .
Reduce the resistance in steps, recording $I$ and $V$ each time. Do not let the current become large, and switch off between readings so that the cell does not run down or heat up.
Plot $V$ against $I$ . Since $V = E - r I$ , the graph is a straight line: the intercept on the $V$ -axis is $E$ , and the gradient is $- r$ .

Worked example 3: reading the graph

In the graph above, the line meets the $V$ -axis at 1.5 V and falls by 0.40 V when the current rises from 0 to 0.50 A. Find $E$ and $r$ .

E = 1.5\ \text{V} \qquad r = -\text{gradient} = \frac{0.40}{0.50} = 0.80\ \Omega

Cells in series and in parallel

	$n$ identical cells in series	$n$ identical cells in parallel
Total emf	$n E$ : the emfs add	$E$ : the same as one cell
Total internal resistance	$n r$	$\frac{r}{n}$
Main advantage	A larger voltage	A larger current with fewer lost volts; the battery lasts longer
Examples	Two 1.5 V cells in a torch give 3.0 V; a 12 V car battery is six 2 V cells in series	Banks of cells in a power bank or a solar battery store, to supply more current for longer

A cell connected the wrong way round in a series chain subtracts its emf: three 1.5 V cells with one reversed give only $1.5 + 1.5 - 1.5 = 1.5\ \text{V}$ .

Worked example 4: series against parallel

Three cells, each of emf 1.5 V and internal resistance 0.20 Ω, supply a 2.4 Ω lamp. Find the current when the cells are (a) in series; (b) in parallel.

\text{(a) } E = 4.5\ \text{V},\ r = 0.60\ \Omega: \quad I = \frac{4.5}{2.4 + 0.60} = 1.5\ \text{A}

\text{(b) } E = 1.5\ \text{V},\ r = \frac{0.20}{3} \approx 0.067\ \Omega: \quad I = \frac{1.5}{2.4 + 0.067} \approx 0.61\ \text{A}

In series the lamp is much brighter; in parallel it is dimmer but the cells share the current, so they last about three times as long as one cell would.

Warning

Never short-circuit a battery. With almost no external resistance, $I = \frac{E}{r}$ is very large: a car battery can deliver hundreds of amperes, enough to melt a spanner, start a fire or make the battery explode. Do not carry loose cells in a pocket with coins or keys, and never mix old and new cells or different types in one device.

Tip

E is what the cell gives; V is what the circuit gets. $E = V + I r$ . The bigger the current, the bigger the lost volts and the smaller the terminal pd.

Finding emf and internal resistance from a V–I graph (The Physics Academy). Watch the worked examples: the intercept gives the emf and the gradient gives the internal resistance.

✦

Practice

Define emf and internal resistance.
A battery of emf 6.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. Find the current and the terminal pd.
A voltmeter reads 1.6 V across a cell on open circuit and 1.4 V when the cell delivers 0.50 A. Find the internal resistance.
Four cells, each of emf 1.5 V and internal resistance 0.25 Ω, are connected in series to a 5.0 Ω resistor. Find the current.
The same four cells are connected in parallel to the 5.0 Ω resistor. Find the current.
A torch takes three 1.5 V cells in series, but one is put in the wrong way round. What emf does the torch get?
A graph of terminal pd against current for a battery meets the $V$ -axis at 9.0 V, and $V$ is 8.0 V when $I = 2.0\ \text{A}$ . Find the emf and the internal resistance.
A solar home system's battery runs a radio well, but its voltage falls sharply when an electric iron is plugged into the inverter. Explain this using internal resistance.

Show answersHide answers

Emf: the energy converted into electrical energy per coulomb of charge passing through the source. Internal resistance: the resistance of the materials inside the source.
$I = \frac{6.0}{5.0 + 1.0} = 1.0\ \text{A}$ ; $V = 1.0 \times 5.0 = 5.0\ \text{V}$ .
Lost volts $= 1.6 - 1.4 = 0.20\ \text{V}$ , so $r = \frac{0.20}{0.50} = 0.40\ \Omega$ .
$E = 6.0\ \text{V}$ , $r = 1.0\ \Omega$ ; $I = \frac{6.0}{6.0} = 1.0\ \text{A}$ .
$E = 1.5\ \text{V}$ , $r = \frac{0.25}{4} = 0.0625\ \Omega$ ; $I = \frac{1.5}{5.0625} \approx 0.30\ \text{A}$ .
$1.5 + 1.5 - 1.5 = 1.5\ \text{V}$ .
$E = 9.0\ \text{V}$ ; $r = \frac{9.0 - 8.0}{2.0} = 0.50\ \Omega$ .
The radio draws a small current, so the lost volts $I r$ are small. An iron draws a very large current, so $I r$ becomes large and the terminal pd $V = E - I r$ falls sharply, too low for the inverter to work properly.

Key points

$E = V + I r = I (R + r)$ ; the lost volts $I r$ grow with the current.
A graph of $V$ against $I$ has intercept $E$ and gradient $- r$ .
Identical cells in series: emf $n E$ , resistance $n r$ . In parallel: emf $E$ , resistance $\frac{r}{n}$ .
A short circuit gives a dangerous current $\frac{E}{r}$ .

Sources and credits

21.2: Electromotive Force: Terminal Voltage, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Video: Higher Physics, Finding EMF and Internal Resistance from a V-I Graph, The Physics Academy, embedded from YouTube.

Written by Ganzaa: the lesson text, the circuit and graph, the table, worked examples and practice exercises.

1Why do the headlights dim?

2Emf and terminal potential difference

3Experiment: measuring E E E and r r r

4Cells in series and in parallel

Why do the headlights dim?

Emf and terminal potential difference

Experiment: measuring $E$ and $r$

Cells in series and in parallel