Direct Current Circuits

Lessons

Senior 4 · Physics

Direct Current Circuits

5 lessons

Lesson 2 of 5

Electromotive Force, Internal Resistance and Combinations of Cells

25 min

Why do the headlights dim?

Switch on a car's headlights and then start the engine: the lights dim for a moment while the starter motor turns. The battery is the same, its chemistry unchanged, yet the voltage across the lamps falls. The reason is that every real source of electrical energy has some resistance inside it. This lesson separates what a battery can give (its electromotive force) from what reaches the circuit (its terminal potential difference), and shows how to measure both.

E.m.f. and potential difference

A cell, a battery, a generator or a solar cell converts energy from some other form (chemical, mechanical, light) into electrical energy. The electromotive force (e.m.f.) $E$ is the energy converted into electrical energy per unit charge that passes through the source:

E = \frac{W}{Q} \qquad (\text{unit: volt, } 1\,\mathrm{V} = 1\,\mathrm{J\,C^{-1}})

The potential difference (p.d.) between two points of a circuit is the electrical energy converted into other forms (heat, light, motion) per unit charge passing between them. Despite its name, e.m.f. is not a force; both e.m.f. and p.d. are measured in volts, but e.m.f. is energy given to each coulomb, and p.d. is energy taken from it.

Definition

The electromotive force $E$ of a source is the energy converted into electrical energy per unit charge passing through it. The internal resistance $r$ is the resistance of the material of the source itself. The terminal potential difference $V$ is the p.d. across the source's terminals when it delivers current; $V = E - I r$ . Example: a cell of e.m.f. 1.5 V gives 1.5 J to every coulomb that passes through it.

Internal resistance and terminal p.d.

Model a real cell as an ideal source of e.m.f. $E$ in series with a small resistance $r$ (Figure 2.1). When it drives a current $I$ through an external resistance $R$ , energy conservation per coulomb gives

E = IR + Ir \quad\Rightarrow\quad I = \frac{E}{R + r}

The p.d. across the terminals, which is also the p.d. across the external circuit, is

V = I R = E - I r

The term $I r$ is sometimes called the "lost volts": energy per coulomb turned into heat inside the cell. When no current flows ( $I = 0$ ), $V = E$ , so a high-resistance voltmeter connected alone across a cell reads very nearly its e.m.f. The larger the current drawn, the lower the terminal p.d.; that is why the headlights dim when the starter motor draws a huge current.

Figure 2.1: Measuring E and r. The voltmeter reads the terminal p.d. V; the ammeter reads the current I.

Worked example 1: a torch cell

A cell of e.m.f. 1.5 V and internal resistance $0.50\,\Omega$ is connected to a $2.5\,\Omega$ lamp. Find the current, the terminal p.d. and the lost volts.

I = \frac{E}{R + r} = \frac{1.5}{2.5 + 0.50} = 0.50\,\mathrm{A}, \qquad V = IR = 0.50 \times 2.5 = 1.25\,\mathrm{V}, \qquad Ir = 0.25\,\mathrm{V}

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

Worked example 2: starting a car

A car battery has $E = 12.6\,\mathrm{V}$ and $r = 0.020\,\Omega$ . The starter motor draws 150 A. Find the terminal p.d. while the engine is being started.

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

The headlights, connected across the same terminals, suddenly receive 9.6 V instead of about 12.6 V, so they dim. A battery's internal resistance must be very small to deliver such currents.

Experiment 2.1: measuring $E$ and $r$

Apparatus: a dry cell, a rheostat (or a set of resistors), an ammeter (0–1 A), a high-resistance voltmeter, a switch, connecting wires.

Connect the circuit of Figure 2.1 with the rheostat at its largest value. Close the switch only while taking readings, so that the cell does not run down or warm up.
Record $V$ and $I$ . Reduce $R$ in steps and repeat, for about six pairs of readings.
Plot $V$ against $I$ . Since $V = - r I + E$ , the graph is a straight line with gradient $- r$ and intercept $E$ on the $V$ axis.

$I$ (A)	0.10	0.20	0.30	0.40	0.50	0.60
$V$ (V)	1.48	1.38	1.31	1.24	1.14	1.07

Table 2.1: Illustrative results for one cell.

Figure 2.2: Terminal p.d. against current for Table 2.1.

Warning

Safety. Never connect a wire straight across the terminals of a cell or battery (a short circuit): the current is limited only by $r$ , so the wire and the battery can become very hot, and a car or lithium battery can burn or explode. Keep the rheostat from reaching zero resistance.

Worked example 3: reading the graph

From Figure 2.2 the line meets the $V$ axis at 1.55 V, so $E = 1.55\,\mathrm{V}$ . Two points on the line: (0.10 A, 1.47 V) and (0.60 A, 1.07 V):

\text{gradient} = \frac{1.07 - 1.47}{0.60 - 0.10} = \frac{-0.40}{0.50} = -0.80\,\Omega, \qquad r = 0.80\,\Omega

The short-circuit current would be $\frac{E}{r} = \frac{1.55}{0.80} = 1.9\,\mathrm{A}$ , where the line would cut the $I$ axis.

Power delivered to the load

The power delivered to the external resistance is

P = I^2R = \frac{E^2 R}{(R + r)^2} = \frac{E^2}{R + 2r + \frac{r^2}{R}}

$P$ is largest when the denominator $R + \frac{r^2}{R}$ is smallest. Since $\left(\sqrt{R} - \frac{r}{\sqrt{R}}\right)^2 \ge 0$ gives $R + \frac{r^2}{R} \ge 2r$ , with equality when $R = r$ , the maximum power is delivered when the load resistance equals the internal resistance:

P_{\max} = \frac{E^2}{4r} \qquad (R = r)

At that point half the energy is wasted inside the source, so the efficiency is only 50%. Matching $R$ to $r$ suits weak signals (aerials, loudspeakers); power stations instead keep $r$ much smaller than the load to work efficiently.

Combining cells

Figure 2.3: Cells in series and in parallel.

In series (Figure 2.3a), each coulomb passes through every cell, so the e.m.f.s add, and so do the internal resistances. A cell connected the wrong way round subtracts its e.m.f. but still adds its resistance. Identical cells in parallel (Figure 2.3b) share the current: the e.m.f. is that of one cell, and the internal resistance is $\frac{r}{n}$ . Parallel cells cannot give a higher voltage, but they can deliver a larger current with less internal loss, and last longer. Cells of different e.m.f. should never be connected in parallel: a current would circulate between them and waste energy.

Worked example 4: four cells, three ways

Four cells, each of e.m.f. 1.5 V and internal resistance $0.40\,\Omega$ , drive current through a $2.0\,\Omega$ resistor. Find the current when they are connected (a) all in series, (b) all in parallel, (c) as two parallel rows of two cells in series.

\text{(a)}\;\; I = \frac{6.0}{2.0 + 1.6} = 1.7\,\mathrm{A} \qquad \text{(b)}\;\; I = \frac{1.5}{2.0 + 0.10} = 0.71\,\mathrm{A} \qquad \text{(c)}\;\; I = \frac{3.0}{2.0 + 0.40} = 1.25\,\mathrm{A}

For a load much larger than $r$ , the series arrangement gives the most current; for a load much smaller than $r$ , the parallel one does.

Warning

Common mistakes. (1) Assuming the voltmeter across a cell always reads its e.m.f.: it reads $E - I r$ while current flows. (2) Forgetting $r$ in $I = \frac{E}{R + r}$ . (3) Dividing the e.m.f. by $n$ for cells in parallel: it is the internal resistance that is divided. (4) Reading the gradient of the $V$ – $I$ graph as $+ r$ : it is negative.

Tip

Think of a cell as "an ideal battery with a small resistor inside it". Draw that resistor in every circuit diagram that mentions internal resistance, and the problem becomes an ordinary series circuit.

Measuring internal resistance, a practical (Physics Online). Watch how the terminal p.d. and current are measured for several load resistances and how the graph gives the e.m.f. and internal resistance.

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Practice

Define e.m.f. and explain how it differs from potential difference.
A battery of e.m.f. 9.0 V and internal resistance $1.0\,\Omega$ is connected to a $17\,\Omega$ resistor. Find the current and the terminal p.d.
A voltmeter reads 1.60 V across a cell on open circuit and 1.20 V when the cell drives 0.50 A through a resistor. Find the internal resistance and the resistance of the resistor.
When a cell is connected to a $4.0\,\Omega$ resistor the current is 0.30 A; with a $9.0\,\Omega$ resistor it is 0.15 A. Find $E$ and $r$ .
A source has $E = 12\,\mathrm{V}$ and $r = 3.0\,\Omega$ . Find the maximum power it can deliver to a load and the load resistance needed.
Six cells, each 1.5 V and $0.30\,\Omega$ , are connected in series, but one is accidentally reversed. Find the total e.m.f. and internal resistance.
A student wants to light a lamp of resistance $0.20\,\Omega$ as brightly as possible using three identical cells (1.5 V, $0.60\,\Omega$ each). Evaluate whether series or parallel is better.

Show answersHide answers

E.m.f. is energy converted into electrical energy per unit charge by a source; p.d. is electrical energy converted into other forms per unit charge between two points. A source's terminal p.d. is less than its e.m.f. when current flows.
$I = \frac{9.0}{18} = 0.50\,\mathrm{A}$ ; $V = 9.0 - 0.50 \times 1.0 = 8.5\,\mathrm{V}$ .
$E = 1.60\,\mathrm{V}$ ; lost volts $= 0.40\,\mathrm{V}$ , so $r = \frac{0.40}{0.50} = 0.80\,\Omega$ ; $R = \frac{1.20}{0.50} = 2.4\,\Omega$ .
$E = 0.30 (4.0 + r) = 0.15 (9.0 + r)$ : $1.2 + 0.30 r = 1.35 + 0.15 r$ , so $r = 1.0\,\Omega$ and $E = 1.5\,\mathrm{V}$ .
$R = r = 3.0\,\Omega$ ; $P_{\max} = \frac{12^2}{4 \times 3.0} = 12\,\mathrm{W}$ .
$E = 5 \times 1.5 - 1.5 = 6.0\,\mathrm{V}$ ; $r = 6 \times 0.30 = 1.8\,\Omega$ .
Series: $I = \frac{4.5}{0.20 + 1.8} = 2.25\,\mathrm{A}$ . Parallel: $I = \frac{1.5}{0.20 + 0.20} = 3.75\,\mathrm{A}$ . Parallel is better here, because the lamp's resistance is small compared with the cells' internal resistance, so reducing $r$ matters more than raising $E$ .

Quick check

A cell of e.m.f. 2.0 V and internal resistance 1.0 Ω drives current through a 3.0 Ω resistor. What is the terminal p.d.?

Key points

E.m.f. = energy converted to electrical per coulomb; p.d. = electrical energy converted to other forms per coulomb.
$E = I (R + r)$ , $V = E - I r$ ; the $V$ – $I$ graph has intercept $E$ and gradient $- r$ .
Maximum power to the load when $R = r$ : $P_{\max} = \frac{E^2}{4r}$ , at 50% efficiency.
Series cells: e.m.f.s and resistances add. $n$ identical parallel cells: e.m.f. $E$ , resistance $\frac{r}{n}$ .

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. Adapted by Ganzaa.
Photo: 18650, AA, and AAA battery with penny.jpg by JayCubby, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Measuring Internal Resistance - PRACTICAL - A Level Physics, Physics Online, embedded from YouTube.

The derivations, the three diagrams, the experiment and its illustrative results, the worked examples and the practice exercises are written by Ganzaa.

1Why do the headlights dim?

2E.m.f. and potential difference

3Internal resistance and terminal p.d.

4Experiment 2.1: measuring E E E and r r r

5Power delivered to the load

6Combining cells