A cell of e.m.f. 1.5\ V and internal resistance 0.5\ drives a current of 0.4\ A. Find the terminal potential difference.

Two resistors of 6.0\ and 12\ are connected in parallel. Find their combined resistance. [2 marks]

Northern IrelandPhysicsSyllabus dot point

How do Kirchhoff's laws and the idea of e.m.f. explain DC circuits?

Kirchhoff's two laws, series and parallel resistor combinations, electromotive force and internal resistance, and the potential divider.

A CCEA A-Level Physics answer on Kirchhoff's current and voltage laws, combining resistors in series and parallel, electromotive force and internal resistance, and how a potential divider splits a voltage.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to state and apply Kirchhoff's two laws, combine resistors in series and parallel, define electromotive force and internal resistance and use them in calculations, and explain how a potential divider works. Multi-resistor network calculations and internal-resistance problems are common.

The answer

Kirchhoff's laws

Combining resistors

E.m.f., internal resistance and potential dividers

Replacing one resistor with a thermistor or light-dependent resistor makes a sensor circuit whose output changes with temperature or light, the basis of many control systems.

Worked example: finding e.m.f. and internal resistance

Two-reading method for a real cell

A cell drives $0.30\ \text{A}$ through an external resistance giving a terminal voltage of $1.44\ \text{V}$ . When the external resistance is reduced so the current rises to $0.60\ \text{A}$ , the terminal voltage falls to $1.38\ \text{V}$ . Find the e.m.f. and internal resistance.

step Write the equation for each reading

V = \varepsilon - Ir \Rightarrow 1.44 = \varepsilon - 0.30r \quad\text{and}\quad 1.38 = \varepsilon - 0.60r.

step Subtract to eliminate the e.m.f.

1.44 - 1.38 = (0.60 - 0.30)r \Rightarrow 0.06 = 0.30r \Rightarrow r = 0.20\ \Omega.

step Substitute back to find the e.m.f.

\varepsilon = 1.44 + 0.30 \times 0.20 = 1.44 + 0.06 = 1.50\ \text{V}.

This is the standard CCEA practical method: plot

V

against

I

, and the intercept is

\varepsilon

while the gradient is

-r

Examples in context

Example 1. A car battery starting an engine. A starter motor draws a huge current, so the lost volts $Ir$ across the battery's small internal resistance become significant, and the terminal voltage sags. This is why the headlights dim for a moment when you turn the key, and why a battery with high internal resistance (an old one) struggles to start the car on a cold morning.

Example 2. A light-sensing security lamp. A potential divider with a light-dependent resistor feeds a transistor or comparator. In daylight the resistor's low resistance gives a low output; at dusk its resistance rises, the output crosses a threshold, and the lamp switches on automatically, all from the divider relation $V_{\text{out}} = \varepsilon R_2 / (R_1 + R_2)$ .

Try this

Q1. A cell of e.m.f. $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$ drives a current of $0.4\ \text{A}$ . Find the terminal potential difference. [2 marks]

Cue. $V = \varepsilon - Ir = 1.5 - (0.4 \times 0.5) = 1.3\ \text{V}$ .

Q2. State Kirchhoff's first law and the conservation principle it follows from. [2 marks]

Cue. Total current into a junction equals total current out; it follows from conservation of charge.

Q3. Two resistors of $6.0\ \Omega$ and $12\ \Omega$ are connected in parallel. Find their combined resistance. [2 marks]

Cue. $\frac{1}{R} = \frac{1}{6.0} + \frac{1}{12} = \frac{3}{12}$ , so $R = 4.0\ \Omega$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20196 marksA cell of e.m.f. 6.0 V and internal resistance 0.50 ohm is connected to a 2.0 ohm resistor in series with a parallel combination of a 6.0 ohm and a 3.0 ohm resistor. Calculate the total resistance of the external circuit, the current drawn from the cell, and the terminal potential difference.

Show worked answer →

First combine the parallel pair:

$\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}$ , so $R_p = 2.0$ ohm.

The external resistance is then the series total:

$R_{\text{ext}} = 2.0 + 2.0 = 4.0$ ohm.

The current from the cell uses the full circuit including internal resistance:

$I = \frac{\varepsilon}{R_{\text{ext}} + r} = \frac{6.0}{4.0 + 0.50} = \frac{6.0}{4.5} = 1.33$ A.

The terminal potential difference is

$V = \varepsilon - Ir = 6.0 - (1.33 \times 0.50) = 6.0 - 0.67 = 5.3$ V.

Markers reward the correct parallel combination, including the internal resistance in the current calculation, and the lost-volts subtraction.

CCEA 20214 marksA potential divider consists of a 4.0 kilohm fixed resistor in series with a thermistor across a 9.0 V supply. The output is taken across the thermistor. At room temperature the thermistor has a resistance of 8.0 kilohm. Calculate the output voltage, and explain how and why the output changes as the temperature rises.

Show worked answer →

The output is taken across the thermistor, so

$V_{\text{out}} = \varepsilon \times \frac{R_{\text{th}}}{R_{\text{fixed}} + R_{\text{th}}} = 9.0 \times \frac{8.0}{4.0 + 8.0} = 9.0 \times \frac{8.0}{12.0} = 6.0$ V.

As the temperature rises the thermistor's resistance falls (it has a negative temperature coefficient). A smaller $R_{\text{th}}$ takes a smaller share of the supply voltage, so the output voltage across the thermistor decreases.

Markers reward the divider ratio with the correct resistor in the numerator, and linking falling thermistor resistance to a falling share of the voltage.

Related dot points

Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)