A cell of electromotive force 1.5\ V drives a current of 0.30\ A through an internal resistance of 0.40\. Find the terminal potential difference. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A cell of electromotive force 1.5\ V and internal resistance 0.50\ is connected to a 2.5\ resistor. Calculate the current in the circuit and the terminal potential difference across the cell.

Total resistance is the external plus internal resistance: R + r = 2.5 + 0.50 = 3.0\. Current: I = R + r = 1.53.0 = 0.50\ A. Terminal potential difference: V = - Ir = 1.5 - (0.50)(0.50) = 1.5 - 0.25 = 1.25\ V (equivalently V = IR = 0.50 × 2.5 = 1.25\ V). Markers reward I = ()/(R+r) = 0.50\ A, using V = - Ir, and the terminal pd 1.25\ V.

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How do Kirchhoff's laws, internal resistance and the potential divider let us analyse any DC circuit?

DC circuits: Kirchhoff's two laws, resistors in series and parallel, electromotive force and internal resistance, the potential divider, and electrical power and energy.

A focused answer to the Eduqas A-Level Physics Component 2 DC circuits content, covering Kirchhoff's two laws as conservation of charge and energy, combining resistors in series and parallel, electromotive force and internal resistance, the potential divider, and electrical power and energy.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

Kirchhoff's first law (charge conservation): the total current into a junction equals the total current out. Kirchhoff's second law (energy conservation): around any closed loop the sum of the electromotive forces equals the sum of the potential differences. Resistors add in series, $R = R_1 + R_2 + \dots$ , and combine in parallel by $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ . A real source has electromotive force $\varepsilon$ and internal resistance $r$ , so the terminal voltage is $V = \varepsilon - Ir$ and $\varepsilon = I(R + r)$ . A potential divider splits a voltage in the ratio of the resistances, $V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}$ . Electrical power is $P = VI = I^2 R = \frac{V^2}{R}$ , and energy is $E = VIt$ .

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What this dot point is asking

Eduqas wants you to state Kirchhoff's two laws and link them to conservation of charge and energy, combine resistors in series and parallel, define electromotive force and internal resistance and use $\varepsilon = I(R + r)$ , analyse a potential divider, and calculate electrical power and energy.

The answer

Kirchhoff's laws

Resistors in series and parallel

Electromotive force and internal resistance

The potential divider

Electrical power and energy

Finding internal resistance from two loads

A battery delivers $0.60\ \text{A}$ to a $2.0\ \Omega$ resistor and $0.40\ \text{A}$ to a $3.5\ \Omega$ resistor. Find the electromotive force and internal resistance.

step 1: Write two terminal-voltage equations

For each load, $\varepsilon = I(R + r)$ . Load 1: $\varepsilon = 0.60(2.0 + r)$ . Load 2: $\varepsilon = 0.40(3.5 + r)$ .

step 2: Set the expressions equal

$0.60(2.0 + r) = 0.40(3.5 + r)$ , so $1.2 + 0.60r = 1.4 + 0.40r$ .

step 3: Solve

$0.20r = 0.2$ , so $r = 1.0\ \Omega$ . Then $\varepsilon = 0.60(2.0 + 1.0) = 1.8\ \text{V}$ .

Examples in context

Kirchhoff's laws are the basis of all circuit analysis, from a torch to a power grid. Internal resistance explains why a car's headlights dim when the starter motor draws a huge current, and why a battery warms in use. Potential dividers set reference voltages and form the front end of countless sensors: light meters, thermostats and the volume control on an amplifier are all potential dividers.

Try this

Q1. State Kirchhoff's first law and the conservation principle behind it. [2 marks]

Cue. The total current into a junction equals the total current out; conservation of charge.

Q2. Two $6.0\ \Omega$ resistors are connected in parallel. Find their combined resistance. [2 marks]

Cue. $\frac{1}{R} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$ , so $R = 3.0\ \Omega$ .

Q3. A cell of electromotive force $1.5\ \text{V}$ drives a current of $0.30\ \text{A}$ through an internal resistance of $0.40\ \Omega$ . Find the terminal potential difference. [2 marks]

Cue. $V = \varepsilon - Ir = 1.5 - (0.30)(0.40) = 1.38\ \text{V}$ .

Exam-style practice questions

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

Eduqas 20194 marksA cell of electromotive force

1.5\ \text{V}

and internal resistance

0.50\ \Omega

is connected to a

2.5\ \Omega

resistor. Calculate the current in the circuit and the terminal potential difference across the cell.

Show worked answer →

Total resistance is the external plus internal resistance: $R + r = 2.5 + 0.50 = 3.0\ \Omega$ .

Current: $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{3.0} = 0.50\ \text{A}$ .

Terminal potential difference: $V = \varepsilon - Ir = 1.5 - (0.50)(0.50) = 1.5 - 0.25 = 1.25\ \text{V}$ (equivalently $V = IR = 0.50 \times 2.5 = 1.25\ \text{V}$ ).

Markers reward $I = \frac{\varepsilon}{R+r} = 0.50\ \text{A}$ , using $V = \varepsilon - Ir$ , and the terminal pd $1.25\ \text{V}$ .

Eduqas 20214 marksA potential divider consists of a

4.0\ \text{k}\Omega

resistor in series with a thermistor, connected across a

6.0\ \text{V}

supply. The output is taken across the thermistor. At a certain temperature the thermistor resistance is

2.0\ \text{k}\Omega

. Calculate the output voltage, and state how it changes if the thermistor is heated.

Show worked answer →

The output is the fraction of the supply voltage across the thermistor: $V_\text{out} = V_\text{in} \dfrac{R_\text{therm}}{R_\text{therm} + R_\text{fixed}} = 6.0 \times \dfrac{2.0}{2.0 + 4.0} = 6.0 \times \dfrac{1}{3} = 2.0\ \text{V}$ .

Heating the thermistor lowers its resistance, so it takes a smaller share of the supply voltage: the output voltage across the thermistor decreases.

Markers reward the potential-divider ratio, the output $2.0\ \text{V}$ , and stating the output falls as the thermistor is heated (its resistance drops).

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)