What is the potential divider?

A potential divider uses two resistors in series to provide a fraction of the supply voltage, V_out = R_2R_1 + R_2V_in. Replacing one resistor with a thermistor or LDR makes the output respond to temperature or light.

A cell of EMF 1.5\,V and internal resistance 0.50\, drives a current of 0.20\,A. Find the terminal voltage. [2 marks]

WJEC WJEC 2020-style practice: A battery of EMF 12\,V and internal resistance 0.80\, is connected to a 5.2\, resistor. Calculate the current, the terminal voltage and the power dissipated in the internal resistance.

Use = I(R + r) to find the current, then the terminal voltage and the internal power loss. Current: I = R + r = 125.2 + 0.80 = 126.0 = 2.0\,A. Terminal voltage: V = - Ir = 12 - 2.0 × 0.80 = 10.4\,V (or V = IR = 2.0 × 5.2 = 10.4\,V). Power in internal resistance: P = I^2 r = (2.0)^2 × 0.80 = 3.2\,W. Markers reward the total resistance including r, the terminal voltage below the EMF, and P = I^2 r for the internal loss.

WalesPhysicsSyllabus dot point

How do Kirchhoff's laws, internal resistance and potential dividers let us analyse a DC circuit?

Kirchhoff's first and second laws, series and parallel resistors, EMF and internal resistance, and the potential divider.

A focused answer to WJEC A-Level Physics Unit 2 DC circuits, covering Kirchhoff's first and second laws, combining resistors in series and parallel, EMF and internal resistance, and the potential divider.

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

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

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

What this dot point is asking

WJEC wants you to apply Kirchhoff's two laws, combine resistors in series and parallel, use EMF and internal resistance, and analyse a potential divider. Circuit analysis is one of the most heavily examined parts of Unit 2, and the internal-resistance and potential-divider sub-topics generate the multi-step calculation questions that separate the grades.

The answer

Kirchhoff's laws

Series and parallel resistors

In series, resistances add: $R = R_1 + R_2 + \ldots$ In parallel, $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \ldots$ , so the combined resistance is less than the smallest. In series the current is common and the voltages add; in parallel the voltage is common and the currents add, which follows directly from Kirchhoff's two laws.

EMF and internal resistance

To find $\varepsilon$ and $r$ , plot terminal voltage against current: the intercept is $\varepsilon$ and the gradient is $-r$ .

Combining resistors then finding the current

A $6.0\,\text{V}$ supply of negligible internal resistance is connected to a $4.0\,\Omega$ resistor in series with a parallel pair of $6.0\,\Omega$ and $3.0\,\Omega$ resistors. Find the total current.

step Combine the parallel pair

$\dfrac{1}{R_{\parallel}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6.0} + \dfrac{2}{6.0} = \dfrac{3}{6.0}$ , so $R_{\parallel} = 2.0\,\Omega$ .

step Add the series resistor

$R_{\text{total}} = 4.0 + 2.0 = 6.0\,\Omega$ .

step Find the current

$I = \dfrac{V}{R_{\text{total}}} = \dfrac{6.0}{6.0} = 1.0\,\text{A}$ drawn from the supply.

The potential divider

A potential divider uses two resistors in series to provide a fraction of the supply voltage, $V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}V_{\text{in}}$ . Replacing one resistor with a thermistor or LDR makes the output respond to temperature or light.

Examples in context

Example 1. Car headlights dimming on start-up: A car battery has a low but non-zero internal resistance. When the starter motor draws a huge current $I$ , the lost volts $Ir$ become large, so the terminal voltage $V = \varepsilon - Ir$ drops and the headlights briefly dim. Once the engine fires and the starter disengages, the current falls and the lights brighten again.
Example 2. A light-sensing switch: A potential divider with a light-dependent resistor (LDR) and a fixed resistor feeds a transistor. In darkness the LDR resistance is high, taking most of the supply voltage; as the output rises past a threshold the transistor switches a street light on. The same divider with the LDR and fixed resistor swapped would switch a light off at dawn instead.
Example 3. Measuring internal resistance in the lab: Connecting a cell to a variable resistor and recording terminal voltage $V$ against current $I$ for several settings produces a straight-line graph. The vertical intercept gives the EMF $\varepsilon$ (the terminal voltage at zero current) and the gradient is $-r$ , the negative of the internal resistance. This is the standard WJEC method for finding the otherwise hidden internal resistance of a real cell, and it reinforces the relation $V = \varepsilon - Ir$ .

Try this

Q1. A cell of EMF $1.5\,\text{V}$ and internal resistance $0.50\,\Omega$ drives a current of $0.20\,\text{A}$ . Find the terminal voltage. [2 marks]

Cue. $V = \varepsilon - Ir = 1.5 - 0.20\times0.50 = 1.4\,\text{V}$ .

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

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

Exam-style practice questions

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

WJEC 20205 marksA battery of EMF

12\,\text{V}

and internal resistance

0.80\,\Omega

is connected to a

5.2\,\Omega

resistor. Calculate the current, the terminal voltage and the power dissipated in the internal resistance.

Show worked answer →

Use $\varepsilon = I(R + r)$ to find the current, then the terminal voltage and the internal power loss.

Current: $I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{5.2 + 0.80} = \dfrac{12}{6.0} = 2.0\,\text{A}$ .

Terminal voltage: $V = \varepsilon - Ir = 12 - 2.0 \times 0.80 = 10.4\,\text{V}$ (or $V = IR = 2.0 \times 5.2 = 10.4\,\text{V}$ ).

Power in internal resistance: $P = I^2 r = (2.0)^2 \times 0.80 = 3.2\,\text{W}$ .

Markers reward the total resistance including $r$ , the terminal voltage below the EMF, and $P = I^2 r$ for the internal loss.

WJEC 20174 marksA potential divider is made from a

4.0\,\text{k}\Omega

fixed resistor in series with a thermistor across a

9.0\,\text{V}

supply, with the output taken across the thermistor. At room temperature the thermistor has resistance

8.0\,\text{k}\Omega

. Calculate the output voltage, and state how it changes as the temperature rises.

Show worked answer →

Apply the potential divider formula with the output across the thermistor.

$V_{\text{out}} = \dfrac{R_{\text{therm}}}{R_{\text{fixed}} + R_{\text{therm}}} V_{\text{in}} = \dfrac{8.0}{4.0 + 8.0} \times 9.0 = \dfrac{8.0}{12.0} \times 9.0 = 6.0\,\text{V}$ .

As temperature rises, the thermistor resistance falls, so it takes a smaller share of the supply. The output voltage across the thermistor therefore decreases.

Markers reward the divider calculation giving $6.0\,\text{V}$ and the correct direction of change as the thermistor resistance drops.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)