Circuits, potential dividers, e.m.f. and internal resistance | AQA A-level Physics

AQA 7408 · Section 3.5.1

Circuits, potential dividers, e.m.f. and internal resistance

Build series and parallel networks, design sensor dividers, and determine e.m.f. and internal resistance using the method from AQA Required Practical 6.

13 specification pointsRequired Practical 6Three active circuit models
Exam-style potential divider sensor circuits and terminal potential difference against current graph

Predict, test, explain

1

Predict

Commit to a direction or value before moving a control.

2

Test

Change one variable and compare the evidence with your prediction.

3

Explain

Use an equation and a physical reason, then attempt the exam questions.

Series and parallel resistor networks

series: RT = R₁ + R₂ + …

Series components carry the same current; p.d.s add to the supply e.m.f.

parallel: 1/RT = 1/R₁ + 1/R₂ + …

Parallel branches share the same p.d.; branch currents add to the total.

Build and compare

Cell combinationsIdentical cells in series add their e.m.f.s. Identical cells in parallel retain one cell’s e.m.f. but can reduce the effective internal resistance. Do not add their e.m.f.s in parallel.

Potential dividers and sensors

Vout = Vin × Routput / (Rtop + Rbottom)

Use the resistance across which the output is measured in the numerator. A variable resistor gives adjustable output; an LDR or thermistor creates an automatic sensor circuit. A potentiometer as a measuring instrument is not required by AQA.

Choose the sensor position

Determine e.m.f. and internal resistance

ε = I(R + r) = V + Ir

E.m.f. is energy supplied per unit charge. Terminal p.d. is energy transferred per unit charge in the external circuit. When current flows, some energy is transferred inside the source: the “lost volts” are Ir.

Method

  1. Connect a cell, switch, ammeter and variable resistor in series; connect a voltmeter across the cell.
  2. Open the switch between readings to limit heating and cell discharge.
  3. Vary resistance and record several pairs of I and terminal V.
  4. Plot V against I. In V = ε − Ir, intercept = ε and gradient = −r.

Evaluation

Use a best-fit line rather than two adjacent points. Quote internal resistance as the magnitude of the gradient. Repeat readings and avoid currents large enough to change the cell temperature.

Generate the practical graph

Questions and answers

1. Find the total resistance of 120 Ω and 220 Ω in parallel. [2 marks]

1/R = 1/120 + 1/220, so R = 77.6 Ω. It is correctly smaller than either branch resistance.

2. A 6.0 V cell has internal resistance 0.80 Ω and supplies a 4.0 Ω load. Find current and terminal p.d. [3 marks]

I = ε/(R+r) = 6.0/4.8 = 1.25 A. V = IR = 5.0 V.

3. A V–I graph has intercept 1.56 V and gradient −0.42 V A⁻¹. State ε and r. [2 marks]

ε = 1.56 V. Internal resistance is the magnitude of gradient: r = 0.42 Ω.

4. Design a divider whose output rises as light level rises. [3 marks]

Use an LDR as the top resistor and take Vout across the fixed bottom resistor. More light lowers LDR resistance, so a larger fraction of the supply p.d. appears across the fixed resistor.

5. State the physical principles behind the junction and loop rules. [2 marks]

The junction rule follows conservation of charge. The loop rule follows conservation of energy.

AQA 7408 coverage

  • 3.5.1.4(a) Calculate total resistance of resistors in series using RT = R1 + R2 + R3 + ...
  • 3.5.1.4(b) Calculate total resistance of resistors in parallel using 1/RT = 1/R1 + 1/R2 + 1/R3 + ...
  • 3.5.1.4(e) Apply current, voltage and resistance relationships in series and parallel circuits.
  • 3.5.1.4(f) Analyse circuits with cells in series and identical cells in parallel.
  • 3.5.1.4(g) Apply conservation of charge and conservation of energy in dc circuits.
  • 3.5.1.5(a) Describe a potential divider used to supply a constant or variable potential difference from a power supply.
  • 3.5.1.5(b) Describe potential dividers using variable resistors, thermistors and light-dependent resistors.
  • 3.5.1.5(c) Design and analyse potential divider circuits to achieve specified outcomes.
  • 3.5.1.5(d) Recognise that use of a potentiometer as a measuring instrument is not required.
  • 3.5.1.6(a) Define electromotive force using ε = E/Q.
  • 3.5.1.6(b) Use ε = I(R + r) or ε = IR + Ir for circuits with internal resistance.
  • 3.5.1.6(c) Define terminal potential difference and electromotive force.
  • 3.5.1.6(d) Perform calculations for circuits where internal resistance of the supply is not negligible.

Written against AQA Physics 7408 section 3.5.1 and Required Practical 6. All questions are original.

Written by: PhysicsUK teaching team

Expertise: Built by a UK A Level Physics teacher and examiner.

Reviewed for: AQA A Level Physics 7408

Last reviewed: 2026-07-15

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