The big idea: Crank a car's starter and the headlights dim for a moment — the battery can't quite hold its full voltage when a big current is drawn.
Why the lights dim: That shortfall is the cell's internal resistance at work. No real cell is perfect.
| Term | What it means |
|---|---|
| Emf (electromotive force) | the energy a cell gives to each coulomb of charge (unit: the volt, V) |
| Internal resistance r | a small resistance inside the cell that uses up some energy as current flows |
| Terminal p.d. | the voltage you actually get out — a bit less than the emf |
A real cell = emf ε with internal resistance r inside it. The same current I flows through r and the load R. The voltmeter across R reads the terminal p.d. V — what's actually delivered to the circuit.
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Spot it: Think of a cell as a perfect battery (emf ε) with a tiny resistor r built in.
The same current flows through both r and the outside circuit, so r quietly 'steals' some volts.
Every cell is a source of emf: it does work on each coulomb of charge, changing another form of energy into electrical energy.
| Source of emf | Energy changed | For | Against |
|---|---|---|---|
| Chemical cell (battery) | chemical → electrical | portable; steady emf; rechargeable types can be reused | runs down; limited energy; disposal and mining harm the environment |
| Solar cell | light → electrical | no fuel and no emissions while running; long life | output depends on light, nothing at night; low emf per cell, so many are needed |
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The emf has to push current through everything — the outside load R and the internal resistance r. Adding the two resistances together gives the data-booklet equation:
- emf of the cell (V) — the energy it gives each coulomb
- current in the circuit (A)
- external (load) resistance (Ω)
- internal resistance of the cell (Ω)
Multiply out the bracket: ε = IR + Ir. The part IR is the voltage across the load — the terminal p.d. you actually use. Rearranging gives:
- terminal p.d. — voltage across the cell's terminals (V)
- emf of the cell (V)
- current in the circuit (A)
- internal resistance of the cell (Ω)
Two things to get right: 1. The lost volts are Ir — the volts used up inside the cell. They grow as the current grows.
2. The terminal p.d. V is what a voltmeter across the cell reads, and it equals IR (the volts across the load).
A cell of emf 1.5 V has an internal resistance of 0.50 Ω. It drives a current of 0.40 A through a circuit. Find the terminal p.d. across the cell.
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Why the terminal pd changes with the load: Lower the load resistance R and the current I rises. The lost volts Ir rise with it, so the terminal pd V = ε − Ir falls. Raise R and V climbs back towards ε.
Finding ε and r from a graph
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How this is tested — internal resistance turns the simple V = IR circuit into a two-resistor problem (R outside, r inside):
Paper 1A
- Why the terminal p.d. drops when more current is drawn.
- The special case r ≈ 0 (terminal p.d. ≈ emf).
Paper 2
- 'Show that' questions — use ε = I(R + r) to find the emf, the internal resistance r, or the current.
The classic trap: Forgetting r, so you use ε = IR and get the current slightly wrong.
Finding r from two readings: A common exam set-up gives you the emf and a terminal p.d. (or a current).
Lost volts = ε − V = Ir, so the internal resistance is r = (ε − V) ÷ I.
The words to know
| Word | What it means |
|---|---|
| emf ε | Work done by the source on each coulomb of charge, in volts |
| Internal resistance r | The resistance of the cell itself |
| Terminal pd | The pd across the cell's terminals: V = ε − Ir |
| Lost volts | Ir: the pd used up inside the cell |
A battery of emf 9.0 V is connected to a 4.0 Ω resistor. The current in the circuit is 2.0 A. Show that the internal resistance of the battery is 0.50 Ω.
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