The big idea: Emf (electromotive force) is the energy a cell gives to each coulomb of charge it pushes round the circuit. Its unit is the volt (V).
No real cell is perfect: it has a small internal resistance r inside it. As current flows, some energy is used up inside the cell itself.
So the voltage you actually get out — the terminal p.d. — is a bit less than the emf.
[Diagram: phys-circuit] - Available in full study mode
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.
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).
Worked example — terminal p.d. of a real cell
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.
Solution
- Start with the given formula and multiply out the bracket:
- The terminal p.d. is the emf minus the lost volts Ir:
- Put in the numbers (ε = 1.5, I = 0.40, r = 0.50):
- Work it out — keep the unit:
Final answer
terminal p.d. = 1.3 V — a little less than the 1.5 V emf, because 0.20 V is lost inside the cell.
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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: quick MCQs — why the terminal p.d. drops when more current is drawn, or 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.
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.
IB-style question — find the internal resistance
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.
Solution
- Start with the given formula:
- Put in the numbers (ε = 9.0, I = 2.0, R = 4.0):
- Divide both sides by 2.0:
- Rearrange for r — keep the unit:
Final answer
r = 0.50 Ω. The cell loses Ir = 2.0 × 0.50 = 1.0 V inside itself, so only 8.0 V reaches the 4.0 Ω resistor.