Topic B.5 — Current & Circuits Paper 1 & 2 ε = I(R + r) ~7 min read

Electromotive Force & Internal Resistance

A brand new 9 V battery sounds like it should give exactly 9 volts every time. In reality, it almost never quite does. Let’s find out why — and it turns out the answer is something we already know a lot about.

📘 What You Need to Know

What emf really means

Emf is the maximum “push” a source can possibly give — the total energy handed to every coulomb of charge before anything gets used up along the way. You can only measure the true emf when no current is flowing at all, using a voltmeter with an enormous resistance connected straight across the cell’s terminals, with nothing else in the circuit.

Why don’t we get the full emf?

Here’s the twist: a cell isn’t just a source of energy, it’s also made of the same kind of material as any other conductor — and we already know conductors resist current a little. So as charge passes through the cell itself, it bumps into ions inside the cell, just like it does everywhere else, and hands over a little energy as heat. That’s internal resistance, and it means a bit of the emf gets “used up” before the charge even leaves the cell.

THE CELL ε internal resistance r load R lost volts (Ir) used up inside the cell terminal p.d. (IR) is what’s left for the rest of the circuit
Picture the cell as a perfect emf source with a small internal resistance built right in. Some of the emf is “spent” inside the cell as lost volts, and the rest reaches the rest of the circuit as the terminal p.d.

Putting a number on it

The full emf always splits into two parts: the p.d. “lost” pushing charge through the cell’s own internal resistance, and the p.d. left over for the rest of the circuit — the terminal p.d. Since current is the same all the way round a single loop, we can write:

EMF equation ε = I(R + r)

where ε is emf in volts (V), I is the current in amps (A), R is the resistance of everything else in the circuit, and r is the cell’s internal resistance. Expanding the brackets splits it neatly into the two parts we just described:

Two parts of the emf ε = IR + Ir = terminal p.d. + lost volts
Quick recap: ε = I(R + r); the “lost volts” (Ir) heat up the cell itself; the terminal p.d. (IR) is all that’s left for the rest of the circuit; a bigger internal resistance means a bigger gap between emf and terminal p.d.
WE 1

A cell has an emf of 9.0 V and an internal resistance of 0.50 Ω. It is connected to an external resistor of 8.5 Ω. Find the current in the circuit and the lost volts.

Find the current first: ε = I(R + r) I = 9.0 ÷ (8.5 + 0.50) = 9.0 ÷ 9.0 I = 1.0 A Now find the lost volts: lost volts = Ir = 1.0 × 0.50 Lost volts = 0.5 V
WE 2

A battery of emf 6.0 V and internal resistance 1.5 Ω is connected to a 4.5 Ω lamp. Find the current and the terminal potential difference.

Find the current: I = 6.0 ÷ (4.5 + 1.5) = 6.0 ÷ 6.0 I = 1.0 A Now find the terminal p.d.: V = IR V = 1.0 × 4.5 Terminal p.d. = 4.5 V That leaves 1.5 V as lost volts, since 4.5 + 1.5 = 6.0 V, exactly the emf.

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Up next: Variable Resistance — now we understand real cells, let’s look at components whose own resistance can be adjusted or changes with conditions, like thermistors and LDRs.

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