All subjects · Physics · WPH12 (Unit 2)

E.m.f. and internal resistance (with core practical 8)

Real cells have internal resistance, so the p.d. at their terminals drops when they supply current. Core practical 8 measures the e.m.f. and internal resistance of a cell.

E.m.f.

The electromotive force (e.m.f.) ε of a source is the energy transferred into electrical energy per unit charge passing through it: ε = W / Q, measured in volts. It equals the terminal p.d. when no current flows.

Internal resistance

Every real source has some internal resistance r. When a current I flows, energy is dissipated inside the source, and the 'lost volts' are Ir. So:

ε = I(R + r) and V = ε - Ir

where V is the terminal p.d. across the external resistance R. The terminal p.d. is less than the e.m.f. whenever current flows.

Core practical 8

Connect the cell in series with a variable resistor and an ammeter, with a voltmeter across the cell's terminals.
Vary the resistance and record pairs of V and I.
Plot V (y-axis) against I (x-axis). From V = -rI + ε, the y-intercept is ε and the gradient is -r.
Open the switch between readings so the cell does not heat up or run down, which would change r and ε. A small fixed resistor in series protects the cell from very large currents.

Worked example

A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current and the terminal p.d.

I = ε / (R + r) = 1.5 / 3.0 = 0.50 A
V = ε - Ir = 1.5 - 0.50 x 0.50

Answer: I = 0.50 A; V = 1.25 V

Key idea: ε = I(R + r); V = ε - Ir. Plot V against I: intercept = ε, gradient = -r.

Check yourself

A battery has e.m.f. 9.0 V and internal resistance 1.0 Ω. It supplies 0.60 A. Find its terminal p.d. in V.
Show answer
8.4 V. V = ε - Ir = 9.0 - 0.60 = 8.4 V.
A 12 V battery with internal resistance 0.40 Ω is connected to a 5.6 Ω resistor. Find the current in A.
Show answer
2 A. I = ε / (R + r) = 12 / 6.0 = 2.0 A.
A cell of e.m.f. 1.50 V has a terminal p.d. of 1.20 V when it supplies 0.60 A. Find its internal resistance in Ω.
Show answer
0.5 Ω. Lost volts = 0.30 V; r = 0.30 / 0.60 = 0.50 Ω.
In core practical 8, the V-I graph has a y-intercept of 1.58 V and passes through (0.80 A, 1.18 V). Find the internal resistance in Ω.
Show answer
0.5 Ω. Gradient = (1.18 - 1.58) / 0.80 = -0.50 Ω, so r = 0.50 Ω.
On a graph of terminal p.d. V against current I for a cell, what does the y-intercept give?
- The internal resistance
- The external resistance
- The maximum current
- The e.m.f.
Show answer
The e.m.f.. When I = 0 there are no lost volts, so V = ε.
Why is the terminal p.d. of a cell less than its e.m.f. when it supplies a current?
- Charge is lost inside the cell
- Energy is dissipated in the cell's internal resistance
- The external resistor creates energy
- The e.m.f. increases with current
Show answer
Energy is dissipated in the cell's internal resistance. Some of the energy per coulomb is transferred inside the cell (lost volts = Ir).
In core practical 8, why is the switch opened between readings?
- To reset the ammeter to zero
- To increase the e.m.f.
- To make the readings bigger
- To stop the cell heating up and running down, which would change r and ε
Show answer
To stop the cell heating up and running down, which would change r and ε. A cell supplying current for long periods gets warm and its e.m.f. falls.
What is the e.m.f. of a source?
- The force that pushes electrons
- The current it supplies per volt
- The energy transferred into electrical energy per unit charge passing through it
- The p.d. across its internal resistance
Show answer
The energy transferred into electrical energy per unit charge passing through it. Despite the name, e.m.f. is energy per charge, measured in volts.

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Last updated 7 November 2026. Spotted a mistake? Tell us.