Potential Difference and Electromotive Force

Potential difference (p.d., or voltage) and electromotive force (e.m.f.) both measure energy per unit charge, but they apply to different parts of a circuit.

Potential difference

The potential difference across a component is the energy transferred per unit charge as charge passes through the component.

$V = \frac{W}{Q}$

Symbol	Quantity	Unit
$V$	Potential difference	V (volts)
$W$	Energy transferred (work done)	J
$Q$	Charge	C

1 volt means 1 joule of energy is transferred per coulomb of charge.

Electromotive force (e.m.f.)

The e.m.f. of a source (battery, cell, generator) is the energy transferred per unit charge by the source. It is the total energy the source provides per coulomb.

$\varepsilon = \frac{W}{Q}$

The difference

• E.m.f. is the energy supplied by the source per coulomb.
• P.d. is the energy used (transferred) by a component per coulomb.

In a simple circuit, the total e.m.f. equals the sum of the potential differences across all components (Kirchhoff’s voltage law).

Measuring p.d.

Potential difference is measured using a voltmeter, connected in parallel across the component.

Worked example

A battery supplies 540 J of energy when 45 C of charge flows. Calculate the e.m.f.

$\varepsilon = \frac{540}{45} = 12\text{ V}$

Common errors and how to correct them

Connecting a voltmeter in series. A voltmeter must be connected in parallel across the component. In series, it would alter the circuit.

Confusing e.m.f. and p.d. The e.m.f. is the total energy per coulomb from the source. The p.d. is the energy per coulomb used by a component. In a real circuit with internal resistance, the terminal p.d. is less than the e.m.f.

How to apply this in an exam

Define both terms using “energy per unit charge.” State that e.m.f. applies to the source and p.d. applies to components. Show the equation and units.