Potential difference and electromotive force

In a nutshell

Potential difference is defined as the energy transferred per unit charge between two points in a circuit. The electromotive force is the energy per unit charge given to the charge carriers by a power source. Voltmeters have to be placed in parallel to measure the potential difference between two points.

Equations

Description	Equation
Potential difference	$V=\dfrac{W}{Q}$
Electromotive force	$\varepsilon =\dfrac{W}{Q}$

Variable definitions

Quantity name	symbol	derived units	alternate units	si base units
$potential\ difference$	$V$	$V$	$J\ C^{-1}$	$kg\ m^2\ s^{-3}\ A^{-1}$
$work\ done$	$W$	$J$	$eV$	$kg\ m^{2}\ s^{-2}$
$charge$	$Q$	$C$		$A\ s$
$electromotive\ force$	$\varepsilon$	$V$	$J\ C^{-1}$	$kg\ m^2\ s^{-3}\ A^{-1}$

Potential difference

Potential difference or voltage is a measure of how much energy is transferred by charge carriers to a circuit component. It is defined as the energy transferred per unit charge between two points in a circuit. It is symbolised by the letter $V$ and is measured in volts $V$ . You can calculate it with the equation:

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

Electromotive force

Potential difference describes the work done by the charge carriers i.e. the energy given to the components by the charge carriers. The electromotive force (e.m.f.) on the other hand, is the work done on the charge carriers or the energy given to the charge carriers by a power source. It is symbolised by the greek letter $\varepsilon$ and is measured in volts $V$ . You can calculate it with the equation:

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

Using a voltmeter

A voltmeter is a component that is used to measure the potential difference between two points in a circuit by placing it in parallel to a component. You can also use it to measure the e.m.f. of a power supply by placing it in parallel to one. A voltmeter has a very high resistance in order to minimise the current flowing through it since current splits in parallel circuits.

Physics; Electrical circuits; KS5 Year 12; Potential difference and electromotive force

Example

In the circuit above the cell provides $20\ C$ of charge carriers with $200\ J$ of energy. $40\ J$ are then transferred to the resistor, calculate the e.m.f of the cell and the potential difference across the resistor.

Firstly, write down what you know:

$\begin{aligned}Q&=20\ C \\W_{cell}&= 200\ J \\W_{resistor} &=40\ J\end{aligned}$ 

Next, write down the equation for the e.m.f.:

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

In this case you have to use the work done by the cell:

$\varepsilon=\dfrac{W_{cell}}{Q}$ 

Substitute the numbers in and calculate the e.m.f:

$\varepsilon=\dfrac{200}{20}=10\ V$ 

Now write down the equation for voltage:

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

Since you are calculating the voltage across the resistor you have to use the work done on the resistor by the charge carriers:

$V=\dfrac{W_{resistor}}{Q}$ 

Substitute the number and calculate the potential difference:

$V=\dfrac{40}{20}=2\ V$ 

The e.m.f. of the cell is $\underline{20\ V}$ and the potential difference across the resistor is $\underline{2\ V}$ .