Know that the p.d. between two points is measured in volts (V)

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – 4.2.3 Electromotive Force and Potential Difference

4.2.3 Electromotive Force and Potential Difference

Objective

Know that the potential difference (p.d.) between two points is measured in volts (V).

Key Definitions

  • Electromotive force (emf) – The energy supplied per unit charge by a source such as a battery or generator. It is the maximum potential difference the source can provide when no current flows.

  • Potential difference (p.d.) – The work done per unit charge in moving a charge between two points in an electric circuit.

  • Internal resistance (r) – The resistance inherent within a source that reduces the terminal p.d. when current flows.

Potential Difference Formula

The potential difference between two points is given by

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

where

  • \$V\$ is the potential difference (volts, V),

  • \$W\$ is the work done (joules, J),

  • \$Q\$ is the charge transferred (coulombs, C).

Units of Electromotive Force and Potential Difference

QuantitySymbolUnit (SI)Equivalent
Electromotive force / Potential difference\$\mathcal{E}\$ or \$V\$volt1 V = 1 J · C⁻¹
Charge\$Q\$coulomb1 C = 1 A·s
Energy\$W\$joule1 J = 1 N·m

Measuring Potential Difference

A voltmeter is used to measure the p.d. between two points. It must be connected in parallel with the component whose p.d. is to be measured. The voltmeter has a very high internal resistance so that it draws only a negligible current and does not affect the circuit.

Relationship Between emf, Internal Resistance, and Terminal p.d.

When a current \$I\$ flows, the terminal p.d. (\$V\$) of a source is reduced by the voltage drop across its internal resistance \$r\$:

\$V = \mathcal{E} - I r\$

where

  • \$\mathcal{E}\$ is the emf of the source,

    • \$I\$ is the current flowing through the circuit,

  • \$r\$ is the internal resistance of the source.

Practical Example

  1. A 12 V battery has an internal resistance of 0.5 Ω.

  2. If a current of 2 A is drawn, the terminal p.d. is

    3. \$V = 12\ \text{V} - (2\ \text{A})(0.5\ \Omega) = 11\ \text{V}\$

  3. The voltmeter placed across the battery terminals will read 11 V, not the emf of 12 V.

Suggested diagram: A simple circuit showing a battery (emf \$\mathcal{E}\$, internal resistance \$r\$), a load resistor \$R\$, and a voltmeter connected across the load.

Key Points to Remember

  • The unit of potential difference is the volt (V).

  • 1 V = 1 J · C⁻¹ (one joule of energy per coulomb of charge).

  • Emf is the maximum p.d. a source can provide; the terminal p.d. is lower when current flows because of internal resistance.

  • Voltmeters must be connected in parallel and have high internal resistance.