Know that e.m.f. is measured in volts (V)

4.2.3 Electromotive Force (e.m.f.) and Potential Difference

Learning objective

State that electromotive force (e.m.f.) is measured in volts (V) and use the relevant equations to relate e.m.f. to work, charge, current, power and internal resistance.

Key definitions (core)

• Electromotive force (e.m.f.) 𝓔 – the work done by a source on each coulomb of charge when no current is flowing.
  

  Equation: \(\displaystyle \mathcal{E}= \frac{W}{Q}\)  (Units J C⁻¹ = V)

• Potential difference (p.d.) V – the work done per coulomb of charge as it moves between two points while a current is flowing.
  

  Equation: \(\displaystyle V = \frac{W}{Q}\)  (Units V)

Symbols and units

Measuring e.m.f. and potential difference

• e.m.f. (open‑circuit voltage) – connect a high‑impedance voltmeter across the terminals of the source with the circuit open (no external load). The reading is the e.m.f. in volts. The high input resistance ensures the source is essentially unloaded.
• Potential difference across a component (live circuit) – connect a voltmeter in parallel with the component whose voltage you wish to measure. Choose a range that is higher than the expected p.d. to avoid over‑loading the meter and to obtain an accurate reading.
• Both analogue and digital voltmeters may be used; digital meters usually display the range automatically, whereas analogue meters require manual range selection.

Supplementary material (optional – for extended syllabus)

• Internal resistance effect: when a current I flows, the terminal voltage of a source is reduced by the drop across its internal resistance \(r_{\text{int}}\).
  

  \(\displaystyle V{\text{terminal}} = \mathcal{E} - I\,r{\text{int}}\)

• Power supplied by an ideal source: \(\displaystyle P = \mathcal{E}\,I\)
• e.m.f. in a generator: in a simple a.c. generator a coil rotates in a magnetic field, inducing an e.m.f. that varies sinusoidally with time. The peak e.m.f. is given by \(\mathcal{E}_{\text{max}} = N B A \omega\) (where N = number of turns, B = magnetic flux density, A = coil area, ω = angular speed).

Worked example – terminal voltage of a cell

A 1.5 V AA cell has an internal resistance of 0.2 Ω. When it supplies a current of 0.5 A, the terminal voltage is:

\[

V{\text{terminal}} = \mathcal{E} - I r{\text{int}} = 1.5\ \text{V} - (0.5\ \text{A})(0.2\ \Omega) = 1.5\ \text{V} - 0.10\ \text{V} = 1.40\ \text{V}

\]

Typical e.m.f. values (core + supplement)

• AA alkaline cell: ≈ 1.5 V (single cell) – 3 V for a 2‑cell pack
• 9‑V battery (six 1.5 V cells in series): ≈ 9 V
• Car battery (lead‑acid): ≈ 12 V (12.6 V when fully charged)
• UK household mains (r.m.s.): ≈ 230 V (peak ≈ 325 V)
• Simple a.c. generator (hand‑crank): peak e.m.f. typically 2–5 V depending on speed

Safety reminder

Never short‑circuit a cell or battery; the resulting large current can cause overheating, leakage or explosion.

Common misconceptions

• e.m.f. is not a mechanical force – it is a potential difference (energy per unit charge).
• Terminal voltage equals e.m.f. only when the current is zero. With current flowing, internal resistance causes a drop.
• Volt (V) measures energy per charge, whereas ampere (A) measures charge per time. They are not interchangeable.
• A voltmeter must be connected in parallel with the element whose p.d. is being measured; connecting it in series would alter the circuit.

Summary

Electromotive force is the maximum potential difference a source can provide and is measured in volts (1 V = 1 J C⁻¹). The open‑circuit e.m.f. is obtained with a high‑impedance voltmeter. When a current flows, the terminal voltage is reduced by \(I r_{\text{int}}\). Knowing \(\mathcal{E}=W/Q\) and the power relation \(P=\mathcal{E}I\) enables students to solve a wide range of IGCSE physics problems involving batteries, generators and other sources. Selecting the correct voltmeter range and connecting it in parallel ensures accurate measurement of potential difference in live circuits.