Cambridge Syllabus Notes

4.2.3 Electromotive Force and Potential Difference

Understand that both electromotive force (emf) and potential difference (p.d.) – commonly called **voltage** – are measured in volts (V).
Be able to define, relate and use the equations for emf, p.d., internal resistance and terminal voltage.
Develop the practical skill of measuring voltage with a voltmeter and interpreting the difference between emf and terminal voltage.

Electromotive force (emf, ℰ) – The electrical work done **by a source** in moving a unit charge once round a complete circuit. It is the maximum voltage the source can deliver when no current flows (open‑circuit condition).
Potential difference (p.d.) or voltage (V) – The work done per unit charge **by the external circuit** in moving a charge between two points.
Internal resistance (r) – The inherent opposition to the flow of charge inside a source (cell, battery, generator). It arises from the material and construction of the source and causes a loss of voltage when current flows.
Terminal voltage – The voltage actually measured across the external terminals of a source while it supplies a current.
Open‑circuit condition – When the external circuit is broken (I = 0). Under this condition the terminal voltage equals the emf (V = ℰ).

Quantity	Symbol	Unit (SI)	Equivalent expression
Electromotive force / Potential difference (voltage)	ℰ or V	volt (V)	1 V = 1 J · C⁻¹ (one joule per coulomb)
Charge	Q	coulomb (C)	1 C = 1 A·s
Energy / Work	W	joule (J)	1 J = 1 N·m
Current	I	ampere (A)	1 A = 1 C · s⁻¹
Resistance	R, r	ohm (Ω)	1 Ω = 1 V · A⁻¹

Potential difference (voltage): $$V = \frac{W}{Q}$$ where W is the work done by the external circuit and Q the charge transferred.
Electromotive force: $$\mathcal{E} = \frac{W_{\text{source}}}{Q}$$ (same form as the p.d. equation but the work is done **by the source**.)
Terminal voltage of a source delivering a current I: $$V = \mathcal{E} - I r$$
‑ The term I r is the voltage drop across the internal resistance, which **reduces** the potential available at the terminals.
Re‑arranged forms useful for AO2 questions:
- Current: $I = \dfrac{\mathcal{E} - V}{r}$
- Internal resistance: $r = \dfrac{\mathcal{E} - V}{I}$

A **voltmeter** is designed to measure the voltage between two points.
It must be connected **in parallel** with the component whose voltage is required.
Its internal resistance is very high (≈ 10 MΩ or more) so that it draws a negligible current and does not appreciably alter the circuit.

The source supplies 6 J of energy to move 2 C of charge when no current flows.
Emf: $$\mathcal{E} = \frac{W}{Q} = \frac{6\ \text{J}}{2\ \text{C}} = 3\ \text{V}$$
The source has an internal resistance of 0.2 Ω and a current of 1 A is drawn.
Terminal voltage: $$V = \mathcal{E} - I r = 3\ \text{V} - (1\ \text{A})(0.2\ \Omega) = 2.8\ \text{V}$$
A voltmeter placed across the terminals will read **2.8 V**, not the emf of 3 V.

A 9 V dry cell has an internal resistance of 0.5 Ω.
If it supplies a current of 0.20 A to a lamp, the terminal voltage is: $$V = 9\ \text{V} - (0.20\ \text{A})(0.5\ \Omega) = 9\ \text{V} - 0.10\ \text{V} = 8.9\ \text{V}$$
The voltmeter across the lamp reads 8.9 V, while the emf of the cell remains 9 V.

Goal: Demonstrate the difference between emf and terminal voltage.

Set up the circuit shown in the diagram below: a battery (unknown emf ℰ, internal resistance r), a variable resistor R, and a digital voltmeter across the battery terminals.
Vary R from a high value (≈ 10 kΩ) down to a low value (≈ 10 Ω) and record the current (using an ammeter in series) and the terminal voltage for each setting.
Plot a graph of V (y‑axis) against I (x‑axis). The straight‑line intercept on the voltage axis gives the emf ℰ; the slope (negative) gives the internal resistance r (since $V = \mathcal{E} - I r$).
Interpret the graph: explain why the measured voltage falls as the current increases, and relate the result to the equation $V = \mathcal{E} - I r$.

Battery (emf ℰ, internal resistance r) → load resistor R → ammeter → back to battery. A voltmeter is connected across the battery terminals.

The unit of both emf and potential difference is the **volt (V)**; 1 V = 1 J · C⁻¹.
Emf is the maximum voltage a source can provide (open‑circuit). When a current flows, the terminal voltage is reduced by the drop $I r$ across the internal resistance.
Internal resistance is a property of the source itself; it is not the same as the external load resistance.
Voltmeters must be connected in parallel and have a very high internal resistance to avoid disturbing the circuit.
For problem‑solving (AO2) you may need to rearrange $V = \mathcal{E} - I r$ to find any of the four quantities ℰ, V, I or r.
For experimental skills (AO3) you can determine ℰ and r from a V‑I graph obtained by varying the external load.