Recall and use the equation for e.m.f. E = W / Q

4.2.3 Electromotive force and potential difference

Learning objective

Recall the definitions and equations for electromotive force (e.m.f.) and potential difference (p.d.), and use them to solve typical IGCSE/A‑Level problems.

Key definitions (Core)

  • Electromotive force (e.m.f.) – the work done per unit charge by a source (battery, generator, etc.) in moving charge round a complete circuit.
    E = W ⁄ Q
  • Potential difference (p.d.) – the work done per unit charge as the charge moves between two specified points of a circuit while current may be flowing.
    V = W ⁄ Q
  • Work (W) – energy transferred to or from the charge, measured in joules (J).
  • Charge (Q) – quantity of electricity, measured in coulombs (C).
  • Terminal potential difference – the voltage you read across the external load when the source is delivering current. It is lower than the e.m.f. if the source has internal resistance.

Simple electron model (Supplement – optional)

In a battery chemical reactions separate electrons from their parent atoms, creating an excess of electrons at the negative terminal and a deficit at the positive terminal. This separation establishes a potential difference that drives electrons round the external circuit when a conductive path is provided.

Units

QuantitySymbolUnit (name)SI unit symbol
Electromotive force / Potential differenceE, VVoltV
Work (energy)WJouleJ
ChargeQCoulombC

Deriving the e.m.f. equation (Core)

  1. By definition, e.m.f. is the work done per unit charge:
    E = W ⁄ Q
  2. If a current I flows for a time t, the charge transferred is
    Q = I t
  3. Substituting (2) into (1) gives an alternative form useful for calculations:
    E = W ⁄ (I t)
  4. If you have already studied power (P = W ⁄ t), the equation can also be written as:
    E = P ⁄ I (optional).

Internal resistance (Supplement – optional)

Real sources possess an internal resistance \(r_{\text{int}}\). When a current \(I\) flows, the terminal p.d. is reduced:

\[ V{\text{terminal}} = E - I\,r{\text{int}} \]

This relationship is part of the supplementary material in the syllabus.

Practical note (Core)

  • To measure e.m.f., connect a voltmeter across the terminals of the source without any external load (open‑circuit).
  • To measure the potential difference across a component, place the voltmeter in parallel with that component while the circuit is operating.
  • Always observe safety: never short‑circuit a battery, handle cells with dry hands, and use appropriate rated voltmeters.

Sample calculation (Core)

Example: A 12 V battery does 0.60 J of work to move 0.05 C of charge from its negative to its positive terminal. Find the e.m.f. of the battery.

  1. Write the e.m.f. formula: E = W ⁄ Q
  2. Substitute the given values: E = 0.60 J ⁄ 0.05 C
  3. Calculate: E = 12 V

The e.m.f. of the battery is 12 V, a realistic value for a school‑level cell.

Common misconceptions

  • e.m.f. ≠ terminal voltage when current flows. e.m.f. is measured with no current (open‑circuit); the terminal p.d. is lower by \(I r_{\text{int}}\).
  • Symbol E does not stand for energy. In this context E represents electromotive force and is expressed in volts.
  • Internal resistance can be ignored. In real cells it always causes a small drop in the terminal voltage when current is drawn.
  • Voltage is “consumed” by a component. Voltage is a difference in electric potential; energy is the product of voltage and charge (or voltage and current‑time).

Practice questions

  1. A galvanometer requires 2.5 J of work to move a charge of 0.01 C through it. What is the e.m.f. of the galvanometer?
  2. A battery has an e.m.f. of 12 V and an internal resistance of 0.5 Ω. If a current of 4 A is drawn, calculate the terminal potential difference.
  3. During an experiment a student measures that a source supplies a current of 0.2 A for 30 s while doing 15 J of work. Determine the e.m.f. of the source.

Answers to practice questions

  1. E = W ⁄ Q = 2.5 J ⁄ 0.01 C = 250 V
  2. Terminal p.d. = \(E - I r_{\text{int}}\) = 12 V − (4 A)(0.5 Ω) = 10 V
  3. Charge transferred: \(Q = I t = 0.2 \text{A} \times 30 \text{s} = 6 \text{C}\).
    e.m.f.: \(E = W ⁄ Q = 15 \text{J} ⁄ 6 \text{C} = 2.5 \text{V}\).

Suggested diagram: A simple circuit showing a battery (e.m.f. E) with internal resistance \(r_{\text{int}}\), an external resistor R, the direction of current I, and a voltmeter across the external resistor to indicate the terminal potential difference.