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

4.2.3 Electromotive Force and Potential Difference

Objective

  • 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.

Key Definitions

  • 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).

Units (SI)

QuantitySymbolUnit (SI)Equivalent expression
Electromotive force / Potential difference (voltage)ℰ or Vvolt (V)1 V = 1 J · C⁻¹ (one joule per coulomb)
ChargeQcoulomb (C)1 C = 1 A·s
Energy / WorkWjoule (J)1 J = 1 N·m
CurrentIampere (A)1 A = 1 C · s⁻¹
ResistanceR, rohm (Ω)1 Ω = 1 V · A⁻¹

Fundamental Equations

  • 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: \(r = \dfrac{\mathcal{E} - V}{I}\)

Measuring Voltage (Potential Difference)

  • 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.

Practical Example 1 – Using the emf Definition

  1. The source supplies 6 J of energy to move 2 C of charge when no current flows.
  2. Emf:

    \$\mathcal{E} = \frac{W}{Q} = \frac{6\ \text{J}}{2\ \text{C}} = 3\ \text{V}\$

  3. The source has an internal resistance of 0.2 Ω and a current of 1 A is drawn.
  4. Terminal voltage:

    \$V = \mathcal{E} - I r = 3\ \text{V} - (1\ \text{A})(0.2\ \Omega) = 2.8\ \text{V}\$

  5. A voltmeter placed across the terminals will read 2.8 V, not the emf of 3 V.

Practical Example 2 – More Typical Battery Values

  1. A 9 V dry cell has an internal resistance of 0.5 Ω.
  2. 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}\$

  3. The voltmeter across the lamp reads 8.9 V, while the emf of the cell remains 9 V.

Suggested Experimental Activity (AO3)

Goal: Demonstrate the difference between emf and terminal voltage.

  1. 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.
  2. 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.
  3. 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 \(V = \mathcal{E} - I r\)).
  4. Interpret the graph: explain why the measured voltage falls as the current increases, and relate the result to the equation \(V = \mathcal{E} - I r\).

Diagram – Simple Circuit

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

Key Points to Remember

  • 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.