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)
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
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.
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.
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 \(V = \mathcal{E} - I r\).
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.
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