Recall and use the equation for p.d. V = W / Q

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – 4.2.3 Electromotive Force and Potential Difference

4.2.3 Electromotive Force (EMF) and Potential Difference

Learning Objective

Recall and use the equation for potential difference:

\$V = \frac{W}{Q}\$

where V is the potential difference (volts), W is the work done (joules) and Q is the charge transferred (coulombs).

Key Definitions

  • Electromotive Force (EMF) – The energy supplied per unit charge by a source (e.g., a battery). It is measured in volts and is denoted by the symbol ε.

  • Potential Difference (p.d.) – The work done per unit charge in moving a charge between two points in a circuit. Also measured in volts and denoted by V.

  • Work (W) – Energy transferred when a charge moves through a potential difference. Measured in joules (J).

  • Charge (Q) – Quantity of electricity. Measured in coulombs (C).

Relationship Between EMF and Potential Difference

In an ideal circuit with no internal resistance, the EMF of the source equals the potential difference across its terminals:

\$\varepsilon = V\$

When internal resistance (r) is present, the terminal p.d. is reduced:

\$V = \varepsilon - I r\$

where I is the current flowing through the circuit.

Units and Symbols

QuantitySymbolUnitUnit Symbol
Potential DifferenceVvoltV
Electromotive ForceεvoltV
Work (Energy)WjouleJ
ChargeQcoulombC
CurrentIampereA
ResistanceRohmΩ

Deriving the Equation

  1. Start with the definition of work done in moving a charge: \$W = V Q\$.

  2. Rearrange to solve for potential difference: \$V = \dfrac{W}{Q}\$.

  3. This shows that the larger the work done for a given charge, the greater the potential difference.

Example Calculation

Problem: A battery does 12 J of work to move 3 C of charge through the circuit. Find the potential difference across the battery terminals.

Solution:

\$V = \frac{W}{Q} = \frac{12\ \text{J}}{3\ \text{C}} = 4\ \text{V}\$

Therefore, the terminal p.d. is 4 V (assuming negligible internal resistance).

Common Misconceptions

  • Confusing EMF with the potential difference measured across a load when internal resistance is present.

  • Thinking that the equation \$V = W/Q\$ only applies to batteries; it applies to any situation where work is done moving charge.

  • Neglecting the sign of work: if work is done by the source, \$W\$ is positive; if work is done on the source, \$W\$ is negative.

Practice Questions

  1. A 9 V battery supplies a current of 0.5 A through a circuit that has an internal resistance of 2 Ω. Calculate the terminal potential difference.

  2. What amount of work is required to move 0.02 C of charge through a potential difference of 15 V?

  3. If a source has an EMF of 12 V and the terminal p.d. measured under load is 10 V, what is the current flowing if the internal resistance is 0.5 Ω?

Answers to Practice Questions

  1. Using \$V = \varepsilon - I r\$: \$V = 9\ \text{V} - (0.5\ \text{A})(2\ \Omega) = 9\ \text{V} - 1\ \text{V} = 8\ \text{V}\$.

  2. Using \$W = V Q\$: \$W = (15\ \text{V})(0.02\ \text{C}) = 0.30\ \text{J}\$.

  3. First find the voltage drop across the internal resistance: \$\varepsilon - V = I r \Rightarrow 12\ \text{V} - 10\ \text{V} = I(0.5\ \Omega)\$. Thus \$2\ \text{V} = 0.5 I\$, giving \$I = 4\ \text{A}\$.

Suggested Diagram

Suggested diagram: A simple circuit showing a battery (EMF ε), internal resistance r, external resistor R, and the direction of current I. Label the terminal potential difference \cdot across the external resistor.

Summary

The potential difference between two points in a circuit is the work done per unit charge, expressed by \$V = \dfrac{W}{Q}\$. EMF represents the source’s ability to do work, and when internal resistance is present the terminal p.d. is reduced according to \$V = \varepsilon - I r\$. Mastery of these relationships enables accurate analysis of electrical circuits in the IGCSE syllabus.