Cambridge IGCSE Physics 0625 – 4.2.3 Electromotive Force and Potential Difference
4.2.3 Electromotive Force (EMF) and Potential Difference (p.d.)
Learning Objective
Define potential difference (p.d.) as the work done by a unit charge passing through a component.
Key Definitions
Potential Difference (p.d.): The amount of work done by an external agent in moving a unit positive charge between two points in a circuit. It is expressed in volts (V).
Electromotive Force (EMF): The work done per unit charge by a source (e.g., a battery) in moving charge from its negative terminal to its positive terminal when no current flows. It is also measured in volts.
Mathematical Formulation
The relationship between work, charge and potential difference is given by:
\$V = \frac{W}{q}\$
where
\$V\$ = potential difference (volts, V)
\$W\$ = work done (joules, J)
\$q\$ = charge moved (coulombs, C)
Similarly, the EMF of a source is:
\$\mathcal{E} = \frac{W_{\text{source}}}{q}\$
When a current \$I\$ flows, the terminal p.d. of the source is reduced by the internal resistance \$r\$:
\$V_{\text{terminal}} = \mathcal{E} - I r\$
Conceptual Explanation
Consider a small positive test charge \$q\$ moving through a component (e.g., a resistor). If an external agent (such as a battery) does \$W\$ joules of work on this charge, the p.d. across that component is the work per unit charge. This definition applies to any component, whether it supplies energy (like a battery) or consumes it (like a resistor).
Comparison of EMF and Terminal p.d.
Aspect
Electromotive Force (EMF)
Terminal Potential Difference
Definition
Work done per unit charge by the source when no current flows.
Work done per unit charge between the terminals when current flows.
Symbol
\$\mathcal{E}\$
\$V\$
Units
Volts (V)
Volts (V)
Effect of Internal Resistance
None (ideal source)
Reduced by \$I r\$ (where \$r\$ is internal resistance).
Worked Example
A 12 V battery has an internal resistance of \$0.5\;\Omega\$. If the circuit draws a current of \$2\;\text{A}\$, calculate the terminal p.d. across the battery.
Solution:
Calculate the voltage drop due to internal resistance: \$I r = 2\;\text{A} \times 0.5\;\Omega = 1\;\text{V}\$.
Apply the formula \$V_{\text{terminal}} = \mathcal{E} - I r\$:
Thus, the p.d. across the external circuit is \$11\;\text{V}\$.
Important Points to Remember
Potential difference is always measured between two points; it is not a property of a single point.
EMF is a special case of p.d. measured when the circuit is open (no current).
Both EMF and p.d. have the same unit (volt), but they differ when internal resistance is present.
When a charge moves through a component, the sign of the work determines whether the component supplies or consumes energy.
Suggested Diagram
Suggested diagram: A simple circuit showing a battery with EMF \$\mathcal{E}\$, internal resistance \$r\$, an external resistor \$R\$, and arrows indicating the direction of current and the work done on a unit charge as it moves from the negative to the positive terminal.
Practice Questions
Define potential difference in your own words and state its SI unit.
A 9 V battery supplies a current of \$0.3\;\text{A}\$ to a circuit. If the internal resistance of the battery is \$2\;\Omega\$, calculate the terminal p.d. across the battery.
Explain why the terminal p.d. of a real battery is always less than its EMF when current flows.
In a circuit, a resistor dissipates \$20\;\text{J}\$ of energy while \$0.5\;\text{C}\$ of charge passes through it. Determine the p.d. across the resistor.