define and use the electromotive force (e.m.f.) of a source as energy transferred per unit charge in driving charge around a complete circuit

Practical Circuits – Electromotive Force (e.m.f.)

Learning Objective

Define and use the electromotive force (e.m.f.) of a source as the energy transferred per unit charge in driving charge around a complete circuit.

Key Concepts

• Electromotive force (e.m.f.) – the work done by a source per unit charge to move charge around a closed circuit. It is a scalar quantity measured in volts (V).
• Internal resistance (\$r\$) – the resistance inherent to the source itself, causing a voltage drop when current flows.
• Terminal voltage (\$V\$) – the voltage measured across the external terminals of the source when a current \$I\$ is flowing.

Definition of e.m.f.

The e.m.f. (\$\mathcal{E}\$) of a source is defined as the energy transferred to each coulomb of charge as it moves through the source:

\$\mathcal{E} = \frac{W_{\text{source}}}{Q}\$

where \$W_{\text{source}}\$ is the work done by the source (in joules) and \$Q\$ is the charge (in coulombs). The unit of e.m.f. is the volt (1 V = 1 J C⁻¹).

Relationship Between e.m.f., Terminal \cdot oltage and Internal Resistance

When a current \$I\$ flows, the source’s internal resistance causes a voltage drop \$Ir\$. The terminal voltage \$V\$ is therefore:

\$V = \mathcal{E} - Ir\$

Re‑arranging gives the practical expression for e.m.f.:

\$\mathcal{E} = V + Ir\$

Measuring e.m.f. in the Laboratory

1. Set up a simple circuit consisting of the source, a variable resistor (or a set of known resistors), an ammeter, and a voltmeter (connected across the source terminals).
2. Record the current \$I\$ and the corresponding terminal voltage \$V\$ for at least three different resistance values.
3. Plot \$V\$ (vertical axis) against \$I\$ (horizontal axis). The graph should be a straight line with a negative gradient.
4. Determine the e.m.f. (\$\mathcal{E}\$) from the \$V\$‑intercept (where \$I = 0\$) and the internal resistance \$r\$ from the magnitude of the gradient (\$-r\$).

Example Calculation

Data obtained from a circuit:

Using two points to find the gradient:

\$\text{Gradient} = \frac{V2 - V1}{I2 - I1} = \frac{1.30 - 1.45}{0.20 - 0.10} = -1.5\ \text{Ω}\$

Thus, internal resistance \$r = 1.5\ \Omega\$.

Extrapolating to \$I = 0\$ (using \$V = \mathcal{E} - Ir\$):

\$\mathcal{E} = V + Ir = 1.45\ \text{V} + (0.10\ \text{A})(1.5\ \Omega) = 1.60\ \text{V}\$

Therefore, the e.m.f. of the source is \$1.60\ \text{V}\$.

Common Sources of Error

• Contact resistance at the terminals of the voltmeter or ammeter.
• Parallax error when reading analog meters.
• Neglecting the resistance of connecting wires, which adds to the measured internal resistance.
• Temperature changes affecting the internal resistance of the source during the experiment.

Safety Precautions

• Ensure all connections are secure before energising the circuit.
• Do not exceed the rated voltage or current of the source to avoid overheating.
• Use appropriately rated meters; overload can damage equipment.
• Disconnect the circuit before adjusting resistances.

Summary Table of Symbols