Know that e.m.f. is measured in volts (V)

4.2.3 Electromotive Force (e.m.f.) and Potential Difference

⚡ Electromotive force (e.m.f.) is the energy supplied per unit charge by a source such as a battery or generator. It is the maximum potential difference that can drive current through a circuit when no current is flowing. The symbol is often ε.

Units of e.m.f.

The unit of e.m.f. is the volt (V). One volt is defined as the potential difference that will move one coulomb of charge through a circuit while doing one joule of work: $$V = \frac{W}{Q}$$ where $W$ is work in joules and $Q$ is charge in coulombs. Thus, $$1\;\text{V} = 1\;\frac{\text{J}}{\text{C}}.$$

Analogy: Water Flow & Battery ⚡

Think of a battery like a water pump. The e.m.f. is the pressure the pump creates, pushing water (electric charge) through a pipe (circuit). The higher the pressure (e.m.f.), the more water can flow. Just as a pump can push water uphill, a battery can push charge against a potential difference.

Potential Difference (Voltage) Across a Component

When current flows, the potential difference $V$ across a component is the energy lost per unit charge: $$V = \frac{\Delta W}{Q}.$$ Unlike e.m.f., this voltage is measured across the component while current is flowing and can be less than the e.m.f. of the source due to internal resistance.

Exam Tips for 4.2.3

• Use the symbol ε for electromotive force.
• Remember: e.m.f. is always in volts (V), not amperes.
• When converting units, keep the relation $1\;\text{V} = 1\;\text{J/C}$ in mind.
• In questions, if a battery supplies a voltage, that is its e.m.f. unless internal resistance is specified.
• Check whether the question asks for the e.m.f. (maximum potential) or the actual potential difference across a component.

Example Problem

A 1.5 V battery supplies a current of 0.5 A to a resistor. Calculate:

1. The work done by the battery on 2 C of charge.
2. The potential difference across the resistor.

Solution:

1. Work $W = ε \times Q = 1.5\,\text{V} \times 2\,\text{C} = 3\,\text{J}$.
2. Potential difference across the resistor $V_R = I \times R$. First find $R = V/I = 1.5\,\text{V}/0.5\,\text{A} = 3\,\Omega$. Then $V_R = 0.5\,\text{A} \times 3\,\Omega = 1.5\,\text{V}$ (equal to the e.m.f. because internal resistance is negligible).

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