What Is EMF and How Is It Different from Voltage?

For IB Physics SL and HL: electromotive force (emf) is the energy supplied by a source per unit charge, whereas potential difference is the energy transferred from electrical energy per unit charge between two points. Both are measured in volts, but they describe different energy transfers.

The Reasoning

The emf of a cell is defined by

$$\varepsilon = \frac{W}{q}$$

where $W$ is the energy supplied by the cell and $q$ is the charge passing through it. An emf of $12\text{ V}$ means the source supplies $12\text{ J}$ to each coulomb of charge.

Potential difference is also energy per unit charge:

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

However, here $W$ usually represents energy transferred from the charges to a component. For example, a $6.0\text{ V}$ potential difference across a lamp means each coulomb transfers $6.0\text{ J}$ to the lamp.

In a complete circuit, the source raises the electrical energy of the charge, while components such as lamps and resistors transfer that energy to thermal, light, or other forms. Energy conservation means that the sum of potential differences around the circuit equals the total emf.

A real cell has internal resistance $r$. When current $I$ flows, some energy is transferred inside the cell, so its terminal potential difference is

$$V = \varepsilon - Ir.$$

The quantity (Ir) is the lost volts, the potential difference across the cell's internal resistance. Increasing current therefore reduces terminal potential difference, even though the emf remains approximately constant.

For a cell with $\varepsilon = 12\text{ V}$, $r = 0.50\,\Omega$, and $I = 2.0\text{ A}$:

$$V = 12 - (2.0)(0.50) = 11\text{ V}.$$

The common misconception is that emf is a force. It is not: despite its name, emf is an energy-per-charge quantity. With no current, $I=0$, so the terminal potential difference equals the emf.

Exam Technique

In B.5 Current and circuits, define both quantities using energy per unit charge, state that both have unit volts, and use $\varepsilon = I(R+r)$ or $V=\varepsilon-Ir$ when internal resistance is present.