E.m.f. and internal resistanceEdexcel International A Level Physics: Revision notes

The electromotive force (e.m.f.) of a source is the electrical energy transferred to each coulomb of charge that passes through it, so $\varepsilon = \dfrac{W}{Q}$ and it is measured in volts. The energy comes from chemical energy in a cell (or other forms in other sources).

Despite the name, e.m.f. is not a force. It is the energy per unit charge supplied by the source, and it is the same however much current is drawn.

A real cell has internal resistance $r$, because charge passing through it meets resistance inside the cell. The terminal potential difference $V$ is the energy per coulomb delivered to the external circuit, which is what a voltmeter across the terminals reads.

Some of the energy per coulomb is dissipated in $r$, so $V$ is less than $\varepsilon$ whenever a current $I$ flows:

$\varepsilon = V + Ir$, so $V = \varepsilon - Ir$

The term $Ir$ is the lost volts. When $I = 0$, $V = \varepsilon$.

• E.m.f.: energy transferred to each coulomb by the source. It is constant for the cell.
• Terminal pd: energy delivered from each coulomb to the external circuit. It depends on the current drawn.

For a circuit with external resistance $R$: $\varepsilon = I(R + r)$. As $I$ increases (smaller $R$), the lost volts $Ir$ grow and $V$ falls. In a short circuit ($R = 0$) the current is greatest, $I = \varepsilon/r$, and $V = 0$.

The power wasted inside the cell is $I^2 r$, so a cell with large $r$ heats up when it supplies a large current.

A cell of e.m.f. 1.60 V is connected across a 4.0 Ω resistor and the terminal pd falls to 1.50 V.

• Current: $I = V/R = 1.50 \div 4.0 = 0.375$ A
• Lost volts: $1.60 - 1.50 = 0.10$ V
• Internal resistance: $r = 0.10 \div 0.375 = 0.27\ \Omega$

Check: $\varepsilon = I(R + r) = 0.375 \times 4.27 = 1.60$ V.

Connect the cell in series with an ammeter, a switch and a variable resistor, with a high-resistance voltmeter across the cell terminals.

1. Close the switch and set a current; record $V$ and $I$.
2. Change the variable resistor and repeat for at least six different currents.
3. Switch off between readings.
4. Plot $V$ against $I$.

From $V = \varepsilon - Ir$: the intercept is the e.m.f. and the gradient is $-r$, so $r$ is the magnitude of the gradient.

Keep currents small and switch off between readings, because a large or continuous current heats the cell, changes $r$ and discharges the cell, so $\varepsilon$ falls.