B.5.14—Emf and internal resistance

Real-cell model

A real cell has emf ε and internal resistance r. With external resistance R and current I,

$$\varepsilon=I(R+r)$$

The internal resistance accounts for energy transferred inside the cell.

Terminal potential difference

The terminal voltage across the external load is

$$V=IR=\varepsilon-Ir$$

As current increases, the internal voltage drop Ir increases and terminal voltage falls.

Use a graph

A graph of terminal V against I has intercept ε and gradient −r. A graph of ε against I with total resistance has slope R+r.

Worked example from the mapped local textbook

A cell has $\varepsilon=1.5\,\mathrm V$, internal resistance $r=0.82\,\Omega$ and load $R=5.6\,\Omega$.

$$I=\frac{\varepsilon}{R+r}=\frac{1.5}{5.6+0.82}=0.23\,\mathrm A$$

$$V_{terminal}=IR=(0.23)(5.6)=1.3\,\mathrm V$$

The loaded terminal voltage is below the emf because energy is also transferred in the internal resistance.

Common trap

The emf is not always the same as the terminal voltage. They are equal only when current is zero or internal resistance is negligible.