What Is Internal Resistance of a Cell and How Does It Affect the Circuit?

A cell's internal resistance is the resistance to current within the cell itself. It reduces the cell's terminal potential difference, limits circuit current, and transfers some energy into thermal energy inside the cell.

This is SL and HL content in B.5 Current and circuits. A real cell is modelled as a source of electromotive force (emf) $\varepsilon$ in series with an internal resistance $r$. The emf is the energy supplied by the cell per unit charge, whereas terminal potential difference $V$ is the energy transferred per unit charge to the external circuit.

When the cell supplies current $I$, the potential difference across its internal resistance is (Ir), called the lost volts. Therefore,

$$V=\varepsilon-Ir$$

If the external resistance is $R$, then $V=IR$. Combining the equations gives

$$\varepsilon=I(R+r), \qquad I=\frac{\varepsilon}{R+r}.$$

For example, a cell has $\varepsilon=6.0\text{ V}$, $r=0.50\,\Omega$, and $R=2.5\,\Omega$:

$$I=\frac{6.0}{2.5+0.50}=2.0\text{ A}$$

Hence $V=IR=5.0\text{ V}$, while the lost volts are $Ir=1.0\text{ V}$.

The energy balance also explains efficiency. In time $t$, the cell supplies energy $\varepsilon It$, the external circuit receives $VIt=I^2Rt$, and the cell dissipates $I^2rt$ internally. Thus the fraction delivered to the load is . Increasing reduces current and internal heating, although it may also reduce the useful power transferred to the load.

Experimentally, plotting $V$ against $I$ gives a straight line with vertical intercept $\varepsilon$ and gradient (-r). This provides a standard way to determine both quantities from data.

A common misconception is that the emf decreases when current flows. In this model, the emf remains $\varepsilon$; it is the terminal potential difference that falls.

Exam technique: State $\varepsilon=I(R+r)$, distinguish emf from terminal potential difference, and show equation, substitution, and answer in a Calculate response.