Potentiometer — Internal Resistance

Internal resistance of a cell measured using a potentiometer. l₁ is balance length with cell in open circuit, l₂ with external resistance R connected. Derived from EMF and terminal voltage comparison.

Setup

Cell under test (EMF $\mathcal{E}$, internal resistance $r$) connected to the potentiometer.

Step 1: Switch $S$ open (cell in open circuit). Balance length $= l_1$. Since no current flows:

$$\mathcal{E} = \phi l_1$$

Step 2: Switch $S$ closed — external resistance $R$ connected across the cell. Balance length $= l_2$. Current $I = \mathcal{E}/(R+r)$ flows; terminal voltage:

$$V = \mathcal{E} - Ir = IR = \phi l_2$$

Derivation

Dividing the two equations:

$$\frac{\mathcal{E}}{V} = \frac{l_1}{l_2} \implies \frac{\mathcal{E}}{IR} = \frac{l_1}{l_2}$$

Also $\mathcal{E} = I(R + r)$, so $\mathcal{E}/IR = (R+r)/R$:

$$\frac{R + r}{R} = \frac{l_1}{l_2} \implies r = R\left(\frac{l_1}{l_2} - 1\right)$$ $$\boxed{r = R\left(\frac{l_1 - l_2}{l_2}\right)}$$