• Tags: potentiometer, EMF, internal-resistance
• Difficulty: Moderate

A potentiometer is a long uniform wire carrying a steady current from a driver cell. The potential drops linearly along its length: V(l) = ($E_{driver}$$\frac{R_wire}{R_wire + r_driver}$) * $\frac{l}{L}$. Comparing EMFs: connect two cells alternately and find null points $l_1$ and $l_2$. Then $E_1$/$E_2$ = $\frac{l_1}{l_2}$. This is exact because at null, no current flows through the test cell — it measures true EMF (unlike a voltmeter). Finding internal resistance: measure null length $l_1$ (cell in open circuit). Then connect resistance R across the cell and find new null $l_2$. V = E - Ir = E$\frac{R}{R+r}$. So E/V = $\frac{l_1}{l_2}$, giving r = R*($l_1$/$l_2$ - 1). Precautions: EMF of driver cell must exceed test cell EMF; wire must be uniform; avoid heating (keep currents low); clean jockey contacts. Sensitivity increases with wire length and driver cell voltage.