Figure 3.34 shows a potentiometer circuit for comparison of two resistances.

Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l1 = 58.3 cm

Current in the potentiometer wire = i

Hence, potential drop across R, E1 = iR

Resistance of the unknown resistor = X

Balance point for this resistor, l2 = 68.5 cm

Hence, potential drop across X, E2 = iX

The relation connecting emf and balance point is,

$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

$\frac{i R}{i X}=\frac{l_{1}}{l_{2}}$

$X=\frac{l_{1}}{l_{2}} \times R$

$=\frac{68.5}{58.3} \times 10=11.749 \Omega$

Therefore, the value of the unknown resistance, X, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.