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,

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.