Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor $R = 10.0 Ω$ is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf $ϵ$ ?

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Solution

Resistance of the standard resistor, $R = 10.0 Ω$
Balance point for this resistance, $l_{1} = 58.3 c m$
Current in the potentiometer wire $= i$
Hence, potential drop across R, $E_{1} = i R$
Resistance of the unknown resistor $= X$
Balance point for this resistor, $l_{2} = 68.5 c m$
Hence, potential drop across X, $E_{2} = i X$
The relation connecting emf and balance point is,
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$
$\frac{i R}{i X} = \frac{l_{1}}{l_{2}}$
$X = \frac{l_{1}}{l_{2}} \times R$
$= \frac{68.5}{58.3} \times 10 = 11.749 Ω$
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.

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