Question

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor $R=10.0\Omega$ 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 $ϵ$?

Answer 1

Resistance of the standard resistor, $R=10.0\Omega$

Balance point for this resistance, $l_1=58.3cm$

Current in the potentiometer wire $=i$

Hence, potential drop across R, $E_1=iR$

Resistance of the unknown resistor $=X$

Balance point for this resistor, $l_2=68.5cm$

Hence, potential drop across X, $E_2=iX$

The relation connecting emf and balance point is,

$\frac{E_1}{E_2}=\frac{l_1}{l_2}$

$\frac{iR}{iX}=\frac{l_1}{l_2}$

$X=\frac{l_1}{l_2}\times R$

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

Therefore, the value of the unknown resistance, X, is $11.75\Omega$.

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.