Question

A potentiometer wire has an length 4 m and resistance $4\Omega$. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient of ${10}^{-3}$ V/cm on the wire?

Answer 1

Length of the potentiometer wire, $I=10m$

Resistance of the potentiometer wire, $R=4\Omega$

Emf of the accumulator connected in series with the wire, $E=2V$

A resistance box in series is connected to the potentiometer wire.

Required potential gradient, $k={10}^{-3}V/cm=0.1V/m$

Potential drop along the potentiometer wire, $V=k.I=0.1\times 10=1=1V$

Therefore, current through the potentiometer wire,

$I=\frac{V}{R}=\frac{1}{4}=0.25A$

Let R'' be the resistance of the resistance box then,

Using the relation

$I=\frac{E}{R}-\frac{V}{R^{\prime }}$

$0.25=\frac{2}{4}-\frac{1}{R^{\prime }}$

$\frac{1}{R^{\prime }}=0.25$

$R^{\prime }=4\Omega$