Question

A $5m$ potentiometer wire having $3\Omega$ resistance per meter is connected to a storage cell of steady emf $2V$ and internal resistance $1\Omega$. A primary cell is balanced against $3.5m$ of it. When a resistance of $\frac{32}{n}\Omega$ is put in series with the storage cell, the null point shifts to the center of the last wire, i.e.$,4.5m$. What is '$n$'?

Answer 1

Resistance of potentiometer wire ,$R=5\times 3=15\Omega$
Current through it is $i=\frac{V}{R+r}=\frac{2}{15+1}=\frac{1}{8}A$
Voltage in primary cell is $E_p=i{R}_p=\left(\frac{1}{8}\right)(3\times 3.5)=\left(\frac{1}{8}\right)(3\times \frac{7}{2})=\frac{21}{16}$
now, $E_p=i^{\prime }{R}_p^{\prime }\Rightarrow \frac{21}{16}=i^{\prime }(3\times 4.5)=\frac{27}{2}$
$i^{{}^{\prime }}=\left(\frac{21}{16}\right)\left(\frac{2}{27}\right)=\frac{7}{72}A$
Let $R^{\prime }$ be the resistance put in series in storage cell.
thus $i^{\prime }=\frac{V}{R+r+R^{\prime }}$
$\frac{7}{72}=\frac{2}{15+1+R^{\prime }}=\frac{2}{16+R^{\prime }}$
$16+R^{\prime }=\frac{144}{7}$
$R^{\prime }=\frac{144}{7}-16=\frac{32}{7}$ thus $n=7$