Question

In the primary circuit of potentiometer the rheostat can be varied from $0$ to $10\Omega$. Initially it is at minimum resistance (zero).The rheostat is put at maximum resistance ($10\Omega$) and the switch $S$ is closed. New balancing length is found to be $8m$. Find the internal resistance $r$ (in ohms) of the $4.5V$ cell.

Answer 1

When deflection is zero $\Rightarrow$ voltage of secondary cell=voltage drop on the balancing length of the wire AB
voltage drop on the balancing length of the wire $\Rightarrow$i$\rho$X
where i= current flowing in primary circuit
$\rho$=resistance per unit length of the wire
X=balancing length of wire AB
when deflection in galvanometer is zero $R_{eq.}=\frac{2r}{r+2}$
$E_{eq.}$ of secondary cell$=\frac{(\frac{E_1}{r_1}+\frac{E_2}{r_2})}{\frac{1}{R_{eq.}}}=\left(\frac{\frac{4.5}{r}+0}{\frac{r+2}{2r}}\right)=\frac{9}{r+2}volt$
when deflection is zero$\Rightarrow \frac{9}{r+2}=\left(\frac{10}{20}\right)\left(\frac{9}{12}\right)\left(8\right)$
$r+2=3\Rightarrow r=1\Omega$