In the figure, the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of emf $\varepsilon$ and internal resistance $r$. The cell $C's$ emf is $\varepsilon/2$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is

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

When there is no any deflection in the galvanometer $\Rightarrow$ voltage of secondary circuit $=$ voltage drop across the wire AJ

$\Rightarrow {\dfrac {E}{2}}=(\dfrac {E}{10r})(\dfrac {9r}{L}){x}$ where $x$ is the length of wire AJ

${x}={\dfrac {5L}{9}}$