A potentiometer wire AB having length $$L$$ and resistance $$12r$$ is joined to a cell D of emf $$\varepsilon$$ and internal resistance $$r$$. A cell C having EMF $$\varepsilon/2$$ and internal resistance $$3r$$ is connected. The length AJ, at which the galvanometer, as shown in the figure, shows no deflection is:

Current in the primary circuit: $$I = \frac{\varepsilon}{R_{AB} + r} = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}$$

Potential difference across the wire AB: $$V_{AB} = I \cdot R_{AB} = \left(\frac{\varepsilon}{13r}\right) \cdot 12r = \frac{12\varepsilon}{13}$$

Potential gradient along the wire: $$k = \frac{V_{AB}}{L} = \frac{12\varepsilon}{13L}$$

At zero deflection, no current flows through cell C: $$V_{AJ} = \frac{\varepsilon}{2}$$

$$V_{AJ} = k \cdot AJ \implies \frac{\varepsilon}{2} = \left(\frac{12\varepsilon}{13L}\right) \cdot AJ$$

$$AJ = \frac{\varepsilon}{2} \times \frac{13L}{12\varepsilon} = \frac{13}{24}L$$