The potentiometer wire AB shown in the figure is $40 c m$ long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?

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Solution

If the wire is connected to the potentiometer wire so that $R_{A B} / R_{D B} = 8 / 12$ , then according to wheat stone's bridge no current will flow through galvanometer.
$R_{A B} / R_{D B} = L_{A B} / L_{B} = 8 / 12 = 2 / 3$ (Acc. To principle of potentiometer).
$\begin{matrix} I_{A B} + I_{D B} = 40 c m I_{D B} (2 / 3) + I_{D B} = 40 c m (2 / 3 + 1) I_{D B} = 40 c m 5 / 3 I_{D B} = 40 \Rightarrow I_{D B} = \frac{40 \times 3}{5} = 24 c m I_{A B} = (40 - 24) c m = 16 c m \end{matrix}$

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The potentiometer wire AB shown in the figure is 40cm long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?

The potentiometer wire AB shown in the figure is $40 c m$ long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?