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Question
The potentiometer wire AB shown in the figure is 40 cm long. Where should the free end of the galvanometer be connected on AB, so that the galvanometer may show zero deflection?
Solution
Let D be the null-point of the potentiometer. Since the bridge is balanced at this point,
\[\frac{R_{AD}}{R_{DB}} = \frac{8}{12}\]
According to the principle of a potentiometer,
\[\frac{R_{AD}}{R_{DB}} = \frac{l_{AD}}{l_{DB}} = \frac{8}{12} = \frac{2}{3} \]
\[ \Rightarrow l_{AD} = \frac{2}{3} l_{DB} \]
\[ l_{AD} + l_{DB} = 40 cm\]
\[ \Rightarrow l_{DB} \frac{2}{3} + l_{DB} = 40 cm\]
\[ \Rightarrow \frac{5}{3} l_{DB} = 40 cm\]
\[ \Rightarrow l_{DB} = 40 \times \frac{3}{5} = 24 cm\]
Hence, the null-point is obtained 24 cm from B.
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