A 6 , V battery of negligible internal resist | vedclass

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(A) The potential gradient $k$ along the wire $AB$ is given by the potential difference across the wire divided by its length.Since the $6 \, V$ batte

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$2/3 \, m$

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(A) The potential gradient $k$ along the wire $AB$ is given by the potential difference across the wire divided by its length.Since the $6 \, V$ battery is connected directly across the $1 \, m$ wire,the potential gradient $k = \frac{6 \, V}{1 \, m} = 6 \, V/m$.When the ammeter shows zero deflection,it means no current flows through the secondary circuit containing the $4 \, V$ battery.In this condition,the potential difference across the length $AC$ of the wire must be equal to the emf of the secondary battery.Let $l_{AC}$ be the length of $AC$. Then,$V_{AC} = k 	imes l_{AC} = 4 \, V$.Substituting the value of $k$,we get $6 	imes l_{AC} = 4$.Therefore,$l_{AC} = \frac{4}{6} \, m = \frac{2}{3} \, m$.

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