In the given potentiometer circuit,the length | vedclass

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(A) $1$. First,calculate the current $I$ flowing through the potentiometer wire $AB$. The total resistance of the primary circuit is $R_{total} = r +

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$2$

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(A) $1$. First,calculate the current $I$ flowing through the potentiometer wire $AB$. The total resistance of the primary circuit is $R_{total} = r + R = 0.5 \, \Omega + 4.5 \, \Omega = 5.0 \, \Omega$.$2$. The current $I$ in the wire $AB$ is $I = \frac{E}{R_{total}} = \frac{5 \, V}{5 \, \Omega} = 1 \, A$.$3$. The potential difference across the wire $AB$ is $V_{AB} = I 	imes R = 1 \, A 	imes 4.5 \, \Omega = 4.5 \, V$.$4$. The potential gradient $x$ along the wire $AB$ is $x = \frac{V_{AB}}{L_{AB}} = \frac{4.5 \, V}{3 \, m} = 1.5 \, V/m$.$5$. For no deflection in the galvanometer,the potential difference across the length $AC$ must be equal to the $EMF$ of the secondary cell $E_1 = 3 \, V$.$6$. Therefore,$x 	imes AC = E_1 \implies 1.5 \, V/m 	imes AC = 3 \, V$.$7$. Solving for $AC$,we get $AC = \frac{3 \, V}{1.5 \, V/m} = 2 \, m$.

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is ................ $m$.

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