In the given potentiometer circuit,the length | vedclass

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(A) The primary circuit consists of a battery of $E = 5 \, V$ and internal resistance $r = 0.5 \, \Omega$ connected in series with the potentiometer w

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(A) The primary circuit consists of a battery of $E = 5 \, V$ and internal resistance $r = 0.5 \, \Omega$ connected in series with the potentiometer wire $AB$ of resistance $R = 4.5 \, \Omega$.The total resistance of the primary circuit is $R_{total} = R + r = 4.5 \, \Omega + 0.5 \, \Omega = 5.0 \, \Omega$.The current flowing through the potentiometer wire $AB$ is $I = \frac{E}{R_{total}} = \frac{5 \, V}{5 \, \Omega} = 1 \, A$.The potential difference across the wire $AB$ is $V_{AB} = I 	imes R = 1 \, A 	imes 4.5 \, \Omega = 4.5 \, V$.The potential gradient along the wire is $k = \frac{V_{AB}}{L_{AB}} = \frac{4.5 \, V}{3 \, m} = 1.5 \, V/m$.For no deflection in the galvanometer,the potential difference across the length $AC$ (let it be $\ell$) must be equal to the $EMF$ of the secondary cell $E_1 = 3 \, V$.Therefore,$V_{AC} = k 	imes \ell = 3 \, V$.$1.5 \, V/m 	imes \ell = 3 \, V$.$\ell = \frac{3}{1.5} = 2 \, m$.

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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