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(A) In a potentiometer circuit,for a balancing point to exist,the potential difference across the wire $AB$ must be greater than the electromotive for

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$2 \, m$

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(A) In a potentiometer circuit,for a balancing point to exist,the potential difference across the wire $AB$ must be greater than the electromotive force $(EMF)$ of the cell in the secondary circuit $(E_1)$.Furthermore,the positive terminal of the primary cell $(E)$ and the positive terminal of the secondary cell $(E_1)$ must be connected to the same point (point $A$).Looking at the circuit diagram,the positive terminal of the primary cell $(E = 5 \, V)$ is connected to point $A$,while the positive terminal of the secondary cell $(E_1 = 3 \, V)$ is also connected to point $A$.First,calculate the current $I$ in the primary circuit: $I = \frac{E}{R + r} = \frac{5}{4.5 + 0.5} = \frac{5}{5} = 1 \, A$.The potential drop across the wire $AB$ is $V_{AB} = I 	imes R = 1 	imes 4.5 = 4.5 \, V$.Since $V_{AB} > E_1$ $(4.5 \, V > 3 \, V)$,a balancing point exists.The potential gradient $k$ along the wire is $k = \frac{V_{AB}}{L} = \frac{4.5 \, V}{3 \, m} = 1.5 \, V/m$.For no deflection in the galvanometer,the potential difference across length $AC$ must equal $E_1$.$V_{AC} = k 	imes AC = E_1$$1.5 	imes AC = 3$$AC = \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

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