A simple potentiometer circuit is a setup as in figure, using uniform wire AB, 1m long, which has a resistance of 2 Ω. The resistance of the 4 V battery is negligible. If the variable resistor R were given a value of 2.4 Ω, what would be the length AC for zero galvanometer deflection?

Given,

Length of wire $AB$ $(l_{AB}) = 1\;m$

Resistannce of wire $(R) = 2.4 \; \Omega$

EMF of cell $(E_1) = 4\; V$

Resistance across $AB$ $(R_{AB}) = 2 \; \Omega$

Length of $AC$ $(l_{AC}) = ?$

For zero galvanometer deflection,

P.D. across $AC = 1.5 \; V$

And, $V = I * R_{AC}$

or, $1.5 = \frac{4}{2.4 \; + \; 2} * R_{AC}$

$R_{AC} = \frac{1.5 \; * \; 4.4}{4}$  $= 1.65 \; \Omega$

Since, $2 \; \Omega$ wire $AB$ has length $1\;m$

$1 \; \Omega$ wire has length $\frac{1}{2}\;m$

$\therefore$ $1.65 \Omega$ wire $AB$ has length $\frac{1}{2} * 1.65 = 0.825 \; m$

$\therefore$ Length of $AC = 0.825 \; m.$

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