A simple potentiometer circuit is setup as in fig., using uniform wire AB, 1.0 m 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?

A simple potentiometer circuit is setup as in fig., using uniform wire AB, 1.0 m 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?
Solution:

Given,
length of potentiometer wire, $L$ = 1.0 m
resistance of potentiometer wire, $R_{AB}$ = 2 Ω
emf of driving cell, $V_0$ = 4 V
variable resistance, $R$ = 2.4 Ω
balancing length, $l$ = ?

For balancing length, potential difference across potentiometer wire is equal to p.d. across cell. So, we first calculate p.d. across wire.
Current through wire is, $$\begin{array}{rl}I& =\frac{V_0}{R_{AB}+R}\\ & =\frac{4}{4.4}\\ \therefore I& =0.91A\end{array}$$ Then p.d. across wire is, $$\begin{array}{rl}{V}_{AB}& =I\times R_{AB}\\ & =0.91\times 2\\ \therefore V_{AB}& =1.82V\end{array}$$ We know,
p.d. across AC = p.d. across cell (for balancing length) Then, $$\begin{array}{rl}{V}_{AC}& =kl\\ 1.5& =\frac{V_{AB}}{L}\times l\\ 1.5& =1.82\times l\\ \therefore l& =0.82m\end{array}$$ Hence, the balancing length is 0.82 m .