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, \[\begin{align*} I&=\frac{V_0}{R_{AB}+R}\\ &=\frac{4}{4.4}\\ \therefore I&=0.91 \hspace{0.1cm} A\\ \end{align*}\] Then p.d. across wire is, \[\begin{align*} V_{AB}&=I \times R_{AB}\\ &=0.91 \times 2\\ \therefore V_{AB}&=1.82 \hspace{0.1cm} V\\ \end{align*}\] We know,
p.d. across AC = p.d. across cell (for balancing length) Then, \[\begin{align*} V_{AC}&=kl\\ 1.5 &=\frac{V_{AB}}{L}\times l\\ 1.5 & = 1.82 \times l\\ \therefore l&=0.82\hspace{0.1cm} m\\ \end{align*}\] Hence, the balancing length is 0.82 m .