Question

Two potentiometer wires $w_{1}$ and $w_{2}$ of equal length $l$, connected to a battery of emf $e_{0}$ and internal resistance $1 \, \Omega$ through two switches $s_{1}$ and $s_{2}$. A battery of emf $\varepsilon$ is balanced on these potentiometer wires one by one. The potentiometer wire $w_{1}$ is of resistance $2 \, \Omega$ and balancing length is $l / 2$ on it, when only $s_{1}$ is closed and $s_{2}$ is open. On closing $s_{2}$ and opening $s_{1}$, the balancing length on $w_{2}$ is found to be $\frac{2 l}{3}$. If the resistance of potentiometer wire $w_{2}$ is given by $\alpha \circ oh$, then find $6 \alpha$.

Answer 1

When the circuit is balanced by $w_{1}$,
balancing length is achieved at $\frac{l}{2}$.
Resistance of $w_{1}=2 \,\Omega$
$\therefore $ Current through the circuit $=\left(\frac{\varepsilon_{o}}{3}\right) A$
At balancing length, $\dot{\varepsilon}=\left(\frac{\varepsilon_{o}}{3}\right) \times 1 $
$\Rightarrow 3 \varepsilon=\varepsilon_{o}$
For wire $w_{2}$, let the resistance be $R$ ohm.
$\therefore $ Current $=\left(\frac{\varepsilon_{o}}{R+1}\right) A$
Also, at balancing point,
$\varepsilon=\left(\frac{\varepsilon_{o}}{R+1}\right) \times \frac{2}{3} R $
$\Rightarrow 3(R+1)=2 R \times 3 $
$\Rightarrow R=1\, \Omega$