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 $\epsilon$ 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 $\frac{2l}{3}$. If the resistance of potentiometer wire $w_2$ is given by $\alpha \circ oh$, then find $6\alpha$.

Answer 1

Answer: 6

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 $=\left(\frac{{\epsilon }_o}{3}\right)A$
At balancing length, $\dot{\epsilon }=\left(\frac{{\epsilon }_o}{3}\right)\times 1$
$\Rightarrow 3\epsilon ={\epsilon }_o$
For wire $w_2$, let the resistance be $R$ ohm.
$\therefore$ Current $=\left(\frac{{\epsilon }_o}{R+1}\right)A$
Also, at balancing point,
$\epsilon =\left(\frac{{\epsilon }_o}{R+1}\right)\times \frac{2}{3}R$
$\Rightarrow 3(R+1)=2R\times 3$
$\Rightarrow R=1\Omega$