Question

An accumulator of emf $2Volt$ and negligible internal resistance is connected across a uniform wire of length $10m$ and resistance $30\Omega$. The appropriate terminals of a cell of emf $1.5Volt$ and internal resistance $1\Omega$ is connected to one end of the wire, and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. If the balancing change when the cell of $1.5volt$ is shunted with a resistance of $5\Omega$ is $\frac{x}{4}m$. Find $x$.

Answer 1

voltage drop on the balancing length of the wire $\Rightarrow$i$\rho$X

where i= currnt flowing in primary circuit

$\rho$=resistance per unit length of the wire

X=balancing length of wire AB when deflection in galvanometer is zero

$\frac{1}{R_{eq.}}\Rightarrow (\frac{1}{1}+\frac{1}{5})=\frac{6}{5}$

$E_{eq.}=\frac{(\frac{E_1}{r_1}+\frac{E_2}{r_2})}{\frac{1}{R_{eq.}}}=\left(\frac{\frac{1.5}{1}+\frac{0}{5}}{\frac{6}{5}}\right)=\frac{5}{4}volt$

when deflection is zero$\Rightarrow \frac{5}{4}=\frac{2}{3}\times \frac{30}{10}\times y$

where $y$ is the length of wire AJ

$\Rightarrow y=\frac{25}{4}m.$

$\Rightarrow x=25$