Question

A battery of emf $E_0=12\text{V}$ is connected across a $4\text{m}$ long uniform wire having resistance $4\Omega /\text{m}$. The cells of small emfs ${\epsilon }_1=2\text{V}$ and ${\epsilon }_2=4\text{V}$ having internal resistance $2\Omega$ and $6\Omega$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point $N$, the distance of point $N$ from the point $A$ is equal to (in $\text{cm}$)

Answer 1

Given circuit can be rearranged as



Resistance of wire AB per unit length is $=4\frac{\Omega }{\text{m}}$
Length of wire AB is $4\text{m}$
So, Resistance of wire AB is $R_{AB}=4\times 4=16\Omega$
Applying KVL to the loop
${\epsilon }_0-R_{AB}i-Ri=0$
$12-16i-8i=0$
$i=0.5\text{A}$

Potential across AB is $V_{AB}=i\left(16\right)=8\text{V}$
Potential gradient across AB is $\frac{V_{AB}}{AB}=\frac{8}{4}=2\Omega /\text{m}$

As cells of small emfs ${\epsilon }_1=2\text{V}$ and ${\epsilon }_2=4\text{V}$ having internal resistance $2\Omega$ and $6\Omega$ are in parallel.(in opposite polarity)

Equivalent emf is ${\epsilon }_{eq}=\frac{\frac{{\epsilon }_1}{r_1}-\frac{{\epsilon }_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$
${\epsilon }_{eq}=\frac{\frac{2}{2}-\frac{4}{6}}{\frac{1}{2}+\frac{1}{6}}=0.5\text{V}$

For galvanometer to show zero deflection, potential difference across $AN$ should be ${\epsilon }_0=0.5\text{V}$

From potential gradient
$V_{AN}=x\left(AN\right)$
$0.5=2\left(AN\right)$
$AN=0.25\text{m}=25\text{cm}$