Question

A battery of emf $E_o=12V$ is connected across a 4 m long uniform wire having resistance 4 $\Omega /m$. Two cells of small emfs ${\epsilon }_1=2Vand{\epsilon }_2=4V$ having internal resistance $2\Omega and6\Omega$ 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

• A. $\frac{5}{3}m$
• B. $\frac{4}{3}m$
• C. $\frac{3}{2}m$
• D. $\frac{25}{24}m$

Answer 1

The correct option is D $\frac{25}{24}m$

Equivalent emf of two batteries ${\epsilon }_{1}and{\epsilon }_2$ is
$\frac{\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{(\frac{2}{2}+\left(\frac{4}{6}\right))}{(\frac{1}{2}+\left(\frac{1}{6}\right))}=2.5VNow,V_{AN}=\epsilon \therefore (I_{AN}\times R_{AN})=\epsilon or\left(\frac{12}{4+4\times 4}\right)\left(4\right)\left(l\right)=2.5$
Solving this equation we get,
$l=\frac{25}{24}m$