Question

A battery of emf $E_0$ = 12V is connected across a 4m long uniform wire having resistance $\frac{4\Omega }{m}$. The cells of small emfs ${\epsilon }_1$ = 2V and ${\epsilon }_2$ = 4V 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, find the distance of point N from the point A.

Answer 1

Given Resistance of AB wire$=\frac{4\Omega }{m}$ (as this is per meter)

but AB $=4m$ So,

Resistance of AB $=16\Omega$

Potential gradient of AB $=\frac{V}{L}$

${\epsilon }_0=i(8+16)$

$=\frac{12}{24}=i$

$i=0.5A$

${\epsilon }_0-4=V_{AB}$

$12-4=V_{AB}$

$V_{AB}=8volt$

Potential Gradient$=\frac{V_{AB}}{L}=\frac{8}{4}=2\frac{V}{m}$

Potential gradient$=2\frac{V}{m}$

Net EMF$={\epsilon }_{eq}=\frac{\frac{{\epsilon }_1}{r_1}-\frac{{\epsilon }_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$

$=\frac{1-\frac{2}{3}}{\frac{1}{2}+\frac{1}{6}}$

$=\frac{\frac{1}{3}}{\frac{4}{6}}$

$=\frac{1}{3}\frac{6}{4}$

$=0.5V$

If at N, there is no deflection

So we know that

$=\epsilon =kl$

k=potential gradient

l=length at null point occurs

$0.5=2\times l$

$l=25cm$