Question

A battery of emf ${\epsilon }_0=10V$ is connected across a $1\text{m}$ long uniform wire having resistance $10\frac{\Omega }{m}$. Two cells of emf ${\epsilon }_1=2V$ and ${\epsilon }_2=4V$ having internal resistances $1\Omega$ and $5\Omega$ respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point $P$, find the distance of point $P$ from the point $A$.

• A. $46.67\text{cm}$
• B. $24.37\text{cm}$
• C. $33.65\text{cm}$
• D. $44.55\text{cm}$

Answer 1

The correct option is A $46.67\text{cm}$

Let $i_1,i_2$ be current in secondary and primary circuit respectively
Resistance of wire AB is $10\times 1=10\Omega$
If galvanometer shows a zero reading,
$i_1=\frac{10}{10+10}=0.5A$

Let $AP=X$, then the potential across AP is
So, $V_{AP}=i_1\times \frac{X}{1}\times 10=5X$

Now current in secondary circuit is
$i_2=\frac{4-2}{5+1}=\frac{1}{3}A$
Using KVL in secondary circuit, potential across AP can be written as
$V_{AP}=2+i_2\times 1=2+\frac{1}{3}=\frac{7}{3}$
Hence $5X=\frac{7}{3}$
$X=0.466m=46.6cm$