Question

A battery of emf ${ϵ}_0=5V$ and internal resistance $5\Omega$ is connected across a long uniform wire $AB$ of length $1m$ and resistance per unit length $5\Omega m^{-1}$. Two cell of ${ϵ}_1=1V$ and ${ϵ}_2=2V$ are connected as shown in the figure.

• A. The null point is at $A$
• B. If the Jockey is touched to point $B$ the current in the galvanometer will be towards $B$
• C. When Jockey is connected to point $A$ no current is flowing through $1V$ battery
• D. The null point is at distance of $8/15m$ from $A$

Answer 1

The correct option is D The null point is at distance of $8/15m$ from $A$

Here the current flowing through the circuit is
$I=\frac{{\epsilon }_0}{r_0+r_1+r_2}$

$I=\frac{5}{5+1+2}=\frac{5}{8}$

Now,
${\epsilon }_2-{\epsilon }_1=\frac{I(r_1+r_2)l_1}{L}$

$2-1=\frac{\frac{5}{8}\times 3\times l_1}{1}$
Hence $l_1=\frac{8}{15}$ m