Question

A $6\, V$ battery of internal resistance $1 \,\Omega$ is connected across a uniform wire $A B$ of length $100 \,cm$. The positive terminal of another battery of EMF $4 \,V$ and internal resistance $1 \,\Omega$ is joined to point $A$ as shown. The distance of point $P$ from $A$ is $\alpha \times 10 \,cm$. Find the value of $\alpha$, for which there is no current through the galvanometer. (resistance of $A B$ wire is $5\, \Omega$ )

Answer 1

Current flowing through wire AB, $I=\frac{6}{5 + 1}=1A$

The potential gradient of the wire $A B$,
$\phi=\frac{V_{A B}}{A B}=\frac{(1 \times 5)}{100}=0.05 V( cm )^{-1}$
For length AP, equating potential difference,
$(A P) \times \phi=4 V $
$\Rightarrow A P=\frac{4}{0.05}=80\, cm =8 \times 10\, cm$