Question

A thin uniform wire $AB$ of length $50cm$ and resistance $1\Omega$ is connected to the terminals of a battery of emf ${\epsilon }_1=2.2V$ and internal resistance $0.1\Omega .$ If the terminals of another cell (assume ideal) are connected to two points 25 $cm$ apart on the wire $AB$ without altering the current in the wire $AB$, then emf ${\epsilon }_2$ of cell in volts is

• A. 0.5 $V$
• B. 1 $V$
• C. 1.2 $V$
• D. 0.8 $V$

Answer 1

The correct option is B 1 $V$

Current in the wire $AB$ can be found by
$i=\frac{{\epsilon }_1}{r+R}=\frac{2.2}{0.1+1}$ = 2 $A$
Resistance of 25 $cm$ wire $=\frac{25}{50}\times 1\Omega =\frac{1}{2}\Omega$
So emf in ${\epsilon }_2=2A\times \frac{1}{2}\Omega =1V$