Question

A cell can be balanced against $110cm$ and $100cm$ of potentiometer wire respectively, with and without being short-circuited through a resistance of $10\Omega$. Its internal resistance is :

• A. $1.0\Omega$
• B. $0.5\Omega$
• C. $2.0\Omega$
• D. Zero

Answer 1

The correct option is A $1.0\Omega$

In potentiometer experiment in which we find internal resistance of a cell, let $E$ be the emf of the cell and $V$ the terminal potential difference, then
$\frac{E}{V}=\frac{l_1}{l_2}$
where $l_1$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance.
Since, $\frac{E}{V}=\frac{R+r}{R}$
$[\because E=I(R+r)andV=IR]$
$\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$
or $1+\frac{r}{R}=\frac{110}{100}$
or $\frac{r}{R}=\frac{10}{100}$
or $r=\frac{1}{10}\times 10=1\Omega$