Q. 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
Explanation
(a) 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,
$\frac{E}{V} = \frac{R + r}{R}\ \ \ \ \left\lbrack \because E = I(R + r)and\ V = IR \right\rbrack$
$\therefore\ \ \ \ \frac{R + r}{R} - \frac{l_{1}}{l_{2}}$
$or\ \ \ \ 1 + \frac{r}{R} = \frac{110}{100}\ or\ \ \frac{r}{R} = \frac{10}{100}$
$or\ \ \ r = \frac{1}{10} \times 10 = 1\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.