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. $1\Omega$
• B. $0.5\Omega$
• C. $2\Omega$
• D. none

Answer 1

The correct option is A $1\Omega$

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

To find the internal resistance

Solution:

This is an example of application of potentiometer.

Let $E$ be the emf of the cell and $V$ the terminal potential difference, then

$\frac{E}{V}=\frac{l_1}{l_2}$

where $I_1$and $I_2$ are the length of the potentiometer wire with and without short-circuited through a resistance.

Since,

$\frac{E}{V}=\frac{R+r}{R}[\because E=I(R+r),V=IR]$

Therefore, $\frac{R+r}{R}=\frac{l_1}{l_2}$

or, $1+\frac{r}{R}=\frac{110}{100}⟹\frac{r}{R}=\frac{11}{10}-1⟹\frac{r}{R}=\frac{1}{10}⟹r=\frac{1}{10}\times 10=1\Omega$

is the required internal resistance.