Question

In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of $2\text{m}$ when the cell is shunted by a $5\Omega$ resistance and is at a length of $3\text{m}$ when the cell is shunted by a $10\Omega$ resistance, then the internal resistance of the cell is

• A. $1.5\Omega$
• B. $10\Omega$
• C. $15\Omega$
• D. $1\Omega$

Answer 1

The correct option is B $10\Omega$

Given:
$l_1=2\text{m},l_2=3\text{m},R_1=5\Omega$
$R_2=10\Omega$

Let $E$ and $r$ be the emf and internal resistance of the cell respectively.

In case of internal resistance measurement by potentiometer,

$\frac{V_1}{V_2}=\frac{l_1}{l_2}$

Since, cell is shunted by $R_1\&R_2$, so

$\frac{\left[\frac{E{R}_1}{(R_1+r)}\right]}{\left[\frac{E{R}_2}{(R_2+r)}\right]}=\frac{R_1(R_2+r)}{R_2(R_1+r)}=\frac{l_1}{l_2}$

putting the given value

$\frac{5(10+r)}{10(5+r)}=\frac{2}{3}$

$\therefore r=10\Omega$

Therefore, option $\left(b\right)$ is the right choice.