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 2m when the cell is shunted by a 5 $\omega$ resistance and is at a length of 3 m when the cell is shunted by a 10 $\omega$ resistance, the internal resistance of the cell is then

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

Answer 1

The correct option is B 10 $\omega$

In case of internal resistance measurement by potentiometer,

$\frac{V_1}{V_2}=\frac{l_1}{l_2}=\frac{\left[E{R}_1/\right(R_1+r\left)\right]}{\left[E{R}_2/\right(R_2+r\left)\right]}=\frac{R_1(R_2+r)}{R_2(R_1+r)}$

Here $l_1=2m,l_2=3m,R_1=5\omega$, and $R_2=10\omega$. So

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