Question

In an experiment we are calculating the internal resistance of a cell using potentiometer, it is found that balancing length is $2m$ when the cell is shunted by $5\Omega$ resistance and when the cell is shunted by $10\Omega$ resistance, it is $3m$ . Then, what is the value of internal resistance of the cell.

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

Answer 1

Answer: (C)

Suppose a cell of emf ' $E$ and internal resistance
$r$ is shunted by a resistance $R$ and balance point is obtained at a distance $l$ of the potentiometer wire. Then, the terminal potential difference of the cell is equal to p.d. across length $l$ of the wire. If $K$
is the potential gradient along the wire, then:
$x\times l=$ $\left(\frac{E}{R + r}\right)R$
For the $1^{\text{st }}$ case, $x\times 2=\left(\frac{E}{5 + r}\right)\times 5$
For the $2^{\text{nd }}$ case, $x\times 3=\left(\frac{E}{10 + r}\right)\times 10$
$\frac{\left(\right. 1 \left.\right)}{\left(\right. 2 \left.\right)}\Rightarrow \frac{x \times 2}{x \times 3}=\frac{5}{10}\times \frac{\left(\right. 10 + r \left.\right)}{\left(\right. 5 + r \left.\right)}\Rightarrow r=10\Omega$ .