Question

The internal resistance of a cell is to be determined using a potentiometer. In an experiment, an external resistance of $\text{R}$ $=$ $60\Omega$ is used across the given cell. When the key $K_1$ is closed and $K_2$ is open, the balance point is found at a length of $72$ $\text{cm}$. The balance length on the potentiometer is found at $60$ $\text{cm}$ when both the keys are closed. Calculate the internal resistance of the cell.

• A. $3\Omega$
• B. $6\Omega$
• C. $8\Omega$
• D. $12\Omega$

Answer 1

The correct option is D $12\Omega$

Given: $R=60\Omega$

Case $1$ : when key $K_1$ is closed and key $K_2$ is open. Then null point is obtained at at $l_1=72\text{cm}$.

Case $2$ : when both key are closed, then null point is at $l_2=60\text{cm}$.

From formula, internal resistance $r$ will be

$r=R(\frac{l_1}{l_2}-1)$

Here, $l_1=72\text{cm};l_2=60\text{cm}$

$r=\left(60\right)(\frac{72}{60}-1)$

$\therefore r=12\Omega$

Hence, option (d) is the correct answer.