Question

Derive the formula for the determination of internal resistance of a cell with the help of a potentiometer.

Hint:

Using a potentiometer avoids error due to current draw, as it measures emf and potential difference by null deflection method.

Answer 1

Step 1: Principle of potentiometer.
The potential difference across any length of the potentiometer wire is directly proportional to its balancing length. \[ V \propto l \]
Step 2: First observation (no external resistance).
- The cell is connected directly to the potentiometer. - Let the balancing length be $l_1$. - Then the emf of the cell is: \[ E \propto l_1. \]
Step 3: Second observation (with external resistance R).
- Now the cell is connected across a resistance $R$. - The potential difference across the cell (terminal voltage) is: \[ V = \frac{E R}{R + r}. \] - Let the new balancing length be $l_2$. - Then, \[ V \propto l_2. \]
Step 4: Ratio of lengths.
\[ \frac{E}{V} = \frac{l_1}{l_2}. \] Substitute $V = \dfrac{E R}{R + r}$: \[ \frac{E}{\dfrac{E R}{R + r}} = \frac{l_1}{l_2}. \] \[ \frac{R + r}{R} = \frac{l_1}{l_2}. \]
Step 5: Solve for $r$.
\[ R + r = R \cdot \frac{l_1}{l_2}. \] \[ r = R \left( \frac{l_1}{l_2} - 1 \right). \] \[ r = R \cdot \frac{l_1 - l_2}{l_2}. \]
Step 6: Conclusion.
The internal resistance of the cell is: \[ r = R \cdot \frac{l_1 - l_2}{l_2}. \]