Question

When a resistance \( R_1 \) is connected across a cell, the current is \( I_1 \) and if the resistance \( R_1 \) is replaced by \( R_2 \), the current is \( I_2 \). Then the internal resistance of the cell is:

• A. \( \frac{I_1 R_1 + I_2 R_2}{I_1 + I_2} \)
• B. \( \frac{I_2 R_2 + I_1 R_1}{I_1 + I_2} \)
• C. \( \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \)
• D. \( \frac{I_1 R_2 - I_1 R_1}{I_1 - I_2} \)

Hint:

When solving for internal resistance, always use Ohm's law and remember the relationship between the potential difference, resistance, and current.

Answer 1

The Correct Option is D

The total resistance of the circuit when \( R_1 \) is connected to the cell is \( R_1 + r \), where \( r \) is the internal resistance of the cell. From Ohm's law, the current is given by: \[ I_1 = \frac{V}{R_1 + r} \] When the resistance is replaced by \( R_2 \), the new current is: \[ I_2 = \frac{V}{R_2 + r} \] Now, solving the above two equations for \( r \), we get: \[ r = \frac{I_1 R_2 - I_1 R_1}{I_1 - I_2} \] Thus, the internal resistance of the cell is \( \frac{I_1 R_2 - I_1 R_1}{I_1 - I_2} \).