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 \( \frac{I_1 R_2 - I_1 R_1}{I_1 - I_2} \).

Answer 2

Step 1: Apply Ohm’s law to the circuit.
The total current through a circuit with internal resistance \( r \) and external resistance \( R \) is:
\[ I = \frac{E}{R + r} \] where \( E \) is the emf of the cell.

Step 2: Set up equations for the two cases.
- Case 1: When resistance \( R_1 \) is connected:
\[ I_1 = \frac{E}{R_1 + r} \Rightarrow E = I_1(R_1 + r) \] - Case 2: When resistance \( R_2 \) is connected:
\[ I_2 = \frac{E}{R_2 + r} \Rightarrow E = I_2(R_2 + r) \]

Step 3: Equate the two expressions for \( E \).
\[ I_1(R_1 + r) = I_2(R_2 + r) \]
Expand both sides:
\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]
Bring like terms together:
\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \Rightarrow r(I_1 - I_2) = I_2 R_2 - I_1 R_1 \]
Step 4: Solve for internal resistance \( r \).
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
Step 5: Final form (multiplying numerator and denominator by -1 for alternate format):
\[ r = \frac{I_1 R_2 - I_1 R_1}{I_1 - I_2} \]

Conclusion:
The internal resistance of the cell is \( \boxed{\frac{I_1 R_2 - I_1 R_1}{I_1 - I_2}} \).