Question

Two cells with the same emf $ E $ and different internal resistances $ r_1 $ and $ r_2 $ are connected in series to an external resistance $ R $. If the potential difference across the first cell is zero then the value of $ R $ is

• A. \( \sqrt{r_1 r_2} \)
• B. \( r_1 + r_2 \)
• C. \( r_1 - r_2 \)
• D. \( \frac{r_1 r_2}{2} \)

Hint:

When two cells are connected in series, the condition that the potential difference across one cell is zero helps determine the external resistance required to balance the internal resistances.

Answer 1

The Correct Option is C

When two cells are connected in series, the total emf is the sum of the individual emfs, and the total resistance is the sum of the internal resistances and the external resistance. 
Let the emf of each cell be \( E \), the internal resistances of the two cells be \( r_1 \) and \( r_2 \), and the external resistance be \( R \). The potential difference across the first cell is zero, which means no current flows through the first cell. 
This implies that the potential drop across the internal resistance of the first cell is exactly equal to the emf of the second cell. 
Using Ohm's law and the conditions provided, we get the relationship: \[ R = r_1 - r_2 \] 
Thus, the correct value of \( R \) is: \[ \text{(3) } r_1 - r_2 \]