Question

Two cells of the same emf but different internal resistances \( r_1 \) and \( r_2 \) are connected in series with a resistance \( R \). The value of resistance \( R \), for which the potential difference across the second cell is zero, is:

• A. \( r_2 - r_1 \)
• B. \( r_1 - r_2 \)
• C. \( r_1 \)
• D. \( r_2 \)

Hint:

When dealing with cells in series, the total emf is the sum of the individual emfs, and the total internal resistance is the sum of the individual internal resistances. The potential difference across a cell can be zero if the voltage drop across its internal resistance equals its emf.

Answer 1

The Correct Option is A

Step 1: Understanding the Problem
Two cells with the same emf (\( E \)) but different internal resistances (\( r_1 \) and \( r_2 \)) are connected in series with an external resistance \( R \). We need to find the value of \( R \) such that the potential difference across the second cell is zero.
Step 2: Analyzing the Circuit
The total emf in the circuit is \( 2E \) (since the cells are in series).
The total internal resistance is \( r_1 + r_2 \).
The total resistance in the circuit is \( R + r_1 + r_2 \).
The current (\( I \)) in the circuit is given by: \[ I = \frac{2E}{R + r_1 + r_2}. \]
Step 3: Potential Difference Across the Second Cell
The potential difference across the second cell is zero, which means the voltage drop across its internal resistance (\( r_2 \)) equals its emf (\( E \)): \[ E = I \cdot r_2. \]
Substitute the expression for \( I \): \[ E = \frac{2E}{R + r_1 + r_2} \cdot r_2. \]
Simplify the equation: \[ 1 = \frac{2r_2}{R + r_1 + r_2}. \] \[ R + r_1 + r_2 = 2r_2. \] \[ R + r_1 = r_2. \] \[ R = r_2 - r_1. \]
Step 4: Matching with the Options
The value of \( R \) is \( r_2 - r_1 \), which corresponds to option (A). Final Answer: The value of resistance \( R \) is \( r_2 - r_1 \).