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: Problem Statement
Two cells with identical emf (\( E \)) but differing internal resistances (\( r_1 \) and \( r_2 \)) are connected in series with an external resistance \( R \). Determine \( R \) such that the potential difference across the second cell is zero.
Step 2: Circuit Analysis
The total emf in the series combination is \( 2E \).
The total internal resistance is \( r_1 + r_2 \).
The total circuit resistance is \( R + r_1 + r_2 \).
The current (\( I \)) flowing through the circuit is calculated as:\[I = \frac{2E}{R + r_1 + r_2}.\]
Step 3: Potential Difference Across Second Cell
A zero potential difference across the second cell implies that the voltage drop across its internal resistance (\( r_2 \)) is equal to its emf (\( E \)):\[E = I \cdot r_2.\]
Substituting the expression for \( I \):\[E = \frac{2E}{R + r_1 + r_2} \cdot r_2.\]
Simplifying the equation yields:\[1 = \frac{2r_2}{R + r_1 + r_2}.\]\[R + r_1 + r_2 = 2r_2.\]\[R + r_1 = r_2.\]\[R = r_2 - r_1.\]
Step 4: Solution Verification
The calculated value of \( R \) is \( r_2 - r_1 \), which matches option (A).Final Answer: The required value of resistance \( R \) is \( r_2 - r_1 \).