Question:

Two cells of the same emf but different internal resistances r1 r_1 and r2 r_2 are connected in series with a resistance R R . The value of resistance R R , for which the potential difference across the second cell is zero, is:

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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.
Updated On: Mar 29, 2025
  • r2r1 r_2 - r_1
  • r1r2 r_1 - r_2
  • r1 r_1
  • r2 r_2
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Problem
Two cells with the same emf (E E ) but different internal resistances (r1 r_1 and r2 r_2 ) are connected in series with an external resistance R R . We need to find the value of R 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 2E (since the cells are in series).
The total internal resistance is r1+r2 r_1 + r_2 .
The total resistance in the circuit is R+r1+r2 R + r_1 + r_2 .
The current (I I ) in the circuit is given by: I=2ER+r1+r2. 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 (r2 r_2 ) equals its emf (E E ): E=Ir2. E = I \cdot r_2.
Substitute the expression for I I : E=2ER+r1+r2r2. E = \frac{2E}{R + r_1 + r_2} \cdot r_2.
Simplify the equation: 1=2r2R+r1+r2. 1 = \frac{2r_2}{R + r_1 + r_2}. R+r1+r2=2r2. R + r_1 + r_2 = 2r_2. R+r1=r2. R + r_1 = r_2. R=r2r1. R = r_2 - r_1.
Step 4: Matching with the Options

The value of R R is r2r1 r_2 - r_1 , which corresponds to option (A). Final Answer: The value of resistance R R is r2r1 r_2 - r_1 .
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