Question

In a potentiometer experiment, the balancing length for a cell is $240\text{ cm}$. On shunting the cell with a resistance of $2\ \Omega$, the balancing length becomes half the initial balancing length. The internal resistance of the cell is

• A. $1.5\ \Omega$
• B. $1\ \Omega$
• C. $0.5\ \Omega$
• D. $2\ \Omega$

Hint:

Whenever the balancing length drops to exactly half of its initial value after being shunted ($L_2 = \frac{1}{2}L_1$), the internal resistance $r$ of the cell is always exactly equal to the external shunt resistance $R$. You can write down the answer $2\ \Omega$ instantly without doing any calculations!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A potentiometer setup is used to determine the internal resistance of an electrochemical cell. Initially, the open-circuit cell balance length is $L_1 = 240\text{ cm}$. When a shunt resistor $R = 2\ \Omega$ is connected across the cell, the new balancing length drops to half its original value ($L_2 = \frac{L_1}{2} = 120\text{ cm}$). We need to find the internal resistance $r$ of the cell.

Step 2: Key Formula or Approach:
The standard potentiometer formula used to evaluate the internal resistance of a cell is: $$r = R \left( \frac{L_1 - L_2}{L_2} \right)$$ where $L_1$ is the open-circuit balancing length and $L_2$ is the short-circuit (shunted) balancing length.

Step 3: Detailed Explanation:
Let's list the parameter values provided in the problem statement: $$\text{Shunt resistance, } R = 2\ \Omega$$ $$\text{Initial balancing length, } L_1 = 240\text{ cm}$$ $$\text{New balancing length, } L_2 = \frac{240}{2} = 120\text{ cm}$$ Substitute these numbers into the internal resistance formula: $$r = 2 \times \left( \frac{240 - 120}{120} \right)$$ $$r = 2 \times \left( \frac{120}{120} \right)$$ $$r = 2 \times 1 = 2\ \Omega$$

Step 4: Final Answer:
The internal resistance of the cell is $2\ \Omega$, which corresponds to option (D).