Question

A Daniel cell is balanced at 240 cm in length of a potentiometer. Now the cell is short-circuited by a resistance of 0.25 ohm and the balance is obtained at 80 cm. The internal resistance of the Daniel cell is

• A. \( \text{0.5 ohm} \)
• B. \( \text{2 ohms} \)
• C. \( \text{0.25 ohm} \)
• D. \( \text{1 ohm} \)

Hint:

For potentiometer problems involving internal resistance, remember the key relationship: \( \frac{E}{V} = \frac{L_1}{L_2} \), where \( E \) is EMF, \( V \) is terminal voltage, and \( L_1, L_2 \) are the corresponding balancing lengths. Also, use the formula \( V = E \cdot \frac{R}{R + r} \). Combining these gives \( \frac{R + r}{R} = \frac{L_1}{L_2} \), which is a direct way to find the internal resistance.

Answer 1

The Correct Option is A

To solve this problem, let's explore the working principle of a Daniel cell and the use of a potentiometer to measure the internal resistance.

1. What is a Daniel Cell?

A Daniel cell is a type of electrochemical cell consisting of a copper half-cell and a zinc half-cell. The cell provides a constant voltage, which can be measured using a potentiometer. The internal resistance of the cell causes a potential drop inside the cell when it is connected to an external circuit.

2. What is a Potentiometer?

A potentiometer is an instrument used to measure the potential difference (voltage) across two points in a circuit without drawing any current. It works by balancing the potential drop across a known length of wire and comparing it with the potential drop across the unknown voltage.

3. Understanding the Problem:

- Initially, the cell is balanced at 240 cm, meaning the potential drop across the potentiometer wire is proportional to the EMF of the Daniel cell. - When the cell is short-circuited by a resistance of \( 0.25 \, \Omega \), the balance is obtained at 80 cm, indicating a change in the potential drop due to the internal resistance of the cell.

4. Formula for Internal Resistance of the Daniel Cell:

The relationship between the EMF of the cell, the internal resistance, and the external resistance can be used to find the internal resistance. Let \( E \) be the EMF of the Daniel cell, \( r \) be the internal resistance, and \( I \) be the current. Initially, without any external resistance, the balance length is 240 cm, corresponding to the EMF \( E \). When the cell is short-circuited with the external resistance \( R = 0.25 \, \Omega \), the balance length decreases to 80 cm. Using the formula for the potential difference in terms of the EMF and the internal resistance, we can set up the ratio: \[ \frac{\text{Initial length}}{\text{Final length}} = \frac{E}{E - I \cdot r} \] Substitute the values for the lengths: \[ \frac{240}{80} = \frac{E}{E - I \cdot r} \] Now, we know that the current \( I \) is given by: \[ I = \frac{E}{R + r} \] Substituting \( R = 0.25 \, \Omega \) into the equation and solving for \( r \), we find that the internal resistance of the Daniel cell is \( 0.5 \, \Omega \).

5. Final Answer:

The internal resistance of the Daniel cell is 0.5 ohm.