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

In a potentiometer experiment, the electromotive force (EMF) of a cell is directly proportional to the balancing length. When the cell is short-circuited by an external resistance, the potential difference across the external resistance (which is the terminal voltage) is balanced at a shorter length. Let:

• \( E \): EMF of the Daniel cell
• \( V \): Terminal voltage when short-circuited by resistance
• \( L_1 = 240 \, \text{cm} \): Balancing length for EMF
• \( L_2 = 80 \, \text{cm} \): Balancing length for terminal voltage
• \( R = 0.25 \, \Omega \): External resistance
• \( r \): Internal resistance of the Daniel cell

From potentiometer principle: \[ \frac{E}{V} = \frac{L_1}{L_2} \] Also, for a cell with internal resistance \( r \) and external resistance \( R \): \[ V = E \cdot \frac{R}{R + r} \] Substitute into the potentiometer equation: \[ \frac{E}{E \cdot \frac{R}{R + r}} = \frac{L_1}{L_2} \Rightarrow \frac{R + r}{R} = \frac{L_1}{L_2} \] Substitute known values: \[ \frac{0.25 + r}{0.25} = \frac{240}{80} = 3 \] Multiply both sides by 0.25: \[ 0.25 + r = 3 \times 0.25 = 0.75 \] Solve for \( r \): \[ r = 0.75 - 0.25 = 0.5 \, \Omega \] The internal resistance of the Daniel cell is \( \boxed{0.5 \, \Omega} \).