The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the ceil is N/10 Ω , where N is an integer then value of N is .............

Solution

Physics sir · 4 steps

The shunt draws current from the cell, so the balance point falls: use 560 cm and 500 cm.

1. Which way it moves. Without the 10 Ω resistor no current flows through the cell at balance, so the potentiometer reads the emf E. With the resistor across it, current flows through the cell. The potentiometer now reads the terminal voltage V = ER/(R+r), which is less than E, so the balancing length drops from 560 cm to 560 - 60 = 500 cm.
2. Ratio of lengths. Potential difference is proportional to balancing length, so E/V = l₁/l₂. Since E/V = (R+r)/R, this gives the formula for internal resistance.

   r = R(l₁/l₂ - 1) = 10(560/500 - 1)

3. Put in numbers. 560/500 = 1.12, so r = 10 × 0.12 = 1.2 Ω = 12/10 Ω.
4. So. N = 12.

The idea to keep. A resistor across the cell always lowers the balancing length, since you now measure terminal voltage, not emf. Then r = R(l₁/l₂ - 1) with l₁ the longer length.

Answer from the NTA final answer key. Solution by PrepWiser.