A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5Ω. When a resistance of 15Ω is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________Ω.

Solution

Physics sir · 4 steps

The shunted cell sends current, so the null point measures ER/(R+r), not the emf.

1. What is balanced. A shunt R across the cell lets current flow, so the potentiometer balances the terminal voltage V = ER/(R+r), not the emf. The balancing length is proportional to V.

   V₁/V₂ = (5/(5+r))/(15/(15+r)) = l₁/l₂ = 200/300

2. Simplify. The left side becomes 5(15+r)/15(5+r) = (15+r)/3(5+r). Setting this equal to 2/3 gives 15 + r = 10 + 2r, so r = 5 Ω.
3. Check. With r = 5 Ω, V₁ = 5E/10 = E/2 and V₂ = 15E/20 = 3E/4. Their ratio is 2/3, which matches 200 : 300.
4. So. The internal resistance of the cell is r = 5 Ω, so the answer is 5.

The idea to keep. When a cell in a potentiometer circuit is shunted, the null point measures its terminal voltage ER/(R+r). A bigger shunt gives a bigger voltage and a longer balancing length.

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