The balance point in a potentiometer for a cell of emf 1.5 V is 60
cm. If a resistance of 5 Ω is connected in parallel with the cell and the new balance point shifts to 50 cm, what is the internal resistance of the cell?

Let E be the emf of the cell and r be its internal resistance. In the first case, the balance point is at l ₁ ​= 60 cm. The potential difference across the balance point is proportional to the emf of the cell: E ∝ l ₁ E = kl­ ₁ , where k is the potential gradient. 1.5 = k × 60 k = 1.5/60 = 1/40 …(1) In the second case, a resistance R=5Ω is connected in parallel with the cell. The effective voltage across the potentiometer wire is now the terminal voltage V of the cell. The internal resistance of the cell is 1Ω.