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

Question

G1 Physics · Accepted Answer

Let the potential gradient be k and emf be E. With shunt R, terminal voltage is ER/(r+R) = kl. Thus E×5/(r+5) = 200k and E×15/(r+15) = 300k. Dividing the two equations gives [5/(r+5)]/[15/(r+15)] = 2/3. Solving gives r = 5 Ω.