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 is ______ Ω .

Step-by-step solution

Let potential gradient be x . For the first case, we can write ε r + 5 × 5 = 200 x ......(1) and after adding resistance, we can write ε r + 15 × 15 = 300 x .....(2) From both equations, we get ⇒ r = 5   Ω