When an external resistance of 5 is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 resistor is replaced by a 2 resistor, a current of 0.5 A flows through it. The internal resistance of the cell is _______ .

Step-by-step solution

Let the EMF of the cell be E and its internal resistance be r . The current I in the circuit is given by I = E R + r . For the first case, R₁ = 5 and I₁ = 0.25 A : 0.25 = E 5 + r E = 0.25(5 + r) For the second case, R₂ = 2 and I₂ = 0.5 A : 0.5 = E 2 + r E = 0.5(2 + r) Equating the expressions for E : 0.25(5 + r) = 0.5(2 + r) 5 + r = 2(2 + r) 5 + r = 4 + 2r r = 1 Answer: 1