Question

A cell supplies a current of 1.2a through two resistors each of 2 ohm connected in parallel. When the resistors are connected in series it supplies a current of 0.4a

Answer 1

parallel R = ½ + ½ = 1 ohm I = 1.2 A ε = I(R + r) = 1.2(1 + r) = 1.2 + 1.2 r In series R = 2+2 = 4 ohm I = 0.4 A ε = I(R + r) = 0.4(4 + r) = 1.6 + 0.4 r It means : 1.2 + 1.2 r = 1.6 + 0.4 r 0.8 r = 0.4 r = 0.4 / 0.8 = ½ = 0.5 ohm (i) Internal resistance r = 0.5 ohm (ii) ε = I(R+r) = 1.2(1+0.5) = 1.8. ANSWER