Question

A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell. 

Answer 1

In 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 V