Question

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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