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

Best Answer

In parallel R = 1 / 2 + 1 / 2 = 1 ohm

I = 1.2 A

ε = I (R + r)

ε = 1.2 (1 + r)

ε = 1.2 + 1.2r

In series

R = 2 + 2

R = 4 ohm

I = 0.4 A

ε = I (R + r)

ε = 0.4 (4 + r)

ε = 1.6 + 0.4r

This means:

1.2 + 1.2r = 1.6 + 0.4r

1.2r – 0.4r = 1.6 – 1.2

0.8r = 0.4

r = 0.4 / 0.8

r = 0.5 ohm

(i) Internal resistance r = 0.5 ohm

(ii) ε = I(R + r)

ε = 1.2 (1 + 0.5)

ε = 1.8 V