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