Question

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

Answer 1

(i) The effective resistance of a circuit with two 20-ohm resistors linked in parallel is R is `1/"R" = 1/2 + 1/2`

∴ R = 1Ω

'r' stands for the internal resistance of the cell that is in series with the circuit.

∴ V = IR

or V = 1.2 (1 + r)  ...(i)

Connected in series, the effective resistance of the two resistors and the cell's internal resistance, denoted as 'r'

R = 2 + 2 + r

= (4 + r) Ω  ...(ii)

V = IR

= 0.4 (4 + R)

(i) = (ii)

1.2 (1 + r) = 0.4 (4 + r)

1.2 + 1.2 r = 1.6 + 0.4 r

or (1.2 - 0.4) r = (1.6 - 1.2)

0.8 r = 0.4

r = `0.4/0.8`

= `1/2`

= 0.5 Ω

put r = 0.5 in (i)

(ii) e.m.f. of cell E = V = 1.2 (1 + 0.5) = 1.8 V