Question

Four cells each of emf 1.5 V and internal resistance
2.0 ohm are connected in parallel. The battery of cells is connected to an external resistance of 2.5 ohm.

Calculate :

(i) the total resistance of the circuit

(ii) the current flowing in the external circuit and

(iii) the drop in potential across the terminals of the cells.

Answer 1

i) As cells are in parallel, the battery has net emf of 1.5 V and its net internal resistance, r is given by

1/r = ½ + ½ + ½ + ½ = 2 => r = 0.5 ohm.

(ii) total resistance, R = 0.5 + 2.5 = 3 ohm

(iii) current flowing in the external circuit is

I = 1.5 / R = 1.5 / 3.0 = 0.5 A

(iv) The potential difference across the terminals of the cells

= the potential difference across the external resistance connected to the battery.

Hence pd = V = 0.5 A × 2.5 Ω = 1.25 V.