Question

Two batteries of emfs 3 V & 6 V and internal resistances 0.2 Ω & 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find:

1. the equivalent emf of the combination
2. the equivalent internal resistance of the combination
3. the current drawn from the combination

Answer 1

Given: For battery 1:

EMF (E₁) = 3 V

Internal resistance (r₁) = 0.2 Ω

For battery 2:

EMF (E₂) = 6 V

Internal resistance (r₂) = 0.4 Ω

External resistor (R) = 4 Ω

i. E_eq = `(E_1r_2 + E_2r_1)/(r_1 + r_2)`

= `(3 xx 0.4 + 6 xx 0.2)/(0.4 + 0.2)`

= `(1.2 + 1.2)/0.6`

= `2.4/0.6`

= 4 V

ii. r_eq = `(r_1r_2)/(r_1 + r_2)`

= `(0.4 xx 0.2)/(0.4 + 0.2)`

= `0.08/0.60`

= `2/15` Ω

= 0.133 Ω

iii. I = `E_(eq)/(r_(eq) + R)`

= `4/(0.133 + 4)`

= 0.968 A