Question

Two cells of emfs 12 V and 6 V are connected in parallel as shown in the figure. Their internal resistances are 1 Ω and 0.5 Ω respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.

Answer 1

Given: ε₁ = 12 V,

r₁ = 1 Ω,

ε₂ = 6 V,

r₂ = 0.5 Ω

Internal Resistance (r_eq):

`1/r_(eq) = 1/r_1 + 1/r_2`

= `1/1 + 1/0.5`

= 1 + 2

`1/r_(eq)` = 3

⇒ r_eq = `1/3` Ω

≈ 0.33 Ω

Equivalent EMF (ε_eq) = `(epsilon_1/r_1 + epsilon_2/r_2)/(1/r_1 + 1/r_2)`

= `r_(eq)[epsilon_1/r_1 + epsilon_2/r_2]`

= `1/3[12/1 + 6/0.5]`

= `1/3[12 + 12]`

= `24/3`

= 8 V

∴ The equivalent emf is 8 V, and the internal resistance is `1/3` Ω.