Question

Two batteries of emf ε₁ and ε₂ (ε₂ > ε₁) and internal resistances r₁ and r₂ respectively are connected in parallel as shown in figure.

पर्याय

• The equivalent emf εeq of the two cells is between ε₁ and ε₂, i.e. ε₁ < ε_eq < ε₂.

• The equivalent emf ε_eq is smaller than ε₁.

• The ε_eq is given by ε_eq = ε₁ + ε₂ always.

• ε_eq is independent of internal resistances r₁ and r₂.

Answer 1

The equivalent emf εeq of the two cells is between ε₁ and ε₂, i.e. ε₁ < ε_eq < ε₂.

Explanation:

The equivalent emf of this combination is given by

ε_eq = `(ε_1/r_1 + ε_1/r_2)/((1/r_1 + 1/r_2)) = (ε_1(1/r_1 + (ε_2/ε_1)/r_2))/((1/r_1 + 1/r_2)) = (ε_2((ε_1/ε_2)/r_1 + 1/r_2))/((1/r_1 + 1/r_2))`

As `ε_2/ε_1 > 1`

⇒ `((1/r_1 + (ε_2/r_1)/r_2))/((1/r_1 + 1/r_2)) > 1 or ε_(eq) > ε_1` also `ε_1/ε_2 < 1`

⇒ `(((ε_1/ε_2)/r_1 + 1/r_2))/((1/r_1 + 1/r_2)) < 1 or ε_(eq) < ε_1`

Hence ε₁ < ε_eq < ε₂.