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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.
Options
The equivalent emf εeq of the two cells is between ε1 and ε2, i.e. ε1 < εeq < ε2.
The equivalent emf εeq is smaller than ε1.
The εeq is given by εeq = ε1 + ε2 always.
εeq is independent of internal resistances r1 and r2.
Solution
The equivalent emf εeq of the two cells is between ε1 and ε2, i.e. ε1 < εeq < ε2.
Explanation:
The equivalent emf of this combination is given by
εeq = `(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 < εeq < ε2.
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