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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.
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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.
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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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