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

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प्रश्न

Two batteries of emf ε1 and ε22 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure.

विकल्प

  • 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.

MCQ
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उत्तर

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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अध्याय 3: Current Electricity - MCQ I [पृष्ठ १६]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 3 Current Electricity
MCQ I | Q 3.02 | पृष्ठ १६

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