Given below are two statements :
Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.
Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.
In the light of the above statements, choose the correct answer from the options given below.

Solution

Physics sir · 4 steps

In parallel the emfs get averaged, not shrunk. Only the internal resistance drops below both values.

1. Equivalent emf. For two cells (E₁, r₁) and (E₂, r₂) in parallel, the combination behaves like one cell with emf E_(eq) and internal resistance r_(eq):

   E_(eq) = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂), r_(eq) = (r₁ r₂)/(r₁ + r₂)

2. Statement-I. E_(eq) is a weighted average of E₁ and E₂, with positive weights 1/r₁ and 1/r₂. So it always lies between the two emfs. For example, E₁ = 2 V and E₂ = 4 V with r₁ = r₂ = 1 Ω give E_(eq) = 3 V, which is larger than 2 V. Statement-I is false.
3. Statement-II. Write r_(eq) = r₁ × r₂/(r₁ + r₂). The fraction is less than 1, so r_(eq) < r₁. In the same way r_(eq) < r₂. Statement-II is true.
4. So. Statement-I is false and Statement-II is true, which is option (a).

The idea to keep. Cells in parallel: the equivalent emf is a weighted average, so it always sits between the individual emfs. The internal resistances combine like parallel resistors, so the result is smaller than each.

Answer from the NTA final answer key. Solution by PrepWiser.