What will be the most suitable combination of three resistors A = 2Ω, B = 4Ω, C = 6Ω so that ( 22/3 )Ω is equivalent resistance of combination?

Solution

Physics sir · 2 steps

22/3 Ω is more than 6 Ω, so a parallel pair must sit in series with one more resistor.

1. Quick filter. A parallel combination is always smaller than its smallest branch. In (c) and (d) the whole circuit ends in a parallel step, so the result stays below 4 Ω and 2 Ω. That cannot reach 22/3 ≈ 7.33 Ω.

   R_∥ = (R₁ R₂)/(R₁ + R₂), Rₛₑᵣᵢₑₛ = R₁ + R₂

   Option	Working	Equivalent resistance
   (a) (A ∥ C) + B	(2 × 6)/8 + 4 = 1.5 + 4	5.5 Ω
   (b) (A ∥ B) + C	(2 × 4)/6 + 6 = 4/3 + 6	22/3 Ω
   (c) (A + C) ∥ B	(8 × 4)/(8 + 4)	8/3 Ω
   (d) (B + C) ∥ A	(10 × 2)/(10 + 2)	5/3 Ω

2. So. Only (b) gives 22/3 Ω: A and B in parallel give 4/3 Ω, and adding C = 6 Ω in series gives 22/3 Ω.

The idea to keep. Before calculating, compare the target with the given resistors. Parallel always gives less than the smallest branch and series always gives more than the largest, which rules options out fast.

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