From the combination of resistors with resistances values R₁=R₂=R₃=5 Ω and R₄=10 Ω, which of the following combination is the best circuit to get an equivalent resistance of 6 Ω ?

Solution

Physics sir · 2 steps

Since 6 is more than any single 5, look for two series branches in parallel: 10 with 15 gives 6.

1. Two rules. In series, resistances add. Two branches in parallel combine as (Rₐ R_b)/(Rₐ + R_b), which is always smaller than the smaller branch.

   Option	Arrangement	Equivalent resistance
   (a)	all four in parallel	1/5+1/5+1/5+1/10 = 7/10, so 10/7 ≈ 1.43 Ω
   (b)	(R₁ + R₂) parallel to (R₃ + R₄)	10 ∥ 15 = 150/25 = 6 Ω
   (c)	R₁, (R₂ + R₃) and R₄ in parallel	1/5+1/10+1/10 = 4/10, so 2.5 Ω
   (d)	(R₁ + R₂ + R₄) parallel to R₃	20 ∥ 5 = 100/25 = 4 Ω

   R_(eq) = (5+5)(5+10)/((5+5)+(5+10)) = 150/25 = 6 Ω

2. So. Only circuit (b) gives exactly 6 Ω, so the answer is (b).

The idea to keep. A parallel combination is always below its smallest branch. To get a target value, first build series branches that sit above it, then put them in parallel and check with product over sum.

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