Which combination of three resistors R1, R2 and R3 has minimum total resistance? (Consider R1 = 1 Ω, R2 = 2 Ω and R3 = 3 Ω)
- A.When all are connected in parallel
- B.When all are connected in series
- C.When R1 and R2 are connected in parallel and R3 is in series of them
- D.When R1 and R3 are connected in parallel and R2 is in series of them
Correct answer
Explanation
The correct answer is A, When all are connected in parallel. In parallel the equivalent resistance is always smaller than the smallest single resistor, so here it falls below 1 Ω.
For resistors in parallel, 1/R = 1/R1 + 1/R2 + 1/R3 = 1/1 + 1/2 + 1/3 = 11/6, so R = 6/11 Ω, about 0.55 Ω. Each extra parallel path gives the current another route, which lowers the total resistance. For resistors in series the values simply add, so the total is always larger than the largest one.
B is wrong because in series R = 1 + 2 + 3 = 6 Ω, the maximum possible. C is wrong because R1 and R2 in parallel give 2/3 Ω, and adding R3 in series makes 3.67 Ω. D is wrong because R1 and R3 in parallel give 3/4 Ω, and adding R2 in series makes 2.75 Ω.
Exam tip: series = maximum resistance, larger than the largest; parallel = minimum resistance, smaller than the smallest.
Practice Questions
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- A.Coffee < Lemon juice < Milk of magnesia < Blood
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