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NEET 1990 (AIPMT) · Combination of Resistors

Combination of Resistors question from NEET 1990 (AIPMT)

Q86NEET 1990 · AIPMTMedium

Combination of Resistors

You are given several identical resistances each of value R=10 ΩR = 10\ \Omega and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5 Ω5\ \Omega which can carry a current of 4 ampere. The minimum number of resistances of the type RR that will be required for this job is

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Answer: (c)

10 Ω10 Ω4 branches × 20 Ω → 5 Ω, 1 A each4 A
  1. Each resistor carries at most 1 A, so 4 A needs 4 parallel branches carrying 1 A each.
  2. 4 equal branches give 5 Ω5\ \Omega if each branch is 4×5=20 Ω4 \times 5 = 20\ \Omega, i.e. two 10 Ω10\ \Omega in series.
  3. Number of resistors =4×2=8= 4 \times 2 = 8
  4. So the answer is (c) 8.
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