Which combinations of ten identical resistors each of 2 Ω gives rise to the resultant of 10.4 Ω?
Correct answer: A. 5 Parallel, 5 Series
- A. 5 Parallel, 5 Series
- B. 4 Parallel, 3 Series
- C. 3 Parallel, 4 Series
- D. 2 Parallel, 5 Series
Explanation
Use Ohm's Law to add parallel resistors. The equation is 1/RTotal = 1/R1 + 1/R2 + 1/R3 + ... + 1/RLast. Hence 5 X ½ = 5/2. Total Resistance through parallel is ⅖. When resistors are in series their individual resistance is added to get the total resistance. So 5 X 2= 10 Ω. Adding this gives 10.4 Ω.
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