Three resistors of resistances 11 Ω, 22 Ω and 33 Ω are connected in parallel. Their equivalent resistance is equal to
- A.66 Ω
- B.22 Ω
- C.12 Ω
- D.6 Ω
Correct answer
Explanation
The correct answer is D, 6 Ω. For resistors in parallel the reciprocals add: 1/11 + 1/22 + 1/33 = 11/66 = 1/6, so the equivalent resistance is 6 Ω.
Taking 66 as the common denominator, 1/11 becomes 6/66, 1/22 becomes 3/66 and 1/33 becomes 2/66. Their sum is 11/66, which simplifies to 1/6, and the reciprocal of that gives 6 Ω. In a parallel connection each resistor offers an extra path for the current, so the combined resistance is always smaller than the smallest single resistor, here 11 Ω. That quick check alone rules out three of the options. Household wiring is in parallel for this reason, so that each appliance gets the full supply voltage.
A is wrong because 66 Ω is the series total, 11 + 22 + 33. B is wrong because 22 Ω is only the middle resistor's value. C is wrong because 12 Ω is still larger than 11 Ω, which a parallel result can never be.
Exam tip: Series: R = R1 + R2 + R3; parallel: 1/R = 1/R1 + 1/R2 + 1/R3, and the result is below the smallest resistor.
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