Two 1.2 kΩ resistors are in series and this series combination is in parallel with a 3.3 kΩ resistor. The total resistance is____________?
Correct answer: B. 1,389 Ω
- A. 138 Ω
- B. 1,389 Ω
- C. 5,700 Ω
- D. 880 Ω
Explanation
The series pair is 1.2 kΩ + 1.2 kΩ = 2.4 kΩ. In parallel with 3.3 kΩ, the total is (2.4 × 3.3)/(2.4 + 3.3) kΩ ≈ 1.389 kΩ, or 1,389 Ω.
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