Inductive reactance of a series–parallel inductor network: Two 10 H inductors are connected in parallel, and this parallel combination is then in series with a third 10 H inductor. What is the approximate total reactance at 7 kHz?
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A219 kΩ
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B66 kΩ
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C660 kΩ
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D1.3 MΩ
Answer
Correct Answer: 660 kΩ
Explanation
Introduction / Context:Series–parallel combinations of inductors reduce to an equivalent inductance that determines the overall reactance X_L = 2π f L_eq. Accurate reduction and unit conversion are core skills for AC network analysis at audio and RF frequencies.
Given Data / Assumptions:
- L1 = 10 H in parallel with L2 = 10 H.
- That parallel pair is in series with L3 = 10 H.
- Frequency f = 7 kHz; ideal inductors (no resistance considered).
Concept / Approach:Two equal inductors in parallel produce L_par = L/2. Then add the series inductor to get L_eq. Finally compute X_L = 2π f L_eq and convert to kilohms for comparison to options.
Step-by-Step Solution:
Parallel of two 10 H: L_par = (10 * 10)/(10 + 10) = 5 H.Equivalent inductance: L_eq = L_par + 10 H = 5 + 10 = 15 H.Reactance: X_L = 2π f L_eq = 2π * 7000 * 15 ≈ 659,734 Ω ≈ 660 kΩ.Verification / Alternative check:Reasonableness: A very large inductance at several kilohertz should yield hundreds of kilohms of reactance; the computed value aligns with that intuition.
Why Other Options Are Wrong:
- 66 kΩ: A factor-of-10 error from dropping a zero in the frequency or inductance.
- 219 kΩ or 1.3 MΩ: Do not match 2πfL with L_eq = 15 H at 7 kHz.
Common Pitfalls:
- Mistaking parallel of equal inductors as 2L (it is L/2).
- Dropping units when converting to kilohms.
Final Answer:660 kΩ