For an inductor of fixed inductance L, what happens to the inductive reactance when the operating frequency is doubled (f → 2f)?
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Ahas no effect on the inductive reactance
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Bdoubles the amount of inductive reactance
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Cmultiplies the inductive reactance by 6.28
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Dcuts the inductive reactance in half
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ENone of the above
Answer
Correct Answer: doubles the amount of inductive reactance
Explanation
Introduction / Context:Inductive reactance governs how strongly an inductor impedes AC. Designers frequently assess how changes in frequency alter circuit impedances for filters and resonant networks.
Given Data / Assumptions:
- Ideal inductor with inductance L.
- Steady-state sinusoidal operation.
- Frequency doubled from f to 2f.
Concept / Approach:Inductive reactance is X_L = 2 * pi * f * L. If frequency doubles, X_L(2f) = 2 * pi * (2f) * L = 2 * (2 * pi * f * L) = 2 * X_L(f). Thus, reactance doubles when frequency doubles.
Step-by-Step Solution:
1) Start with X_L = 2 * pi * f * L.2) Replace f with 2f.3) Simplify to obtain 2 * X_L.Verification / Alternative check:Numerical example: If X_L = 100 Ω at f, then at 2f it becomes 200 Ω, confirming the doubling behavior.
Why Other Options Are Wrong:
No effect / half: Contradict linear proportionality to f.Multiply by 6.28: 2 * pi is the proportionality constant, not the change factor for doubling f.None of the above: Not applicable because doubling is correct.Common Pitfalls:Mixing inductive with capacitive reactance (which halves when frequency doubles).
Final Answer:doubles the amount of inductive reactance.