Difficulty: Easy
Correct Answer: Incorrect — increasing resistance lowers Q
Explanation:
Introduction / Context:The quality factor Q expresses how under-damped or selective a resonant circuit is. For a series RLC at resonance, Q relates the reactive energy exchange to real power dissipation. Coil resistance is a major contributor to series loss, so its effect on Q is fundamental to RF design and filters.
Given Data / Assumptions:
Concept / Approach:For series resonance, Q = X_L / R at f0 (equivalently Q = ω0L / R). Increasing R in the denominator reduces Q. Lower Q widens the bandwidth (BW ≈ f0/Q) and reduces peak voltage magnification across L and C at resonance.
Step-by-Step Solution:
Start with Q = ω0L / R.Increase R → denominator increases → Q decreases.Result: broader bandwidth, lower selectivity, lower tank magnification.Verification / Alternative check:Network analyzer measurements show increased 3 dB bandwidth and lower peak response as series resistance rises. Simulation likewise confirms Q ∝ 1/R for fixed L, C at f0.
Why Other Options Are Wrong:
Claims that higher R raises Q contradict the formula.Frequency or ESR caveats do not reverse the basic series relation Q = X_L/R at resonance.Common Pitfalls:Mixing up series and parallel definitions of Q; overlooking that parallel-resonant “Rp” enters Q_p = Rp / X_L, a different configuration.
Final Answer:Incorrect — increasing resistance lowers Q
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