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:
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:
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
Discussion & Comments