Difficulty: Medium
Correct Answer: 147.5 kHz to 152.5 kHz
Explanation:
Introduction:
For narrowband resonant RLC networks, bandwidth and quality factor are tightly related. This problem tests your ability to convert a given resonant frequency and Q into the half-power (3 dB) frequencies f1 and f2 that bound the passband where current remains at least 70.7% of its peak (series) or where impedance criteria are met (parallel).
Given Data / Assumptions:
Concept / Approach:
The fundamental relation is BW = f0 / Q. Once BW is known, the half-power frequencies are centered about f0 so that f1 = f0 - BW/2 and f2 = f0 + BW/2. This symmetry is an accurate approximation for practical Q values of 10 or more.
Step-by-Step Solution:
1) Compute bandwidth: BW = f0 / Q = 150 kHz / 30 = 5 kHz.2) Split BW evenly around f0: BW/2 = 2.5 kHz.3) Lower half-power corner: f1 = 150 kHz - 2.5 kHz = 147.5 kHz.4) Upper half-power corner: f2 = 150 kHz + 2.5 kHz = 152.5 kHz.
Verification / Alternative check:
For Q = 30, relative bandwidth BW/f0 = 1/30 ≈ 3.33%. This is small enough that the symmetric approximation about f0 is valid, reinforcing the computed f1 and f2 values.
Why Other Options Are Wrong:
100.0 kHz to 155.0 kHz: Bandwidth is far larger than f0/Q; does not straddle f0 symmetrically.
4500 kHz to 295.5 kHz: Nonsensical range; not centered on 150 kHz.
149,970 Hz to 150,030 Hz: Implies BW = 60 Hz, which would correspond to Q ≈ 2500, not 30.
140 kHz to 160 kHz: BW = 20 kHz, four times the correct 5 kHz.
Common Pitfalls:
Forgetting BW = f0 / Q, mis-centering the range about f0, or confusing series versus parallel definitions. At moderate-to-high Q, the same BW formula applies and the 70.7% criterion leads to the quoted half-power points.
Final Answer:
147.5 kHz to 152.5 kHz
Discussion & Comments