A resonant circuit has lower critical frequency f1 = 7 kHz and upper critical frequency f2 = 13 kHz. What is the bandwidth (BW) of the circuit?

Difficulty: Easy

Correct Answer: 6 kHz

Explanation:


Introduction / Context:
Bandwidth measures the span between the two half-power (−3 dB) frequencies of a resonant network. It captures how wide the useful passband is for filters and tuned circuits, independent of the absolute center frequency.


Given Data / Assumptions:

  • Lower critical frequency f1 = 7 kHz.
  • Upper critical frequency f2 = 13 kHz.
  • Standard definition BW = f2 − f1.


Concept / Approach:

By definition, bandwidth is the simple arithmetic difference between the upper and lower cutoff frequencies. No further computation is needed unless a quality factor or center frequency is required.


Step-by-Step Solution:

BW = f2 − f1.Substitute: BW = 13 kHz − 7 kHz.Compute: BW = 6 kHz.


Verification / Alternative check:

Center frequency f0 could be estimated as √(f1 * f2) ≈ √(7 * 13) kHz ≈ 9.54 kHz, and Q = f0 / BW ≈ 9.54 / 6 ≈ 1.59 (reasonable), reinforcing that the bandwidth arithmetic is correct.


Why Other Options Are Wrong:

13 kHz and 7 kHz are the individual cutoff frequencies, not the span between them. 20 kHz adds instead of subtracting, which is not the definition of bandwidth.


Common Pitfalls:

Confusing center frequency, arithmetic mean, or geometric mean with bandwidth; subtracting in the wrong order.


Final Answer:

6 kHz

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