A resonance curve peaks at 500 kHz, and its half-power points are 41.67 kHz apart. Find the quality factor Q.
LC resonance, Q factor & bandwidth worksheet with answers
1.Easy
f0 = 500 kHzBW = 41.67 kHz
Q =
2.Easy
A 47 nF capacitor charged to 6 V is switched across an ideal 4.7 mH inductor. Find the peak current Imax in the circuit, in mA.
C = 47 nFV0 = 6 VL = 4.7 mH
Imax = mA
3.Medium
A series RLC circuit has R = 220 Ω, L = 2.2 mH and C = 10 nF. Find its bandwidth BW in kHz.
R = 220 ΩL = 2.2 mHC = 10 nF
BW = kHz
4.Medium
A series RLC circuit has R = 100 Ω, L = 100 µH and C = 2.2 nF. Find its bandwidth BW in kHz.
R = 100 ΩL = 100 µHC = 2.2 nF
BW = kHz
5.Hard
The half-power frequencies of a resonance curve are 2 kHz and 2.4 kHz. Find the resonance frequency as their geometric mean, and then the quality factor Q.
f1 = 2 kHzf2 = 2.4 kHz
Q =
6.Hard
A series resonant circuit uses a 1 mH inductor. Which total series resistance R gives a bandwidth of 1 kHz?
L = 1 mHBW = 1 kHz
R = Ω
Answer key
- 1. Q = 12
- 2. Imax = 18.97 mA
- 3. BW = 15.92 kHz
- 4. BW = 159.2 kHz
- 5. Q = 5.477
- 6. R = 6.283 Ω
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