A 470 nF capacitor charged to 2 V is switched across an ideal 100 µH inductor. Find the peak current Imax in the circuit, in mA.
LC resonance, Q factor & bandwidth worksheet with answers
1.Easy
C = 470 nFV0 = 2 VL = 100 µH
Imax = mA
2.Easy
A resonance curve peaks at 455 kHz, and its half-power points are 91 kHz apart. Find the quality factor Q.
f0 = 455 kHzBW = 91 kHz
Q =
3.Medium
A series RLC circuit resonates at 500 kHz with Q = 1.5. Find the lower half-power frequency f1 exactly, in kHz.
f0 = 500 kHzQ = 1.5
f1 = kHz
4.Medium
A series RLC circuit has R = 10 Ω, L = 4.7 mH and C = 1 µF. Find its quality factor Q.
R = 10 ΩL = 4.7 mHC = 1 µF
Q =
5.Hard
A series resonant circuit uses a 4.7 mH inductor. Which total series resistance R gives a bandwidth of 5 kHz?
L = 4.7 mHBW = 5 kHz
R = Ω
6.Hard
A series RLC circuit (R = 2 Ω, L = 22 µH, C = 47 nF) is driven at resonance with a current amplitude of 100 mA. Find the energy Ecycle dissipated in R during one cycle, in nJ.
R = 2 ΩL = 22 µHC = 47 nFIm = 100 mA
Ecycle = nJ
Answer key
- 1. Imax = 137.1 mA
- 2. Q = 5
- 3. f1 = 360.4 kHz
- 4. Q = 6.856
- 5. R = 147.7 Ω
- 6. Ecycle = 63.89 nJ
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