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The example, worked out
Given
Arrangement
R, L and C in series
Resistance R
10 Ω
Inductance L
1 mH
Capacitance C
1 µF
Frequency of interest
5 kHz
Working
the same in series and in parallel
a small R lets it ring
at 5 kHz
Results
Resonance f05.03 kHz
Quality factor Q3.16
Bandwidth f0 / Q1.59 kHz
Damping ζ0.158: under-damped
|Z| at f010 Ω
|Z| at 5 kHz10 Ω
Z0 = √(L/C)31.6 Ω
The formulas
f0 = 1 / (2π √(L C))the same for both
Q = Z0 / R (series), R / Z0 (parallel)the resistor damps one and loads the other
BW = f0 / Qbetween the two half-power points
The usual traps
In series the impedance is lowest at f0; in parallel it is highest.
Q above 0.5 (ζ below 1) means it rings on a step: the overshoot grows fast with Q.
The coil's own resistance counts as part of R.
Questions
How do I calculate the resonance and Q of an RLC circuit?
f0 = 1 / (2π √(L C)): the same for both. Q = Z0 / R (series), R / Z0 (parallel): the resistor damps one and loads the other. BW = f0 / Q: between the two half-power points.
In series the impedance is lowest at f0; in parallel it is highest. Q above 0.5 (ζ below 1) means it rings on a step: the overshoot grows fast with Q. The coil's own resistance counts as part of R.
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Open the calculatorOpens in the app on 1 December, with Circuit Notebook 1.0