The Series RLC Circuit — Question 5

Section 31-8 · Physics pages 5120–5121

Question 5 Bandwidth measurements determine inductance and capacitance A series RLC circuit with R=9.00 ohms has half-power angular frequencies 400 and 625 radians per second. ∼ vs(t) R C L q i = ˙q Schematic; labels, not drawing lengths, specify the data. (a) Infer L, C, and the resonance angular frequency. (b) Find its quality factor and distinguish arithmetic from geometric center.
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Question 5 - Solution Bandwidth measurements determine inductance and capacitance ∼ vs(t) R C L q i = ˙q (a) Method and calculation ∆ω = 625 −400 = 225 s−1 ⇒L = R/∆ω = .0400 H . ω0 = √ 400 · 625 = 500 s−1, C = 1/(Lω2 0) = 100 µF . (b) Interpretation and further calculation Q = ω0/∆ω = 500/225 = 2.22222. ωarithmetic = 512.5 s−1̸ = ω0 . Physical check A broad resonance makes the midpoint error visible even with exact measurements.

Question 5

Bandwidth measurements determine inductance and capacitance

A series RLC circuit with R=9.00 ohms has half-power angular frequencies 400 and 625 radians per second.

Diagram shown in the original SVG worksheet.

  1. Infer L, C, and the resonance angular frequency.

  2. Find its quality factor and distinguish arithmetic from geometric center.

Question 5 - Solution

Bandwidth measurements determine inductance and capacitance

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δω=625−400=225s−1⇒L=R/Δω=.0400H.\Delta\omega=625-400=225\,s^{-1}\Rightarrow\boxed{L=R/\Delta\omega=.0400\,H}.ω0=400⋅625=500s−1,C=1/(Lω02)=100μF.\boxed{\omega_0=\sqrt{400\cdot625}=500\,s^{-1},\quad C=1/(L\omega_0^2)=100\,\mu F}.

(b) Interpretation and further calculation

𝒬=ω0/Δω=500/225=2.22222.\mathcal Q=\omega_0/\Delta\omega=500/225=2.22222.ωarithmetic=512.5s−1≠ω0.\boxed{\omega_{\mathrm{arithmetic}}=512.5\,s^{-1}\ne\omega_0}.

Physical check

A broad resonance makes the midpoint error visible even with exact measurements.