Damped Oscillations in an RLC Circuit — Question 2

Section 31-4 · Physics pages 5034–5035

Question 2 Successive charge extrema identify damping and the undamped frequency Positive charge maxima in an underdamped RLC loop are .0100 s apart and decrease by a factor of two each time. L=.200 H. ωt q/Q0 Schematic; labels, not drawing lengths, specify the data. (a) Infer alpha, R, and the damped angular frequency. (b) Find the capacitance, retaining the damping correction.
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Question 2 - Solution Successive charge extrema identify damping and the undamped frequency ωt q/Q0 (a) Method and calculation e−αTd = 1/2 ⇒α = ln 2/.0100 = 69.3147 s−1. R = 2Lα = 27.7259 Ω, ωd = 2π/.0100 = 628.319 s−1 . (b) Interpretation and further calculation ω2 0 = ω2 d + α2, C = [L(ω2 d + α2)]−1. C = 12.5129 µF . Physical check The extrema are separated by the damped period, not the undamped period.

Question 2

Successive charge extrema identify damping and the undamped frequency

Positive charge maxima in an underdamped RLC loop are .0100 s apart and decrease by a factor of two each time. L=.200 H.

Diagram shown in the original SVG worksheet.

  1. Infer alpha, R, and the damped angular frequency.

  2. Find the capacitance, retaining the damping correction.

Question 2 - Solution

Successive charge extrema identify damping and the undamped frequency

Diagram shown in the original SVG worksheet.

(a) Method and calculation

e−αTd=1/2⇒α=ln⁡2/.0100=69.3147s−1.e^{-\alpha T_d}=1/2\Rightarrow\alpha=\ln2/.0100=69.3147\,s^{-1}.R=2Lα=27.7259Ω,ωd=2π/.0100=628.319s−1.\boxed{R=2L\alpha=27.7259\,\Omega,\quad\omega_d=2\pi/.0100=628.319\,s^{-1}}.

(b) Interpretation and further calculation

ω02=ωd2+α2,C=[L(ωd2+α2)]−1.\omega_0^2=\omega_d^2+\alpha^2,\quad C=[L(\omega_d^2+\alpha^2)]^{-1}.C=12.5129μF.\boxed{C=12.5129\,\mu F}.

Physical check

The extrema are separated by the damped period, not the undamped period.