LC Oscillations, Quantitatively — Question 4

Section 31-3 · Physics pages 5018–5019

Question 4 Peak measurements and period identify both circuit elements An ideal LC oscillator has measured peak capacitor voltage 12.0 V, peak current .0300 A, and angular frequency 2500 radians per second. C L q i = ˙q Schematic; labels, not drawing lengths, specify the data. (a) Infer C and L without knowing the initial phase. (b) Find maximum charge and total energy.
Show solutionHide solution
Question 4 - Solution Peak measurements and period identify both circuit elements C L q i = ˙q (a) Method and calculation Imax = ωCVmax ⇒C = Imax/(ωVmax) = 1.00 µF . L = Vmax/(ωImax) = .160 H . (b) Interpretation and further calculation Q = CVmax = 12.0 µC. U = 1 2CV 2 max = 72.0 µJ . Physical check The ratio of voltage to current gives the impedance scale, while frequency determines the LC product.

Question 4

Peak measurements and period identify both circuit elements

An ideal LC oscillator has measured peak capacitor voltage 12.0 V, peak current .0300 A, and angular frequency 2500 radians per second.

Diagram shown in the original SVG worksheet.

  1. Infer C and L without knowing the initial phase.

  2. Find maximum charge and total energy.

Question 4 - Solution

Peak measurements and period identify both circuit elements

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Imax=ωCVmax⇒C=Imax/(ωVmax)=1.00μF.I_{\max}=\omega CV_{\max}\Rightarrow\boxed{C=I_{\max}/(\omega V_{\max})=1.00\,\mu F}.L=Vmax/(ωImax)=.160H.\boxed{L=V_{\max}/(\omega I_{\max})=.160\,H}.

(b) Interpretation and further calculation

Q=CVmax=12.0μC.Q=CV_{\max}=12.0\,\mu C.U=12CVmax⁡2=72.0μJ.\boxed{U=\tfrac12CV_{\max}^2=72.0\,\mu J}.

Physical check

The ratio of voltage to current gives the impedance scale, while frequency determines the LC product.