Three Simple Circuits — Question 8

Section 31-7 · Physics pages 5106–5107

Question 8 A pure capacitor returns as much energy as it absorbs each cycle An ideal capacitor has v = Vp cos ωt and i = C ˙v. Define Umax = CV 2 p /2. ∼ v i C Schematic; labels, not drawing lengths, specify the data. (a) Find instantaneous power and its average. (b) Find total positive energy absorbed and total magnitude of returned energy during one full voltage cycle.
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Question 8 - Solution A pure capacitor returns as much energy as it absorbs each cycle ∼ v i C (a) Method and calculation p = vi = −ωCV 2 p cos ωt sin ωt = −ωUmax sin 2ωt. ⟨p⟩= 0 . (b) Interpretation and further calculation UC = Umax cos2 ωt rises from 0 to Umax twice per T. W+ = 2Umax, |W−| = 2Umax . Physical check Zero average power means reversible exchange, not zero instantaneous power.

Question 8

A pure capacitor returns as much energy as it absorbs each cycle

An ideal capacitor has v=Vpcos⁡ωtv=V_p\cos\omega t and i=Cv̇i=C\dot v. Define Umax=CVp2/2U_{\max}=CV_p^2/2.

Diagram shown in the original SVG worksheet.

  1. Find instantaneous power and its average.

  2. Find total positive energy absorbed and total magnitude of returned energy during one full voltage cycle.

Question 8 - Solution

A pure capacitor returns as much energy as it absorbs each cycle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

p=vi=−ωCVp2cos⁡ωtsin⁡ωt=−ωUmaxsin⁡2ωt.p=vi=-\omega CV_p^2\cos\omega t\sin\omega t=-\omega U_{\max}\sin2\omega t.⟨p⟩=0.\boxed{\langle p\rangle=0}.

(b) Interpretation and further calculation

UC=Umaxcos⁡2ωtrises from 0 to Umax twice per T.U_C=U_{\max}\cos^2\omega t\quad\text{rises from }0\text{ to }U_{\max}\text{ twice per }T.W+=2Umax,|W−|=2Umax.\boxed{W_+=2U_{\max},\quad |W_-|=2U_{\max}}.

Physical check

Zero average power means reversible exchange, not zero instantaneous power.