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Physics / Chapter 27: Circuits / Section 27-8

RC Circuits — Question 7

Section 27-8 · Physics pages 4464–4465

Question 7 Connecting a precharged capacitor to a lower source returns energy while producing heat A capacitor C initially at Vi is connected through positive resistance R to an ideal source E of the same reference polarity. Allow the transient to finish. Define source work positive when delivered to the circuit. + − E R C + − Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Derive source work, capacitor energy change and total resistor heat for arbitrary Vi and E. (b) Evaluate them for C=200 microfarad, Vi=8.00 V and E=3.00 V, and verify energy conservation.
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Question 7 - Solution Connecting a precharged capacitor to a lower source returns energy while producing heat + − E R C + − (a) Method and calculation ∆q = C(E −Vi), Ws = E∆q = CE(E −Vi), ∆UC = 1 2C(E2 −V 2 i ). QR = Ws −∆UC = 1 2C(E −Vi)2 ≥0. (b) Interpretation and further calculation ∆q = −1.00 mC, Ws = −3.00 mJ, ∆UC = −5.50 mJ. QR = 2.50 mJ, −3.00 = −5.50 + 2.50 (mJ). Physical check The source absorbs energy in this case, while the resistor still dissipates positive heat.

Question 7

Connecting a precharged capacitor to a lower source returns energy while producing heat

A capacitor C initially at Vi is connected through positive resistance R to an ideal source E of the same reference polarity. Allow the transient to finish. Define source work positive when delivered to the circuit.

Diagram shown in the original SVG worksheet.

  1. Derive source work, capacitor energy change and total resistor heat for arbitrary Vi and E.

  2. Evaluate them for C=200 microfarad, Vi=8.00 V and E=3.00 V, and verify energy conservation.

Question 7 - Solution

Connecting a precharged capacitor to a lower source returns energy while producing heat

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Δq=C(E−Vi),Ws=EΔq=CE(E−Vi),ΔUC=12C(E2−Vi2).\Delta q=C(E-V_i),\quad W_s=E\Delta q=CE(E-V_i),\quad\Delta U_C=\tfrac12C(E^2-V_i^2).QR=Ws−ΔUC=12C(E−Vi)2≥0.Q_R=W_s-\Delta U_C=\tfrac12C(E-V_i)^2\ge0.

(b) Interpretation and further calculation

Δq=−1.00mC,Ws=−3.00mJ,ΔUC=−5.50mJ.\Delta q=-1.00\,\mathrm{mC},\quad W_s=-3.00\,\mathrm{mJ},\quad\Delta U_C=-5.50\,\mathrm{mJ}.QR=2.50mJ,−3.00=−5.50+2.50(mJ).Q_R=2.50\,\mathrm{mJ},\quad-3.00=-5.50+2.50\quad\text{(mJ)}.

Physical check

The source absorbs energy in this case, while the resistor still dissipates positive heat.

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