LC Oscillations, Qualitatively — Question 9

Section 31-1 · Physics pages 4988–4989

Question 9 A small resistance selects energy loss without changing its sign A nearly ideal LC oscillator has a small series resistance R. Use L˙i + Ri + q/C = 0 and i = ˙q. R C L q i = ˙q Schematic; labels, not drawing lengths, specify the data. (a) Derive the instantaneous total-energy rate. (b) Compare energy loss at maximum charge with energy loss at zero charge during a cycle.
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Question 9 - Solution A small resistance selects energy loss without changing its sign R C L q i = ˙q (a) Method and calculation ˙U = (q/C)i + Li˙i = i(q/C + L˙i). ˙U = −Ri2 ≤0 . (b) Interpretation and further calculation |q| at a turning point : i = 0 ⇒˙U = 0. q = 0, i̸ = 0 : ˙U < 0 . Physical check A zero instantaneous loss at a turning point does not mean that the full cycle is lossless.

Question 9

A small resistance selects energy loss without changing its sign

A nearly ideal LC oscillator has a small series resistance R. Use Li̇+Ri+q/C=0L\dot i+Ri+q/C=0 and i=q̇i=\dot q.

Diagram shown in the original SVG worksheet.

  1. Derive the instantaneous total-energy rate.

  2. Compare energy loss at maximum charge with energy loss at zero charge during a cycle.

Question 9 - Solution

A small resistance selects energy loss without changing its sign

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U̇=(q/C)i+Lii̇=i(q/C+Li̇).\dot U=(q/C)i+Li\dot i=i(q/C+L\dot i).U̇=−Ri2≤0.\boxed{\dot U=-Ri^2\le0}.

(b) Interpretation and further calculation

|q|at a turning point:i=0⇒U̇=0.|q|\ \text{at a turning point}:\ i=0\Rightarrow\dot U=0.q=0,i≠0:U̇<0.\boxed{q=0,\ i\ne0:\ \dot U<0}.

Physical check

A zero instantaneous loss at a turning point does not mean that the full cycle is lossless.