Three Simple Circuits — Question 9

Section 31-7 · Physics pages 5108–5109

Question 9 A triangular capacitor voltage produces a square current waveform An ideal capacitor C is driven by a continuous symmetric triangular voltage between -Vp and +Vp with period T. Its rises and falls each occupy T/2. ωt v Schematic; labels, not drawing lengths, specify the data. (a) Find the current in each half-cycle and its rms value. (b) Find the charge transferred during a rising half-cycle and the peak stored energy.
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Question 9 - Solution A triangular capacitor voltage produces a square current waveform ωt v (a) Method and calculation ˙v = ± 2Vp T/2 = ±4Vp/T. i = ±4CVp/T, Irms = 4CVp/T . (b) Interpretation and further calculation ∆q = C∆v = 2CVp. Umax = CV 2 p /2 . Physical check Voltage is continuous at the corners, so current has finite jumps rather than impulse spikes.

Question 9

A triangular capacitor voltage produces a square current waveform

An ideal capacitor C is driven by a continuous symmetric triangular voltage between -Vp and +Vp with period T. Its rises and falls each occupy T/2.

Diagram shown in the original SVG worksheet.

  1. Find the current in each half-cycle and its rms value.

  2. Find the charge transferred during a rising half-cycle and the peak stored energy.

Question 9 - Solution

A triangular capacitor voltage produces a square current waveform

Diagram shown in the original SVG worksheet.

(a) Method and calculation

v̇=±2VpT/2=±4Vp/T.\dot v=\pm\frac{2V_p}{T/2}=\pm4V_p/T.i=±4CVp/T,Irms=4CVp/T.\boxed{i=\pm4CV_p/T,\quad I_{\mathrm{rms}}=4CV_p/T}.

(b) Interpretation and further calculation

Δq=CΔv=2CVp.\Delta q=C\Delta v=2CV_p.Umax=CVp2/2.\boxed{U_{\max}=CV_p^2/2}.

Physical check

Voltage is continuous at the corners, so current has finite jumps rather than impulse spikes.