Alternating Current — Question 4

Section 31-5 · Physics pages 5058–5059

Question 4 An asymmetric two-level waveform can have zero mean without equal amplitudes In every period, a source is +A for fraction d and -B for fraction 1 −d, with positive A and B. Require zero mean. Let A=12.0 V and d=1/4. ωt v A −B Schematic; labels, not drawing lengths, specify the data. (a) Find B and the rms voltage. (b) Find the ratio of rms voltage to the positive amplitude for general d.
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Question 4 - Solution An asymmetric two-level waveform can have zero mean without equal amplitudes ωt v A −B (a) Method and calculation dA −(1 −d)B = 0 ⇒B = dA/(1 −d) = 4.00 V. V 2 rms = dA2 + (1 −d)B2 ⇒Vrms = √ 48 = 6.92820 V . (b) Interpretation and further calculation V 2 rms = A2 [ d + d2 1 −d ] = A2 d 1 −d. Vrms/A = √ d/(1 −d) . Physical check Zero mean constrains areas under the waveform, not equal time above and below zero.

Question 4

An asymmetric two-level waveform can have zero mean without equal amplitudes

In every period, a source is +A for fraction d and -B for fraction 1−d1-d, with positive A and B. Require zero mean. Let A=12.0 V and d=1/4.

Diagram shown in the original SVG worksheet.

  1. Find B and the rms voltage.

  2. Find the ratio of rms voltage to the positive amplitude for general d.

Question 4 - Solution

An asymmetric two-level waveform can have zero mean without equal amplitudes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dA−(1−d)B=0⇒B=dA/(1−d)=4.00V.dA-(1-d)B=0\Rightarrow B=dA/(1-d)=4.00\,V.Vrms2=dA2+(1−d)B2⇒Vrms=48=6.92820V.V_{\mathrm{rms}}^2=dA^2+(1-d)B^2\Rightarrow\boxed{V_{\mathrm{rms}}=\sqrt{48}=6.92820\,V}.

(b) Interpretation and further calculation

Vrms2=A2[d+d21−d]=A2d1−d.V_{\mathrm{rms}}^2=A^2\left[d+\frac{d^2}{1-d}\right]=A^2\frac d{1-d}.Vrms/A=d/(1−d).\boxed{V_{\mathrm{rms}}/A=\sqrt{d/(1-d)}}.

Physical check

Zero mean constrains areas under the waveform, not equal time above and below zero.