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Physics / Chapter 31: Electromagnetic Oscillations and Alternating Current / Section 31-10

Transformers — Question 10

Section 31-10 · Physics pages 5170–5171

Question 10 Transformer flux requires balanced volt-seconds even for nonsinusoidal drive A primary with N=50 turns and core area A = 1.00 × 10−4 square meters is driven at f=10.0 kHz. Each period applies +20.0 V for fraction d=.400, then a constant negative voltage for the rest. Neglect winding drops and use v = NA ˙B. ωt v +20 V V− Schematic; labels, not drawing lengths, specify the data. (a) Find the negative voltage required for periodic flux and the peak-to-peak flux-density swing. (b) For a centered flux waveform find peak magnitude; if the negative level were instead -20.0 V, find the net field change per cycle.
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Question 10 - Solution Transformer flux requires balanced volt-seconds even for nonsinusoidal drive ωt v +20 V V− (a) Method and calculation 20d + V−(1 −d) = 0 ⇒V−= −20(.4)/.6 = −13.3333 V . ∆Bpp = 20d/(fNA) = .160 T . (b) Interpretation and further calculation Bmax = ∆Bpp/2 = .0800 T for centered flux . ∆Bcycle = 20(.4) −20(.6) fNA = −.0800 T (V−= −20 V ) . Physical check Zero volt-second imbalance permits periodic flux but does not by itself fix its DC offset.

Question 10

Transformer flux requires balanced volt-seconds even for nonsinusoidal drive

A primary with N=50 turns and core area A=1.00×10−4A=1.00\times10^{-4} square meters is driven at f=10.0 kHz. Each period applies +20.0 V for fraction d=.400, then a constant negative voltage for the rest. Neglect winding drops and use v=NAḂv=NA\dot B.

Diagram shown in the original SVG worksheet.

  1. Find the negative voltage required for periodic flux and the peak-to-peak flux-density swing.

  2. For a centered flux waveform find peak magnitude; if the negative level were instead -20.0 V, find the net field change per cycle.

Question 10 - Solution

Transformer flux requires balanced volt-seconds even for nonsinusoidal drive

Diagram shown in the original SVG worksheet.

(a) Method and calculation

20d+V−(1−d)=0⇒V−=−20(.4)/.6=−13.3333V.20d+V_-(1-d)=0\Rightarrow\boxed{V_-=-20(.4)/.6=-13.3333\,V}.ΔBpp=20d/(fNA)=.160T.\boxed{\Delta B_{pp}=20d/(fNA)=.160\,T}.

(b) Interpretation and further calculation

Bmax=ΔBpp/2=.0800Tfor centered flux.\boxed{B_{\max}=\Delta B_{pp}/2=.0800\,T\quad\text{for centered flux}}.ΔBcycle=20(.4)−20(.6)fNA=−.0800T(V−=−20V).\boxed{\Delta B_{\mathrm{cycle}}=\frac{20(.4)-20(.6)}{fNA}=-.0800\,T\quad(V_-=-20\,V)}.

Physical check

Zero volt-second imbalance permits periodic flux but does not by itself fix its DC offset.

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