Transformers — Question 3

Section 31-10 · Physics pages 5156–5157

Question 3 The applied volts per turn determine sinusoidal core flux density A transformer primary has N turns around uniform core area A. Its flux density is sinusoidal with peak Bmax. Neglect winding voltage drop and leakage. Applied rms voltage is V at frequency f. Np Ns Vp Vs core area A, peak field Bmax Schematic; labels, not drawing lengths, specify the data. (a) Derive the relation among V, f, N, A, and Bmax from Faraday induction. (b) Find the minimum integer turns for V=230 V, f=50.0 Hz, A = 4.00 × 10−4 square meters, and allowed peak Bmax=1.20 T.
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Question 3 - Solution The applied volts per turn determine sinusoidal core flux density Np Ns Vp Vs core area A, peak field Bmax (a) Method and calculation Φ = ABmax sin ωt, |v|max = NωABmax. V = 2π √ 2fNABmax ≈4.44288fNABmax . (b) Interpretation and further calculation N ≥ 230 √ 2 2π(50)(4.00 × 10−4)(1.20) = 2157.01. Nmin = 2158 turns . Physical check The integer choice must round upward because it enforces an upper bound on peak flux density.

Question 3

The applied volts per turn determine sinusoidal core flux density

A transformer primary has N turns around uniform core area A. Its flux density is sinusoidal with peak Bmax. Neglect winding voltage drop and leakage. Applied rms voltage is V at frequency f.

Diagram shown in the original SVG worksheet.

  1. Derive the relation among V, f, N, A, and Bmax from Faraday induction.

  2. Find the minimum integer turns for V=230 V, f=50.0 Hz, A=4.00×10−4A=4.00\times10^{-4} square meters, and allowed peak Bmax=1.20 T.

Question 3 - Solution

The applied volts per turn determine sinusoidal core flux density

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φ=ABmaxsin⁡ωt,|v|max=NωABmax.\Phi=AB_{\max}\sin\omega t,\quad |v|_{\max}=N\omega AB_{\max}.V=2π2fNABmax≈4.44288fNABmax.\boxed{V=\frac{2\pi}{\sqrt2}fNAB_{\max}\approx4.44288fNAB_{\max}}.

(b) Interpretation and further calculation

N≥23022π(50)(4.00×10−4)(1.20)=2157.01.N\ge\frac{230\sqrt2}{2\pi(50)(4.00\times10^{-4})(1.20)}=2157.01.Nmin=2158turns.\boxed{N_{\min}=2158\ \text{turns}}.

Physical check

The integer choice must round upward because it enforces an upper bound on peak flux density.