Transformers — Question 2

Section 31-10 · Physics pages 5154–5155

Question 2 A transformer can match a resistive load to a source resistance An AC source has 12.0 V rms open-circuit voltage and series resistance 50.0 ohms. It drives an 8.00-ohm load through an ideal transformer. Let a = Np/Ns. Np Ns Vp Vs Schematic; labels, not drawing lengths, specify the data. (a) Choose a for maximum real load power. (b) Find that power and secondary rms voltage and current.
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Question 2 - Solution A transformer can match a resistive load to a source resistance Np Ns Vp Vs (a) Method and calculation Rreflected = a2RL, PL = V 2 oca2RL (Rs + a2RL)2 . a = √ Rs/RL = 2.50 . (b) Interpretation and further calculation PL,max = V 2 oc/(4Rs) = 144/200 = .720 W. Vs = Voc/(2a) = 2.40 V, Is = Vs/RL = .300 A . Physical check Maximum transfer here dissipates equal power in source resistance and load; it is not maximum efficiency.

Question 2

A transformer can match a resistive load to a source resistance

An AC source has 12.0 V rms open-circuit voltage and series resistance 50.0 ohms. It drives an 8.00-ohm load through an ideal transformer. Let a=Np/Nsa=N_p/N_s.

Diagram shown in the original SVG worksheet.

  1. Choose a for maximum real load power.

  2. Find that power and secondary rms voltage and current.

Question 2 - Solution

A transformer can match a resistive load to a source resistance

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Rreflected=a2RL,PL=Voc2a2RL(Rs+a2RL)2.R_{\mathrm{reflected}}=a^2R_L,\quad P_L=\frac{V_{\mathrm{oc}}^2a^2R_L}{(R_s+a^2R_L)^2}.a=Rs/RL=2.50.\boxed{a=\sqrt{R_s/R_L}=2.50}.

(b) Interpretation and further calculation

PL,max=Voc2/(4Rs)=144/200=.720W.P_{L,\max}=V_{\mathrm{oc}}^2/(4R_s)=144/200=.720\,W.Vs=Voc/(2a)=2.40V,Is=Vs/RL=.300A.\boxed{V_s=V_{\mathrm{oc}}/(2a)=2.40\,V,\quad I_s=V_s/R_L=.300\,A}.

Physical check

Maximum transfer here dissipates equal power in source resistance and load; it is not maximum efficiency.