Transformers — Question 7

Section 31-10 · Physics pages 5164–5165

Question 7 An unloaded real core can draw both magnetizing and loss current Use a no-load primary model consisting of an ideal magnetizing inductance Lm in parallel with core-loss resistance Rc. Let Lm=.800 H, Rc=2400 ohms, V=120 V rms, and f=50.0 Hz. Np Ns Vp Vs secondary open; primary model Rc ∥Lm Schematic; labels, not drawing lengths, specify the data. (a) Find in-phase current, quadrature magnetizing current, and total rms current. (b) Find no-load real power and power factor.
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Question 7 - Solution An unloaded real core can draw both magnetizing and loss current Np Ns Vp Vs secondary open; primary model Rc ∥Lm (a) Method and calculation Ic = V/Rc = .0500 A, Im = V/(2πfLm) = .477465 A. I0 = √ I2c + I2m = .480076 A . (b) Interpretation and further calculation P0 = V 2/Rc = 6.00 W. pf0 = Ic/I0 = .104150 lagging . Physical check Open secondary terminals do not eliminate the current needed for the supplied nonideal core model.

Question 7

An unloaded real core can draw both magnetizing and loss current

Use a no-load primary model consisting of an ideal magnetizing inductance Lm in parallel with core-loss resistance Rc. Let Lm=.800 H, Rc=2400 ohms, V=120 V rms, and f=50.0 Hz.

Diagram shown in the original SVG worksheet.

  1. Find in-phase current, quadrature magnetizing current, and total rms current.

  2. Find no-load real power and power factor.

Question 7 - Solution

An unloaded real core can draw both magnetizing and loss current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ic=V/Rc=.0500A,Im=V/(2πfLm)=.477465A.I_c=V/R_c=.0500\,A,\quad I_m=V/(2\pi fL_m)=.477465\,A.I0=Ic2+Im2=.480076A.\boxed{I_0=\sqrt{I_c^2+I_m^2}=.480076\,A}.

(b) Interpretation and further calculation

P0=V2/Rc=6.00W.P_0=V^2/R_c=6.00\,W.pf0=Ic/I0=.104150lagging.\boxed{\mathrm{pf}_0=I_c/I_0=.104150\quad\text{lagging}}.

Physical check

Open secondary terminals do not eliminate the current needed for the supplied nonideal core model.