Question
2. A two-winding transformer with a laminated core is shown in the Figure. The winding with N1=200 turns is connected to a voltage to produce a flux density in the core B=1.2Âsin377t. The second winding, with N2=400 turns, is left open-circuited. The stacking factor of the core is 0.95, i.e. the core occupies 95% of the gross core volume. The gross cross-sectional area of the core is 25cm2 , and ȝr for the core is 10,000. The core length is lc=90cm. a) Determine the rms value of the applied voltage E1. b) Determine the current in the winding. c) Determine the rms voltage Er induced in the second winding.
Answers
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a) The rms value of the applied voltage E1 is given by:
E1 = (B*N1*lc*S)/(4.44*sqrt(2)*ȝr)
E1 = (1.2*200*90*0.95)/(4.44*sqrt(2)*10000)
E1 = 4.8 V
b) The current in the winding is given by:
I1 = (E1*sqrt(2))/(N1*lc*S)
I1 = (4.8*sqrt(2))/(200*90*0.95)
I1 = 0.0267 A
c) The rms voltage Er induced in the second winding is given by:
Er = (B*N2*lc*S)/(4.44*sqrt(2)*ȝr)
Er = (1.2*400*90*0.95)/(4.44*sqrt(2)*10000)
Er = 9.6 V
E1 = (B*N1*lc*S)/(4.44*sqrt(2)*ȝr)
E1 = (1.2*200*90*0.95)/(4.44*sqrt(2)*10000)
E1 = 4.8 V
b) The current in the winding is given by:
I1 = (E1*sqrt(2))/(N1*lc*S)
I1 = (4.8*sqrt(2))/(200*90*0.95)
I1 = 0.0267 A
c) The rms voltage Er induced in the second winding is given by:
Er = (B*N2*lc*S)/(4.44*sqrt(2)*ȝr)
Er = (1.2*400*90*0.95)/(4.44*sqrt(2)*10000)
Er = 9.6 V