Lecture 16: TransformersVijay Singh∗April 10, 2003AbstractMutual inductance, transformers.1 MotivationTransmission line voltage is 500 kV.Voltage of home appliances is 208 V or 120 V.Need a step-down transformer.i1(t)v1(t)+−−+v2(t)i2(t)Figure 1: Transformer.2 EquationsBy Faraday’s law,v1(t) = N1dΦ1dt(1)where,N1= Number of turns of the coil,Φ1= Flux in Coil 1= Φ11+ Φ12(2)Φ11= Flux in Coil 1 due to current in Coil 2Φ12= Flux in Coil 1 due to current in Coil 2So, v1(t) = N1dΦ11dt+ N1dΦ12dt(3)∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky, Lex-ington, KY, USA. E-mail: [email protected]. Document prepared using LATEX and figures created usingMetagraf by Ramprasad Potluri ([email protected]).1Figure 2: .But, we know thatΦ11= N1i1P11(4)Φ12= N1i1P12(5)where permiances P11and P12depend upon the magnetic paths taken by the flux components.Therefore,v1(t) = N21P11di1dt+ N1N2P12di2dt(6)Constant N21P11= L11= Self Inductance (7)Constant N1N2P12= L12= Mutual Inductance (8)i1Figure 3: .Thus,v1(t) = L11di1dt+ L12di2dt(9)Similarly,v2(t) = N22P22di2dt+ N1N2P21di1dtor, v2(t) = L22di2dt+ L21di1dt(10)If the medium is linear, thenP12= P21= Mv1(t) = L1di1dt+ Mdi2dt(11)v2(t) = Mdi1dt+ L2di2dt(12)2Figure 4: .If currents enter both dotted terminals of Figure 4, then fluxes produced by both currents willadd.Current entering the dot of coil A marks the dotted end of coil B “positive” by virtue of mutualinductance (Figure 5.A BFigure 5: .3 ExamplePlease see Figure 6.Coil 1 Coil 2va(t)+−i1(t) i2(t)L1L2vc(t)−+PQMNvd(t)−+vb(t)+−Figure 6: .In coil 1:va(t) = L1di1dt+ Mdi2dtHere, va(t) = VP Q, and the “+” sign is because current leaving the dot in coil 2 marks thedotted end (Q) of coil 1 negative.3Thus,(VQP)inducedis negative Mdi2dt(VP Q)inducedis positive Mdi2dtSimilarly, in coil 2:vb(t) = −L2di2dt− Mdi1dtHere, vb(t) = VMN. Of course, vc= −vaand vd= −vb.i1(t)L1PQFigure 7: .VP Q= L1di1dtVMN= −L2di2dti2(t)L2MNFigure 8: .4 ExampleIn Figure 9, find VO.VOI2j2+−j21 Ω1 Ω2 Ω1060 Vj2I1+−Figure 9: .10 = I1(2) + I1(j2) + I2(j2) (13)= I1(2 + j2) + I2(2j)0 = I2(j2 + 1 + 1) + I1(j2) (14)From Equation (14), we have:I1= −I2(2 + j2)j2= −I2(1 − j1)4Substitute in Equation (1):10 = I2[j2 − {1 − j1}(2 + j2)]= I2[j2 − 2 − 2 + j2 − j2]= I2[−4 + j2]I2=10−4 + j2=5−2 + j1= 2.246− 153.43◦VO= I2(1) = 2.246− 154.43◦V5 ExamplejωL1R1V11jωC+−I3R2jωMjωL2I1I2Figure 10: .Write mesh equations:V1= I1(R1+ jωL1) − jωL1I2+jωM I30 = −jωL1I1+ I2(jωL1+ R2−jωC1−I3(jωM ) − I3R20 = +jωM I1− (R2+ jωM )I2+ I3(R2+ jωL2)6 ExampleIn Figure 11, VO=?Reflect the primary circuit of transformer T1to its seconary side (Figure 12).n1=12Reflect the primary circuit of transformer T2to the secondary side (Figure 13).n2=41= 4VO= −481616 + 32 − j8=−966 − j1= 15.786189.46◦V52 : 1−j4 Ω2460 V+−T1VO16 Ωj1 Ω2 Ω1 : 4−j8 ΩT2+−Figure 11: .16 Ωj1 Ω2 Ω1 : 4−j8 ΩT2¡−j14× 4¢Ω+−12(2460)= 1260Figure 12: .16 Ω−j8 Ω4 × 1260= 4860−+VO+−2 × (4)2Dot conventionFigure 13:
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