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UK EE 221 - Lecture 18: Transformers

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Lecture 18: TransformersVijay Singh∗April 13, 2003AbstractMutual inductance, transformers.1 Ideal TransformerIn Figure 1, assume:1. Ideal core: same flux links both coils.2. Wire resistance is negligible.Φ1= Φi1(t) i2(t)v1(t) v2(t)+ +− −Figure 1: .Then,v1(t) = N1dΦ1dt= N1dΦdtv2(t) = N2dΦ2dt= N2dΦdtTherefore,v1v2=N1N2(1)Also, by Ampere’s law:IH · dl = ienclosed= N1i1+ N2i2(2)∗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]).1For an ideal magnetic core, µ −→ ∞ and H −→ 0. Therefore,N1i1+ N2i2= 0 (3)That is,i1i2= −N2N1(4)1.1 PowerFrom Equation (4),i1= −N2i2N1Multiply by v1. Then,v1i1+ v1i2N1N2= 0or,v1i1+ i2v2= 0 (5)Total power into the ideal transformer is thus zero. It is lossless.1.2 Dot ConventionAs shown in Figure 1,v1=N1N2v2(6)where both voltages are referenced positive at the dots.Also,N1i1+ N2i2= 0 (7)when both currents are entering the dots.1.3 IllustrationV2+−ZLN1: N2I2V1+−I1Iron core transformerFigure 2: .Using dot convention,V1V2=N1N2(8)2Here, both voltages are referenced positive at the dot.N1I1− N2I2= 0 (9)Here, one current is leaving the dot.I1I2=N2N1(10)Complex Power = S1= V1I∗1=µN1N2V2¶µN2N1I2¶∗= V2I∗2= S2(11)Input Impedance = Z1=V1I1=N1N2V2N2N1I2=V2I2µN1N2¶2= ZLµN1N2¶2(12)V1=V2nI1= nI2S1= S2Z1=ZLn2Here,n =N2N1= Turns Ratio1.4 ExampleThis is Problem 8.43 from J.David Irwin, Basic Engineering Circuit Analysis, 7-th edition.GivenL1= L2= 2 H, k = 0.8Then,M = kpL1L2= 1.6 Hi1(t) = 10 cos(377t − 30◦)i2(t) = 20 cos(377t − 45◦)w(t) =12i1(t)2+12L2i2(t)2− Mi1(t)i2(t)At t = 1 ns, w = 144.25 µJ.i2(t)i1(t)Figure 3: .31.5 ExampleZL= 1 + 2||(j2) = 2 − j1n =14=N2N1Z1=ZLn2= 32 − j16Zin= 1 + (Z1||(−j1)) = 1 +−16 − j3232 − j17=16 − j4932 − j17= 1.426− 43.9◦Ω1.6 Example+−4 Ω 1 ΩIB2 Ω26180◦+−V2160I2Figure 4: .I2= 6µ57¶= 0.429 AIB= I2− 160 = −0.571V2= IB(1) + I2(2) = 0.287V1= −V2n= 0.1446180◦I1= IBn = 1.1426180◦1.7 Problem-solving techniques1. Remove the ideal transformer from the circuit byreflecting it on the primary or the secondary side.If the unknown voltage or current that we want todetermine is on the secondary side, then reflect theprimary circuit to the secondary side.If . . ..2. Use the circuit solving techniques (KCL etc) learnedearlier to solve the circuit.PrimarySecondaryFigure 5: .1.8 Examplen =14Since we want I1, we reflect Z2to the primary side:Z1= (4)2(2 + j1)4+−2 Ωj118 Ω−j4120604 : 1ABI1Figure 6: .+−18 Ω−j412060ABI1Z1Figure 7: .1.9 Example+−3 Ω11060 VV1+−+−I122060 V−j80 ΩAB32 ΩV2+−1 : 4I2Figure 8: .Find I1, I2, V1, V2.Finding I2To get rid of theideal transformer, we “reflect” the primary circuit to the secondary side.n =N2N1= 4; Using dot conventionV1V2=14andI1I2= −4;Z1Z2=1165+−22060 V−j80 ΩA32 ΩI2B+−(16 × 3) Ω4 × 11060 VFigure 9: .I2=440 − 22048 + 32= 2.7560 AV2= 220 + 32I2= 30860 VV1=V24= 7760I1= −nI2= −4I2= −1160 = 116180◦A1 : 4V1V2+−+−ABFigure 10:


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