1. Transformer Model where r = diag [r 1 r 2 ], a diagonal matrix, and The resistances r 1 and r 2 and the flux linkages l 1 and l 2 are related to coils 1 and 2, respectively. Because it is assumed that 1 links the equivalent turns of coil 1 and 2 links the equivalent turns of coil 2, the flux linkages may be written as Voltage Equation of a transformer in matrix form is: Where

2. Linear Magnetic System Reluctance is impossible to measure accurately, could be determined using:

3. Flux Linkages The coefficients of the 1 st two terms on the right–hand side depend upon the turns of coil 1 and the reluctance of the magnetic system; i.e. independent of coil 2. Similar situation exist in equation for 2

4. Self Inductances  From the previous equations one can define Self-Inductances: Where L l1 and L l2 are leakage inductances of coil 1 and 2 respectively. Lm1 and Lm2 are the magnetizing inductances of coils 1 and 2 respectively.

5. Magnetizing Inductances  The two magnetizing Inductances are related as: Where  m the Magnetizing Reluctance being common for both coils. The mutual Inductances are defined:

6. Mutual Inductances Mutual Reluctance being common for both Circuits; Mutual Inductances are related to Magnetizing Inductances too:

7. Flux Linkages  Flux Linkages may be written as: = Li Where

8. Flux Linkages  The Flux Linkage may also be derived based on self and mutual inductances:

9. Example 1A It is instructive to illustrate the method of deriving an equivalent T circuit from open- and short-circuit measurements. For this purpose let us assume that when coil 2 of the two-winding transformer shown in Fig. is open-circuited, the power input to coil 1 is 12 W with an applied voltage is 100 V (rms) at 60 Hz and the current is 1 A (rms). When coil 2 is short-circuited, the current flowing in coil 1 is 1 A when the applied voltage is 30 V at 60 Hz. The power during this test is 22 W. If we assume L l1 = L’ l2, an approximate equivalent T circuit can be determined from these measurements with coil 1 selected as the reference coil.

10. Equivalent T Circuit of Transformer Where i’ 2 =(N 2 /N 1 )i 2

11. Example 1….. The Power may be expressed: Where are phasor and  is the phase angle between them. Solving for  during the open circuit test, we have:

12. Example 1…… as a reference phasor and in an inductive circuit of the transformer I phasor would lag behind by the angle of  =83.7 0 Z, the impedance may therefore be determined by: That suggests that X l1 +X m1 = 109.3 , while r 1 =12  V I 

13. Example 1……. For short circuit test i 1 = -I’ 2 because transformers are designed so that X m1 >> |r’ 2 +jX’ 12 |. Hence using phase angle equation: In this case input Impedance is (r 1 +r’ 2 )+j(X l1 +X’ l2 ) and that is determined by: That means r’ 2 = 10  and X l1 =X’ l2 both are 10.2 

14. Example 1…. That leads to conclusion that: X m1 = 109.3 -10.2 =99.1  Hence other parameters are: r 1 = 12  L m1 = 262.9mHr’ 2 = 10  L l1 = 27.1mHL’ l2 = 27.1mH V1V1 I1I1 V’ 2 E1E1 

15. Phasor Diagram

16. T circuit ref. to Primary

17. Equivalent Circuit ref. to Primary

18. Active & Reactive Power

19. Magnetic Laws

20. Flux Linkage of a Coil Fig. 1 shows a coil of N turns. All these N turns link flux lines of Weber resulting in the N flux linkages. In such a case: Where N is number of turns in a coil; e is emf induced, and  is flux linking to each coil

21. Design of Transformer  Let's try to proportion a transformer for 120 V, 60 Hz supply, with a full-load current of 10 A. The core material is to be silicon-steel laminations with a maximum operating flux density B max = 12,000 gauss. This is comfortably less than the saturation flux density, B sat. The first requirement is to ensure that we have sufficient ampere-turns to magnetize the core to this level with a permissible magnetizing current I0A.

22. Design of Transformer….  Let's choose the magnetizing current to be 1% of the full-load current, or 0.1 A. The exact value is not sacred; this might be thought of as an upper limit. Here, we will assume a simple, uniform magnetic circuit for simplicity. If l is the length of the magnetic circuit, H is 0.4πN(√2I0)/l, and the magnetization curve for the core iron gives the H required for the chosen B max. From this, we can find the number of turns, N, required for the primary.

23. Design of Transformer We could also estimate the ampere-turns required by using an assumed permeability μ. Experience will furnish a satisfactory value. It is not taken from the magnetization curve, but from the hysteresis loop. Let's take μ = 1000. Then, N = B max l / 0.4π√2 μI 0. If we estimate l = 20 cm, the number of primary turns required is N = 1350. The rms voltage induced per turn is determined from Faraday's Law: √2 e = (2πf)B max A x 10 -8. Now, e must be 120 / 1350 = 0.126 V/turn, f is 60, and B max = 12,000 gauss. We know everything but A, the cross-sectional area of the core. We find A = 2.8 cm 2.

24. Design of Transformer Powdered iron and ferrite cores have low B sat and permeability values. A type 43 ferrite has B sat = 2750 gauss, but a maximum permeability of 3000, and is recommended for frequencies from 10 kHz to 1 MHz. Silicon iron is much better magnetically, but cannot be used at these frequencies. The approximate dimensions of an FT-114 ferrite core (of any desired material) are OD 28 mm, ID 19 mm, thickness 7.5 mm. The magnetic dimensions are l = 74.17 mm, A = 37.49 mm 2, and volume 2778 mm 3. Similar information is available for a wide range of cores. There are tables showing how much wire can be wound on them, and even the inductance as a function of the number of turns.

25. Design Parameters l average length A: X-section Area N number of turns B =  /A Laminated Core

26. Laminations

27. Flux Density & MMF B max

28. Hysteresis loop

29. Hysteresis Loop