2. Mutual Inductance & Transformer Electric Circuit Chairul Hudaya, M.Sc Electric Power and Energy Studies (EPES) Department of Electrical Engineering Universitas Indonesia http://www.ee.ui.ac.id/epes/hudaya

3. CIRCUIT WITH MUTUAL INDUCTANCE PASSIVE SIGN CONVENTION : Arrows for I1 and I2 point into plus end of V1 and V2 DOT CONVENTION : If both current reference arrows point into dotted ends or both into undotted ends of inductor, use plus sign for both mutual inductance, otherwise use the minus sign i1i1 i2i2 Coil 1 Coil 2 M L1L1 L2L2 v1v1 v2v2

4. Dot Convention The dots are placed in such a manner that the currents entering (or leaving) both the dotted terminals will produce adding magnetic flux. In this case the mutual flux linkages will add to the self flux linkages.(Case I) Conversely, if current enters through one dotted terminal and leaves through the other, they produce opposing flux. In this case mutual flux linkages subtract from self linkages.(Case II)

5. i1i1 i2i2 Coil 1 Coil 2 M L1L1 L2L2 Case I v1v1 v2v2

6. Case 1: jMjM I1I1 I2I2 ++ _ V1V1 V2V2 V 1 = j  L 1 I 1 + j  MI 2 jL1jL1 jL2jL2 V 2 = j  L 2 I 2 + j  MI 1 _

7. Case II i1i1 i2i2 Coil 1 Coil 2 M L1L1 L2L2 v1v1 v2v2

8. Case 2: jMjM V 1 = j  L 1 I 1 + j  MI 2 V 2 = j  L 2 I 2 + j  MI 1 jL1jL1 jL2jL2 I1I1 I2I2 + + _ _ V1V1 V2V2

9. Circuits with Mutual Inductance

10. Example 1: I1I1 I2I2 22 j8  -j4  j10  j6  6 6  + _ + _ v a (t)v b (t) v a (t) = 50cos(400t + 30) V v b = 80cos(400t – 40) V V a = 50  30 0 VV b = 80  -40 0 V

11. I1I1 I2I2 2  j8  -j4  j10  j6  6  + _ + _ EXAMPLE 1: Continued 50  30 V 80  -40 V Solve for I 1 and I 2 (2 + j10)I 1 + j8I 2 = 50  30 j8I 1 + (j6 – j4 + 6)I 2 = - 80  -40 (2+j10) j8 I 1 50  30 j8 (6+j4) I 2 -80  -40 = Matrix Form Mesh 1 Mesh 2

12. Example 2: 20  0 V + _ 8  j10  -j4  j8  12  6  j5  I1I1 I2I2 j3  Solve for I 1 and I 2

13. Example 2: Continued (8 + j10 + j5 + 6)I 1 - (j5 + 6 + j3)I 2 = 20  0 -(6 + j5 + j3)I 1 + (6 + j5 + j8 – j4 + 12 + j3)I 2 = 0 (14+j15) -(6+j8) I 1 20  0 -(6+j8) (18+j12) I 2 0 = 20  0 V + _ 8  j10  -j4  j8  12  6  j5  I1I1 I2I2 j3  Mesh 1: Mesh 2 Matrix

14. BASIC TRANSFORMER From figure V2=ZL.I2 so : Substitute I2 we have :

15. Reflected impedance Z R Simplify Circuit – one loop circuit

16. SECONDARY TO PRIMARY RATIO CURRENT RATIO VOLTAGE RATIO

17. IDEAL TRANSFORMER

18. n= N2/N1 called turn ratio of transformer Voltage Ratio Current Ratio

19. IDEAL TRANSFORMER LAWS Define the primary variables (I1,V1) to satisfy and the secondary variable to violate the passive sign convention. Then if I1 or I2 point into dotted end while the other point into undotted end, use plus sign in both i-v laws, otherwise use minus sign

20. EXAMPLE

21. IMPEDANCE REFLECTED  REFLECTED LOAD INTO PRIMARY

23. Contoh  Suatu generator menyuplai beban, spesifikasi rating generator tersebut adalah :  V rat = 500 volt S rat = 100 kVA

24. SISTEM PER-UNIT Besaran yang dijadikan basis utama adalah Daya Semu (S) dan Tegangan (V)

26. Ubah komponen dalam Unit

27. Besarnya I sebenarnya untuk tiap area dapat diperoleh dengan mengalikan nilai I perunit dengan nilai I basisnya di area tersebut