Mutual Inductance & TransformerElectric Circuit

Chairul Hudaya, M.ScElectric Power and Energy Studies (EPES)

Department of Electrical EngineeringUniversitas Indonesia

http://www.ee.ui.ac.id/epes/hudaya

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

i1 i2

Coil 1 Coil 2

M

L1 L2

v1 v2

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 leavesthrough the other, they produce opposing flux. In this case mutual flux linkages subtract from self linkages.(Case II)

i1 i2

Coil 1 Coil 2

M

L1 L2

Case I

1 1 1 2 1 21 1

( ) d d L i Mi di di

v L Mdt dt dt dt

2 2 2 1 2 12 2

( ) d d L i Mi di di

v L Mdt dt dt dt

v1 v2

Case 1:

jM

I1 I2

• •

+ +

_V1 V2

V1 = jL1I1 + jMI2

jL1 jL2

V2 = jL2I2 + jMI1

_

Case IIi1 i2

Coil 1 Coil 2

M

L1 L2

1 1 1 2 1 21 1

( ) d d L i Mi di di

v L Mdt dt dt dt

2 2 2 1 2 12 2

( ) d d L i Mi di di

v L Mdt dt dt dt

v1 v2

Case 2:

jM

V1 = jL1I1 + jMI2

V2 = jL2I2 + jMI1

jL1jL2

I1 I2

•

•

+

+_

_

V1 V2

Circuits with Mutual Inductance

1 1 1 1 1 2 2( ) { ( ) (0 )} { ( ) (0 )} v s sL I s L i sMI s Mi

2 1 1 2 2 2 2( ) { ( ) (0 )} { ( ) (0 )} v s sMI s Mi sL I s L i

1 1 1 2( ) ( ) ( ) v s sL I s sMI s

2 1 2 2( ) ( ) ( ) v s sMI s sL I s

Example 1:

I1 I2

•

•

2j8 -j4

j10 j6

6

+_

+

_va(t) vb(t)

va(t) = 50cos(400t + 30) V vb = 80cos(400t – 40) V

Va = 50300 V Vb = 80-400 V

I1 I2

•

•

2 j8 -j4

j10 j6

6

+

_

+

_

EXAMPLE 1: Continued

5030 V 80-40 V

Solve for I1 and I2

(2 + j10)I1 + j8I2 = 5030

j8I1 + (j6 – j4 + 6)I2 = - 80-40

(2+j10) j8 I1 5030

j8 (6+j4) I2 -80-40=Matrix Form

Mesh 1

Mesh 2

Example 2:

200 V

+

_

8 j10 -j4 j8

12

6

j5

I1 I2

• •j3

Solve for I1 and I2

Example 2: Continued

(8 + j10 + j5 + 6)I1 - (j5 + 6 + j3)I2 = 200

-(6 + j5 + j3)I1 + (6 + j5 + j8 – j4 + 12 + j3)I2 = 0

(14+j15) -(6+j8) I1 200

-(6+j8) (18+j12) I2 0

=

200 V

+

_

8 j10 -j4 j8

12

6

j5

I1 I2

• •j3

Mesh 1:

Mesh 2

Matrix

BASIC TRANSFORMER

From figure V2=ZL.I2 so :

Substitute I2 we have :

Reflected impedance ZR

Simplify Circuit – one loop circuit

SECONDARY TO PRIMARY RATIO

CURRENT RATIO

VOLTAGE RATIO

IDEAL TRANSFORMER

NLi

NV

LiV

l

iANBA

l

iNHB

l

iNH

NLi 2

2

.

Nl

AL

Nl

iALi

l

iANNLi

NLi

22

1

22

1

2 nN

N

L

L n= N2/N1 called turn ratio of transformer

Voltage Ratio

Current Ratio

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

EXAMPLE

IMPEDANCE REFLECTED

REFLECTED LOAD INTO PRIMARY

1 2 2

4LZZn n

2

2 2

41

10nZ

Contoh

Suatu generator menyuplai beban , spesifikasi rating generator tersebut adalah :

V rat = 500 volt S rat = 100 kVA

G

1 : 100 50 : 1Zline = 50 + j100

4520loadZ

I gen I line I load

1 2

3

SISTEM PER-UNIT

basisNilai

sebenarnyaNilaiunitperNilai

basebasebase IVS base

basebase V

SI

base

basebase S

VZ

2

Besaran yang dijadikan basis utama adalahDaya Semu (S) dan Tegangan (V)

G

1 : 100 50 : 1Zline = 50 + j100

4520loadZ

I gen I line I load

1 2

3

AV

VA

V

SI

basis

basisbasis 200

500

000.100

1

1

5,2200

500

1

1

2

11 A

V

I

V

S

VZ

basis

basis

basis

basisbasis

kVa

VV basisbasis 50

01,0

50012

AV

VAI basis 2

000.50

000.1002

kA

VZ basis 25

2

000.502

VV

a

VV basisbasis 1000

50

000.5023

AV

VAI basis 100

1000

000.1003

10100

10003 A

VZ basis

kVAS

VV

basis

basis

100

500

Ubah komponen dalam Unitpu

VVgen

01

480

0480puj

jpuZline 004,0002,0

000.25

10050

pupuZload

452

10

4520

G

puj 004,0002,0

pu452I II III

I genI line

I load

pu 45,04-0,499

04,45004,2

1

)414,1414,1()004,0002,0(

1

jjZtot

VIloadIlineIgen

pu

pu

e45,04amper-0,998 2 . 04,45499,0. 2 basispulineactualline iiI

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