m utual inductance & transformer
DESCRIPTION
M utual 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. M. i 1. i 2. v 1. v 2. L 1. L 2. Coil 2. Coil 1. - PowerPoint PPT PresentationTRANSCRIPT
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