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MAGNETICALLY COUPLED NETWORKS
LEARNING GOALS
Mutual Inductance
Behavior of inductors sharing a common magnetic field
The ideal transformer
Device modeling components used to change voltage and/or
current levels
BASIC CONCEPTS – A REVIEW
Magnetic field
Total magnetic
flux linked by Nturn coil
Ampere’s Law
(linear model)
Faraday’s
Induction Law
Assumes constant L
and linear models!
Ideal Inductor
MUTUAL INDUCTANCE
Overview of Induction Laws
Magnetic
flux
Induced links
on second
coil
( )
2
Total flux linkage
  N (webers)
If linkage is created by a current flowing
through the coils…
  Li (Ampere’s Law)
The voltage created at the terminals of
the components is
vL
di
dt
(Faraday’s Induction Law)
One has the effect of mutual inductance
TWO-COIL SYSTEM
Self-induced
(both currents contribute to flux)
Mutual-induced
Linear model simplifying
notation
THE ‘DOT’ CONVENTION
COUPLED COILS WITH DIFFERENT WINDING CONFIGURATION
Dots mark reference
polarity for voltages
induced by each flux
THE DOT CONVENTION REVIEW
LEARNING EXAMPLE
Currents and voltages follow
passive sign convention

( v 2 ( t ))

 i2 ( t )
 i1 ( t )
Flux 2 induced
voltage has + at dot
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
For other cases change polarities or
current directions to convert to this
basic case
 di 
 di 
v1 ( t )  L1   1   M   2 
 dt 
 dt 
 di 
 di 
 v 2 (t )  M   1   L2   2 
 dt 
 dt 
di1
di
M 2
dt
dt
di
di
v2  M 1  L2 2
dt
dt
v1   L1
LEARNING EXAMPLE
Mesh 1
LEARNING EXAMPLE - CONTINUED
Mesh 2
Voltage Terms
PHASORS AND MUTUAL INDUCTANCE
di1
di
(t )  M 2 (t )
dt
dt
di
di
v2 (t )  M 1 (t )  L2 2 (t )
dt
dt
v1 (t )  L1
Assuming complex
exponential sources
V1  jL1I1  jMI 2
V2  jMI 1  jL2 I 2
Phasor model for mutually
coupled linear inductors
LEARNING EXAMPLE
CASE I
I
 V1 
The coupled inductors can be connected in four different ways.
Find the model for each case
Currents into dots
I
 V2 
V  V1  V2
V1  jL1I  jMI
V2  jMI  jL2 I
V  j( L1  L2  2M ) I  j Leq I
CASE 2
Currents into dots
I
I
 V1 
 V2 
V  V1  V2
V1  j L1I  j MI
V2  j MI  j L2 I
V  j ( L1  2 M  L2 ) I
Leq
Leq  0 imposes a physical constraint
on the value of M
CASE 3
I1
Currents into dots
I2
V 
V 
I  I1  I 2
 I 2  I  I1
V  jL1I1  jMI 2
V  jMI 1  jL2 I 2
V  jL1I1  jM ( I  I1 )
V  jMI 1  jL2 ( I  I1 )
V  j ( L1  M ) I1  jMI
 /( L2  M )
V   j ( L2  M ) I1  jL2 I  /( L1  M )
L1L2  M 2
V  j
I
L1  L2  2 M
CASE 4
( L1  L2  2 M )V  j  M ( L2  M )  L2 ( L1  M ) I
Currents into dots
 I2
I1
V 
 ( V ) 
I  I1  I 2
V  jL1I1  jMI 2
 V  jMI 1  jL2 I 2
L1L2  M 2
V  j
I
L1  L2  2M
LEARNING EXAMPLE
FIND THE VOLTAGEV0
 I2
I1

V1
1. Coupled inductors. Define their
voltages and currents

V2


KVL : 2430  2 I1  V1
VS
KVL : - V2  j 2 I 2  2 I 2  0
MUTUAL INDUCTANCECIRCUIT
V1  j 4 I1  j 2( I 2 )
V0  2I 2
V2  j 2 I1  j 6( I 2 )
VS  (2  j 4) I1  j 2 I 2
 / j2
0   j 2 I1  (2  j 2  j 6) I 2  / 2  j 4


j 2VS  4  (2  j 4)2 I 2
j 2VS
2VS
j
I2 


 8  j16  j 16  8 j
VS
2430
V0  2 I 2 
 5.373.42

4  2 j 4.4726.57
2. Write loop equations
in terms of coupled
inductor voltages
3. Write equations for
coupled inductors
4. Replace into loop equations
and do the algebra
LEARNING EXAMPLE
I1  I 2
Write the mesh equations
3. Write equations for coupled inductors
I 2  I3

V1

V2


V1  jL1 ( I1  I 2 )  jM ( I 2  I 3 )
V2  jM ( I1  I 2 )  jL2 ( I 2  I 3 )
4. Replace into loop equations and
rearrange terms
1. Define variables for coupled inductors

1 
 I1
V   R1  jL1 
j

C

1

1 
 I 2  jMI 3
  jL1  jM 
j

C

1
2. Write loop equations in terms of coupled
inductor voltages

V  R1 I1  V1 
I1  I 2
jC1
 V1  R2 I 2  V2  R3 ( I 2  I 3 ) 
 V2 
1 
 I1
0   jL1  jM 
j

C

1
I 2  I1
0
jC1
I3
 R4 I 3  R3 ( I 3  I 2 )  0
jC 2

1 
 I 2
  jL2  jM  R2  jM  jL2  R3 
j

C

1
  jM  jL2  R3 I 3
0   jMI 1   jL2  jM  R3 I 2


1
  jL2 
 R4  R3  I 3
j C 2


LEARNING EXAMPLE
DETERMINE IMPEDANCE SEEN BY THE SOURCE Z i 
Z S  3  j1()
I1
Z L  1  j1()

V1

V2


 I2
jL1  j 2()
jL2  j 2()
jM  j1()
1. Variables for coupled inductors
2. Loop equations in terms of coupled
inductors voltages
Z S I1  V1  VS
 V2  Z L I 2  0
3. Equations for coupled inductors
V1  jL1I1  jM ( I 2 )
V2  jMI 1  jL2 ( I 2 )
4. Replace and do the algebra
 Z S  jL1 I1  ( jM ) I 2  VS
VS
I1
 /( Z L  jL2 )
 ( jM ) I1  ( Z L  jL2 ) I 2  0  / jM
( Z
S

 jL1 )( Z L  jL2 )  ( jM ) 2 I1
 ( Z L  jL2 )VS
VS
( jM )2
Zi 
 ( Z S  jL1 ) 
I1
Z L  jL2
( j1) 2
1
1 j
Z i  3  j3 
 3  j3 

1  j1
1 j 1 j
1 j
 3.5  j 2.5()
2
Z i  4.3035.54()
Z i  3  j3 
THE IDEAL TRANSFORMER
1  N1
2  N 2
Insures that ‘no magnetic flux
d 
goes astray’
v1 (t )  N1 (t )  v
N
dt
 1  1 First ideal transformer

d
v2 (t )  N 2
(t ) v2 N 2 equation
dt 
v1 (t )i1 (t )  v2 (t )i2 (t )  0 Ideal transformer is lossless
i1
N
  2 Second ideal transformer
i2
N1 equations
Circuit Representations
v1 N1

;
v2 N 2
i1 N 2

i2 N1
REFLECTING IMPEDANCES
For future reference
*
 N  N 
S1  V1I1*  V2 1  I 2 2   V2 I 2*  S2
 N 2  N1 
n
V1 N1

(both  signs at dots)
V2 N 2
Phasor equations for ideal transformer
I1 N 2

(Current I2 leaving transformer)
I 2 N1
V2  Z L I 2 (Ohm's Law)
N
N
V1 2  Z L I1 1
N1
N2
N2
 turns ratio
N1
V2
n
I1  nI 2
V1 
ZL
n2
S1  S2
Z1 
2
N 
V1   1  Z L I1
 N2 
2
N 
V1
 Z1   1  Z L
I1
 N2 
Z1  impedance, Z L , reflected
into the primary side
LEARNING EXAMPLE
Determine all indicated voltages and currents
n  1/ 4  0.25
I1 
1200
1200

 2.33  13.5
50  j12 51.4213.5
V1  Z1I1  (32  j16)  2.33  13.5
 83.3613.07
Strategy: reflect impedance into the
primary side and make transformer
“transparent to user.”
ZL
Z1 
n2
ZL
CAREFUL WITH POLARITIES AND
CURRENT DIRECTIONS!
I2  
I1
 4 I1 (current into dot)
n
V2  nV1  0.25V1 ( is opposite to dot)
Z1  32  j16
USING THEVENIN’S THEOREM TO SIMPLIFY CIRCUITS WITH IDEAL TRANSFORMERS
Replace this circuit with its Thevenin
equivalent
I2  0 
  I1  0  V1  VS1
I1  nI 2 
V1  VS1 
  VOC  nVS1
V2  nV1 
To determine the Thevenin impedance...
Reflect impedance into
secondary
ZTH
ZTH  n2 Z1
Equivalent circuit with transformer
“made transparent.”
One can also determine the Thevenin
equivalent at 1 - 1’
USING THEVENIN’S THEOREM: REFLECTING INTO THE PRIMARY
ZTH 
Z2
n2
Find the Thevenin equivalent of
this part
In open circuit I1  0 and I 2  0
VOC 
Equivalent circuit reflecting
into primary
VS 2
n
Thevenin impedance will be the the
secondary mpedance reflected into
the primary circuit
Equivalent circuit reflecting
into secondary
LEARNING EXAMPLE
Draw the two equivalent circuits
n2
Equivalent circuit reflecting
into secondary
Equivalent circuit reflecting
into primary
LEARNING EXAMPLE
Find Vo
Thevenin equivalent of this part
To compute Vo is better to reflect into secondary
But before doing that it is better to simplify the primary using Thevenin’s Theorem

Vd

ZTH
 j16 8  j8  j16
 2

4  j4
4  j4
ZTH 
2  j6 1  j 8  j 4


1 j 1 j
2
VOC  Vd  4  90
Vd 
 j4
24  90
240 
4  j4
1 j
VOC  14.42 33.69(V )
ZTH  2  (4 ||  j 4)
ZTH  4  j 2()
This equivalent circuit is now transferred to
the secondary
LEARNING EXAMPLE (continued…)
n2
Thevenin equivalent of primary side
Equivalent circuit reflecting
into secondary
Circuit with primary transferred to secondary
Vo 
2
2  28.84  33.69
28.84  33.69 
20  j 5
20.62  14.04
LEARNING EXAMPLE
Find I1, I 2 ,V1,V2
2V1  V2  2 I1  100
 I1  50
 V1  (1  j )V2  2 I 2  0
n2
V2  2V1
I1  2 I 2
I 2  2.50
 V1  (1  j )( 2V1 )  50
V1 
Nothing can be transferred. Use
transformer equations and circuit
analysis tools
Phasor equations for ideal transformer
V2
n
I1  nI 2
V  100 V1  V2
@ Node 1 : 1

 I1  0
2
2
V V V
@Node 2 : 2 1  2  I 2  0
2
j2
V1 
4 equations in 4 unknowns!
50
50

 563.43
1  j 2 2.24  63.43
V2  2 563.43