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LITAR ELEKTRIK II
EET 102/4
SILIBUS LITAR ELEKTRIK
II
Mutual Inductance
 Two port Network
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Pengenalan Jelmaan Laplace
Kaedah Jelmaan Laplace Dlm Analisis Litar
Sambutan Frekuensi Litar AC
Siri Fourier
Jelmaan Fourier
MUTUAL INDUCTANCE
Self inductance
 Concept of mutual inductance
 Dot convention
 Energy in a coupled circuit
 Linear transformer
 Ideal transformer

MUTUAL INDUCTANCE
INTRODUCTION

magnetically coupled

When two loops with or without contacts
between them, affect each other through
magnetic field generated by one of them – they
are said to be magnetically coupled.
Example of device using this concepttransformer.

Transformer



Use magnetically coupled coils to transfer
energy from one circuit to another.
Key circuit element where it is used for stepping
down or up ac voltages or currents.
Also used in electronic circuits such as radio and
tv receiver.
Consider a single inductor with N turns.
When current i, flow through coil, magnetic flux 
is produced around it.
Faraday’s Law
 Induced
voltage, v in the coil is
proportional to number of turns N and
the time rate of change of magnetic
flux, .
d
vN
dt
we know that the flux  is
produce by current i, thus any change
in the current will change in flux  as
well.
 But
d di
di
vN
vL
@
di dt
dt
 The
inductance L of the inductor is
thus given by
d
LN
di
Self-Inductance
Self Inductance
 Inductance
that relates the induced
voltage in a coil with a time-varying
current in the same coil.
Mutual Inductance
 When
two inductors or coils are in
close proximity to each other,
magnetic flux caused by current in
one coil links with the other coil,
therefore producing the induced
voltage.
Mutual Inductance

Magnetic flux 1 originating from coil 1 has
2 components:
1  11  12

Since entire flux 1 links coil 1, the voltage
induced in coil 1 is:
d1
v1  N1
dt

Only flux 12 links coil 2, so the voltage
induced in coil 2 is:
d12
v2  N 2
dt

As the fluxes are caused by current i1
flowing in coil 1, equation v1 can be
written as:
Self
d1 di1
di1
v1  N1
 L1
inductance of
di1 dt
dt
coil 1

Similarly for equation v2:
di1
d12 di1
 M 21
v2  N 2
dt
di1 dt
Mutual inductance of coil 2
With respect to coil 1
Coil 2

Magnetic flux 2 comprises of 2
components:
2  21  22
entire flux 2 links coil 2, so the voltage
induced in coil 2 is:
 The
d2
d2 di2
di2
v2  N 2
 N2
 L2
dt
di2 dt
dt
Self-inductance of coil 2

Since only flux 21 links with coil 1, the
voltage induced in coil 1 is:
d21
d21 di2
di2
v1  N1
 N1
 M 12
dt
di2 dt
dt
Mutual inductance of coil 1
with respect to coil 2
 For
simplicity, M12 and M21 are
equal:
M12  M 21  M
Mutual inductance between
two coils
Reminder


Mutual coupling exists when inductors or coils
are in close proximity and circuit are driven by
time-varying sources.
Mutual inductance is the ability of one inductor
to induce voltage across a neighboring inductor,
measured in henrys (H).
Dot Convention
A
dot is placed in the circuit at one end
of each of the two magnetically
coupled coils to indicate the direction
of magnetic flux if current enters that
dotted terminal of the coil.
Dot convention is stated as follows:


If a current enters the dotted terminal of one
coil, the reference polarity of mutual voltage
in second coil is positive at dotted terminal of
second coil.
If a current leaves the dotted terminal of one
coil, the reference polarity of mutual voltage
in second coil is negative at dotted terminal
of second coil.
Dot convention for coils in series
L  L1  L2  2M
L  L1  L2  2M
Example 1
Example 1
Coil 1:
di1
di2
v1  i1 R1  L1
M
dt
dt
Coil 2:
di2
di1
v2  i2 R2  L2
M
dt
dt
In frequency domain..
V1  ( R1  jL1 ) I1  jMI 2
V2  jMI1  ( R2  jL2 ) I 2
Example 2
Example 2
V  ( Z1  jL1 ) I1  jMI 2
0   jMI1  ( Z L  jL2 ) I 2
Example 3
Solution..

For coil 1, we use KVL:
 12  ( j 4  j5) I1  j3I 2  0
jI1  j3I 2  12

For coil 2,
 j3I1  (12  j 6) I 2  0
(12  j 6) I 2
I1 
 (2  j 4) I 2
j3

Substitute equation 1 into 2:
( j 2  4  j3) I 2  (4  j) I 2  12
12
o
I2 
 2.9114.04 A
4 j

Solve for I1:
I1  (2  j 4) I 2
 (4.372  63.43 )( 2.9114.04 )
o
 13.01  49.39 A
o
o
Energy in a coupled circuit

Energy stored in an inductor is given by:
1 2
w  Li
2

Now, we want to determine energy stored in
magnetically coupled coils.
Circuit for deriving energy stored in
a coupled circuit

Power in coil 1:
di1
p1 (t )  v1i1  i1 L1
dt

Energy stored in coil 1:
w1   p1dt  L1 
I1
0
1
2
i1di1  L1 I1
2

Maintain i1 and we increase i2 to I2. So,
the power in coil 2 is:
p2 (t )  i1v1  i 2v 2
di2
 i2 v2
 i1M 12
dt
di2
di2
i 2 L2
 I1M 12
dt
dt
 Energy
stored in coil 2:
w2   p2 dt
I2
I2
 M 12 I1  di2  L2  i2 di2
0
1
2
 M 12 I1 I 2  L2 I 2
2
0

Total energy stored in the coils when
both i1 and i2 have reached constant
values is:
w  w1  w2
1
1
2
2
 L1 I1  L2 I 2  M 12 I1I 2
2
2
 Since
M12=M21=M, thus
1
1
2
2
w  L1 I1  L2 I 2  MI1 I 2
2
2

Generally, energy stored in magnetically
coupled circuit is:
1
1
2
2
w  L1 I1  L2 I 2  MI1 I 2
2
2
Coupling coefficient, k

A measure of the magnetic coupling
between two coils; 0 ≤ k ≤ 1
M
k
L1 L2
Linear Transformer




Transformer is generally a four-terminal device
comprising two or more magnetically coupled
coils.
Coil that is directly connected to voltage source
is primary winding.
Coil connected to the load is called secondary
winding.
R1 and R2 included to calculate for losses in
coils.
Linear Transformer
Primary winding
Secondary winding
 Obtain
input impedance, Zin as seen
from source because Zin governs the
behaviour of primary circuit.
 Apply KVL to the two loops:
V  ( R1  jL1 ) I1  jMI 2
0   jMI1  ( R2  jL2  Z L ) I 2

Input impedance Zin:
V
Z in 
I1
 M
 R1  jL1 
R2  jL2  Z L
2
ZR
2
Equivalent circuit of linear
transformer
Equivalent T circuit
Equivalent ∏ circuit
Voltage-current relationship for
primary and secondary coils give
the matrix equation:
V1   jL1

V   jM
 2 
jM   I1 



jL2   I 2 
 By
matrix inversion, this can be
written as:
L2

 I1   j ( L1 L2  M 2 )
I   

M
 2 
 j ( L1 L2  M 2 )
M

2  V
j ( L1 L2  M )  1 
 
L1
 V2 
2
j ( L1 L2  M ) 
Matrix equation for equivalent T
circuit:
V1   j ( La  Lc
V    jL
c
 2 
jLc
  I1 



j ( Lb  Lc )  I 2 

If T circuit and linear circuit are the
same, then:
La  L1  M
Lb  L2  M
Lc  M
For ∏ network, nodal analysis gives
the terminal equation as:
1
 1

 I1   jLA jLC
I   
1

 2


jLC
1


jLC  V1 
 
1
1  V2 

jLB jLC 
Equating terms in admittance matrices
of above, we obtain:
L1 L2  M
LA 
L2  M
2
L1 L2  M
LB 
L1  M
L1 L2  M
LC 
M
2
2
IDEAL TRANSFORMER
Properties of ideal transformer:



Coils have very large reactances (L1, L2, M→∞)
Coupling coefficient is equal to unity (k=1)
Primary and secondary winding are lossless
(R1=0=R2)
Ideal transformer is a unity-coupled, lossless
transformer where primary and secondary
coils have infinite self-inductance.
Transformation ratio

We know that:
d
v1  N1
dt

d
v2  N 2
dt
Divide v2 with v1, we get:
v2 N 2

n
v1 N1

Energy supplied to the primary must equal
to energy absorbed by secondary since no
losses in ideal transformer.
v1i1  v2i2

Transformation ratio is:
I1 V2
 n
I 2 V1
Types of transformer:

Step-down transformer
One whose secondary voltage is less than its
primary voltage.

Step-up transformer
One whose secondary voltage is greater than its
primary voltage.
Typical circuits in ideal
transformer
Complex Power

From:

Complex power in primary winding for ideal txt:
V2
V1  @ I1  nI 2
n
V2
*
*
S1  V I  (nI 2 )  V2 I 2  S 2
n
*
1 1
Input impedance

We know that:

Since V2 / I2 = ZL , thus
V1 1 V2
Z in   2
I1 n I 2
ZL
Z in  2
n
Reflected
impedance
Example

Find I1 dan I2 for given circuit:
Solution…
 1st:
Find input impedance
2
Z in  10  2
n
ZR
1
n
3
Z in  10  18  28

Therefore, solve for I1
140
I1 
 0.5 A
28
o
I1
0.5
I2  
 1.5 A
n (  13 )
THE END