CHAPTER 7:
MAGNETICALLY
COUPLED CIRCUIT
1
OBJECTIVES
• To understand the basic concept of self inductance
and mutual inductance.
• To understand the concept of coupling coefficient
and dot determination in circuit analysis.
2
SUB - TOPICS
7-1 SELF AND MUTUAL INDUCTANCE.
7-2 COUPLING COEFFICIENT (K)
7-3 DOT DETERMINATION
3
7-1 SELF AND MUTUAL
INDUCTANCE
• When two loops with or without contacts between them
affect each other through the magnetic field generated
by one of them, it called magnetically coupled.
• Example: transformer
 An electrical device designed on the basis of the
concept of magnetic coupling.
 Used magnetically coupled coils to transfer energy
from one circuit to another.
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a) Self Inductance
• It called self inductance because it relates the voltage
induced in a coil by a time varying current in the same coil.
• Consider a single inductor with N number of turns when
current, i flows through the coil, a magnetic flux, Φ is
produces around it.
+
i(t)
V
Φ
_
Fig. 1
5
• According to Faraday’s Law, the voltage, v induced in the coil is
proportional to N number of turns and rate of change of the
magnetic flux, Φ;
d
vN
.......(1)
dt
• But a change in the flux Φ is caused by a change in current, i.
d d di
Hence;
dt

di dt
.......(2)
• Thus, (2) into (1) yields;
d di
vN
.......(3) or
di dt
vL
di
.......(4)
dt
• From equation (3) and (4) the self inductance L is define as;
LN
d
di
H........(5)
The unit is in Henrys (H)
6
b) 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 is the ability of one inductor to induce a
voltage across a neighboring inductor.
7
Consider the following two cases:
• Case 1:
two coil with self – inductance L1 and L2 which are in
close proximity which each other (Fig. 2). Coil 1 has N1
turns, while coil 2 has N2 turns.
+
i1(t)
L1
Φ12
L2
V1
_
+
V2
Φ11
N1 turns
Fig. 2
_
N2 turns
• Magnetic flux Φ1 from coil 1 has two components;
* Φ11 links only coil 1.
* Φ12 links both coils.
Hence;
Φ1 = Φ11 + Φ12 ……. (6)
8
Thus, voltage induces in coil 1:
d11 di1
di1
v1  N1
 L1 .......(7)
di1 dt
dt
Voltage induces in coil 2
v2  N 2
d12 di1
di
 M 21 1 .......(8)
di1 dt
dt
Subscript 21 in M21 means the
mutual inductance on coil 2
due to coil 1
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• Case 2:
Same circuit but let current i2 flow in coil 2.(Fig. 3)
+
L1
Φ21
L2 +
V1
_
N1 turns
V2
Φ22
_
i2(t)
Fig. 3
N2 turns
• The magnetic flux Φ2 from coil 2 has two components:
* Φ22 links only coil 2.
* Φ21 links both coils.
Hence; Φ2 = Φ21 + Φ22 ……. (9)
10
Thus; voltage induced in coil 2
d22 di2
di2
v2  N 2
 L2
.......(10)
di2 dt
dt
Voltage induced in coil 1
d21 di2
di2
v1  N1
 M 12
.......(11)
di2 dt
dt
Subscript 12 in
M12 means the
Mutual
Inductance on
coil 1 due to coil 2
Since the two circuits and two current are the same:
M 21  M12  M
Mutual inductance M is measured in Henrys (H)
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7-2 COUPLING COEFFICIENT,
(k)
• It is measure of the magnetic coupling between two coils.
• Range of k : 0 ≤ k ≤ 1
 k = 0 means the two coils are NOT COUPLED.
 k = 1 means the two coils are PERFECTLY COUPLED.
 k < 0.5 means the two coils are LOOSELY COUPLED.
 k > 0.5 means the two coils are TIGHTLY COUPLED.
• k depends on the closeness of two coils, their core, their
orientation and their winding.
• The coefficient of coupling, k is given by;
k
M
L1 L2
or
M  k L1L2
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7-3 DOT DETERMINATION
• Required to determine polarity of “mutual” induced
voltage.
• A dot is placed in the circuit at one end of each of the two
magnetically coupled coils to indicate the direction of the
magnetic flux if current enters that dotted terminal of
the coil.
Fig. 4
13
•
•
•
•
Dot convention is stated as follows:
if a current ENTERS the dotted terminal of one coil, the
reference polarity of the mutual voltage in the second coil is
POSITIVE at the dotted terminal of the second coil.
Alternatively;
if a current LEAVES the dotted terminal of one coil, the
reference polarity of the mutual voltage in the second coil is
NEGATIVE at the dotted terminal of the second coil.
The following dot rule may be used:
i.
when the assumed currents both entered or both leaves a
pair of couple coils by the dotted terminals, the signs on
the L – terms.
ii. if one current enters by a dotted terminals while the
other leaves by a dotted terminal, the sign on the M –
terms will be opposite to the signs on the L – terms.
Once the polarity of the mutual voltage is already known, the
circuit can be analyzed using mesh method.
14
Application of the dot convention is illustrated in the four pairs of
mutual coupled coils. (Fig. a,b,c,d)
The sign of the mutual voltage v2 is
determined by the reference polarity
for v2 and the direction of i1. Since i1
enters the dotted terminal of coil 1
and v2 is positive at the
dotted terminal of coil 2, the mutual
voltage is +M di1/dt. (Fig. a)
Current i1 enters the dotted terminal
of coil 1 and v2 is negative at the
dotted terminal of coil 2. The mutual
voltage is –M di1/dt. (Fig. b)
15
Same reasoning with Fig. a
and fig. b are applies to the
coil in Fig. c and Fig. d.
16
Dot convention for coils in series
Series –
aiding
connection
M
i
i
L1
(+)
L2
L  L1  L2  2M
Series –
opposing
connection
M
i
i
L1
(-)
L2
L  L1  L2  2M
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Below are examples of the sets of equations
derived from basic configurations involving
mutual inductance
M
• Circuit 1
R1
ja
Vs
+

I1
jb
R2
I2
R3
Solution:
KVL I1 : ( R1  R2  ja) I1  MI 2  Vs .......(1)
KVL I 2 :  R2 I1  ( R2  R3  jb) I 2  MI1  0.......(2)
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• Circuit 2
ja
R1
R2
M
Vs
+

I1
jb
I2
-jc
Solution:
KVL I1 : ( R1  ja  jb) I1  jbI 2  M ( I1  I 2 )  MI1  Vs .......(1)
KVL I 2 :  jbI1  ( R2  jb  jc) I 2  MI1  0.......(2)
19
• Circuit 3
R1
jb
M
Vs
+

I1
ja
I2
R2
Solution:
KVL I1 : ( R1  ja) I1  jaI 2  MI 2  Vs .......(1)
KVL I 2 :  jaI1  ( R2  ja  jb) I 2  MI 2  M ( I 2  I1 )  0.......(2)
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• Circuit 4
-jc
R1
Vs
+

I1
R2
ja
I2
M
jb
Solution:
KVL I1 : ( R1  R2  jc) I1  R2 I 2  Vs .......(1)
KVL I 2 :  R2 I1  ( R2  ja  jb) I 2  2MI 2  0.......(2)
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Circuit 5
M3
R1
ja
Vs
+

jb
jc
M1
M2
-jd
I1
R2
I2
Solution:
KVL I1 : ( R1  R2  ja  jc) I1  ( R2  jc) I 2  M 3 I 2  M 1 ( I1  I 2 )  M 1 I1  M 2 I 2  Vs .......(1)
KVL I 2 : ( R2  jc) I1  ( R2  jc  jb  jd ) I 2  M 1 I1  M 2 I 2  M 3 I1  M 2 ( I 2  I1 )  0.......(2)
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Example 1
Calculate the mesh currents in the circuit shown below.
-j3Ω
4Ω
j8Ω
j2Ω
100V
+

I1
j6Ω
I2
5Ω
Solution:
KVL I1 : (4  j 3) I1  j 6 I 2  j 2 I 2  100
(4  j3) I1  j8I 2  100.......(1)
KVL I 2 :  j 6 I1  (5  j14) I 2  j 2 I 2  j 2( I 2  I1 )  0
 j8I1  (5  j18) I 2  0.......(2)
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In matrix form;
4  j3  j8   I1  100
  j8 5  j18  I    0 

 2   
The determinants are:

4  j3
 j8
 j8
5  j18
100
 j8
0
5  j18
1 
2 
Hence:
1
 I1 
 20.33.5 A

2
 I2 
 8.719 A

4  j 3 100
 j8
0
 30  j87
 500  j1800
 j800
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Example 2
Determine the voltage Vo in the circuit shown below.
5Ω
10V
+

j3Ω
j2Ω
I1
+
j6Ω
Vo
_
I2
-j4Ω
Solution:
KVL I1 : (5  j 9) I1  j 6 I 2  j 2( I1  I 2 )  j 2 I1  10
(5  j 5) I1  j 4 I 2  10.......(1)
KVL I 2 :  j 6 I1  j 2 I 2  j 2 I1  0
 j 4 I1  j 2 I 2  0.......(2)
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In matrix form:
5  j 5
  j4

 j 4  I1  10
 



j 2  I 2   0 
Answers:
I1  1.47  j 0.88
I 2  2.94  j1.76
V0  j 6( I1  I 2 )  j 2 I1 or
Vo   j 6( I 2  I1 )  j 2 I1  or
Vo   j 4 I 2
hence,
Vo  7.04  j11.76
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Example 3
Calculate the phasor currents I1 and I2 in the circuit below.
j3Ω
-j4Ω
120 V
+

I1
j5Ω
j6Ω
I2
12Ω
Solution:
For coil 1, KVL gives;
Or
-12 + (-j4+j5)I1 – j3I2 = 0
jI1 – j3I2 = 12
1
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For coil 2, KVL gives;
Or
-j3I1 + (12 + j6)I2 = 0
I1 = (12 + j6)I2 = (2 – j4)I2
j3
2
Substituting 2 into 1 :
(j2 + 4 – j3)I2 = (4 – j)I2 = 12 Or
I2 
12
 2.91 14.04 A
4- j
3
From eqn. 2 and 3 :
I1  (2 - j4)I 2  (4.472  - 63.43 ) (2.9114.04 )
 13.01  - 49.39 A
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