coupled circuit

1. 1
CHAPTER 7:
MAGNETICALLY COUPLED
CIRCUIT

2. 2
SUB - TOPICS
 SELF AND MUTUAL INDUCTANCE.
 COUPLING COEFFICIENT (K)
 DOT DETERMINATION

3. 3
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.

4. 4
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.

5. 5
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)
Φ
+
V
_
Fig. 1

6. 6
 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;
 But a change in the flux Φ is caused by a
change in current, i.
Hence;
)1.......(
dt
d
Nv
φ
=
)2.......(
dt
di
di
d
dt
d φφ
=

7. 7
Thus, (2) into (1) yields;
From equation (3) and (4) the self inductance L is
define as;
The unit is in Henrys (H)
)4.......(
or
)3.......(
dt
di
Lv
dt
di
di
d
Nv
=
=
φ
[ ] )5........(H
di
d
NL
φ
=

8. 8
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.

9. 9
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)
Φ12+
V1
_
+
V2
_
Φ11
L2L1
N1 turns N2 turns
Fig. 2

10. 10
 Magnetic flux Φ1 from coil 1 has two
components;
* Φ11 links only coil 1.
* Φ12 links both coils.
Hence; Φ1 = Φ11 + Φ12 ……. (6)
Thus;
Voltage induces in coil 1
)7.......(1
1
1
1
11
11
dt
di
L
dt
di
di
d
Nv ==
φ

11. 11
Voltage induces in coil 2
)8.......(1
21
1
1
12
22
dt
di
M
dt
di
di
d
Nv ==
φ
Subscript 21 in M21
means the mutual
inductance on coil 2
due to coil 1

12. 12
 Case 2:
Same circuit but let current i2 flow in coil 2.
 The magnetic flux Φ2 from coil 2 has two
components:
* Φ22 links only coil 2.
* Φ21 links both coils.
Hence; Φ2 = Φ21 + Φ22 ……. (9)
i2(t)
Φ21
+
V1
_
+
V2
_Φ22
L2L1
N1 turns N2 turns
Fig. 3

13. 13
Thus;
Voltage induced in coil 2
Voltage induced in coil 1
)10.......(2
2
2
2
22
22
dt
di
L
dt
di
di
d
Nv ==
φ
)11.......(2
12
2
2
21
11
dt
di
M
dt
di
di
d
Nv ==
φ
Subscript 12 in M12
means the Mutual
Inductance on coil 1
due to coil 2

14. 14
 Since the two circuits and two current are the
same:
 Mutual inductance M is measured in Henrys
(H)
MMM == 1221

15. 15
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.

16. 16
 k depends on the closeness of two coils, their
core, their orientation and their winding.
 The coefficient of coupling, k is given by;
or
21LL
M
k =
21LLkM =

17. 17
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.

18. 18
Φ12
Φ21
Φ22Φ11
Coil 2Coil 1

19. 19
 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.
 Conversely, Dot convention may also be stated
as follow:
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.

20. 20
 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.

21. 21
 Once the polarity of the mutual voltage is
already known, the circuit can be analyzed using
mesh method.
 Application of the dot convention
 Example 1
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
i1(t)
+
V1
_
+
V2 (t) = M di1/dt
_
L2
L1
M

22. 22
 Example 2
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
i1(t)
+
V1
_
+
V2 (t) = -M di1/dt
_
L2
L1
M

23. 23
 Same reasoning applies to the coil in example 3
and example 4.
 Example 3
 Example 4
i2(t)
+
V1= -M di2/dt
_
+
V2 (t)
_
L2
L1
M
i2(t)
+
V1= M di2/dt
_
+
V2 (t)
_
L2
L1
M

24. 24
Dot convention for coils in series
MLLL 221 ++=
i
L2L1
M
i
(+)
i
L2L1
M
i
(-)
MLLL 221 −+=
Series –
aiding
connection
Series –
opposing
connection

25. 25
Below are examples of the sets of equations
derived from basic configurations involving
mutual inductance
 Circuit 1
Solution:
+
−
M
ja
R2
R3
R1
jb
Vs
I1 I2
)2.......(0)(:IKVL
)1.......()(:IKVL
1232122
21211
=−+++−
=−++
MIIjbRRIR
VMIIjaRR s

26. 26
 Circuit 2
Solution:
+
−
M
ja R2
-jc
R1
jb
Vs
I1 I2
)2.......(0)(:IKVL
)1.......()()(:IKVL
12212
1212111
=−−++−
=+−+−++
MIIjcjbRjbI
VMIIIMjbIIjbjaR s

27. 27
 Circuit 3
Solution:
R2
+
−
M
ja
R1 jb
Vs
I1
I2
)2.......(0)()(:IKVL
)1.......()(:IKVL
1222212
22111
=−−−+++−
=+−+
IIMMIIjbjaRjaI
VMIjaIIjaR s

28. 28
 Circuit 4
Solution:
+
−
M
ja
R2
-jcR1
jbVs
I1
I2
)2.......(02)(:IKVL
)1.......()(:IKVL
222122
221211
=−+++−
=−−+
MIIjbjaRIR
VIRIjcRR s

29. 29
 Circuit 5
Solution:
+
−
M2
ja
R2
jc
R1
-jd
Vs
I1
I2
M1
M3
jb
)2.......(0)()()(:IKVL
)1.......()()()(:IKVL
12213221122122
221121123221211
=−−−−+−++++−
=+−−−−+−+++
IIMIMIMIMIjdjbjcRIjcR
VIMIMIIMIMIjcRIjcjaRR s

30. 30
Example 1
Calculate the mesh currents in the circuit shown below
+
−
4Ω
100V
I1
I2
-j3Ω
j8Ω
j2Ω
5Ωj6Ω

31. 31
Solution
In matrix form;
)1.......(1008)34(
10026)34(:IKVL
21
2211
=−+
=−−+
IjIj
IjIjIj
)2.......(0)185(8
0)(22)145(6:IKVL
21
122212
=++−
=−++++−
IjIj
IIjIjIjIj






=











+−
−+
0
100
1858
834
2
1
I
I
jj
jj

32. 32
The determinants are:
800
08
10034
1800500
1850
8100
8730
1858
834
2
1
j
j
j
j
j
j
j
jj
jj
=
−
+
=∆
+=
+
−
=∆
+=
+−
−+
=∆
AI
AI
°∠=
∆
∆
=∴
°∠=
∆
∆
=∴
197.8
5.33.20
2
2
1
1Hence:

33. 33
Example 2
Determine the voltage Vo in the circuit shown below.
+
−
5Ω
10V
I1
I2
j3Ω
j2Ω
-j4Ω
+
Vo
_
j6Ω

34. 34
Solution
In matrix form;
)1.......(104)55(
102)(26)95(:IKVL
21
121211
=−+
=−−−−+
IjIj
IjIIjIjIj
)2.......(024
0226:IKVL
21
1212
=+−
=++−
IjIj
IjIjIj






=











−
−+
0
10
24
455
2
1
I
I
jj
jj

35. 35
Answer:
[ ]
76.1104.7
hence,
4
or2)(6
or2)(6
76.194.2
88.047.1
2
112
1210
2
1
jV
IjV
IjIIjV
IjIIjV
jI
jI
o
o
o
−=
−=
+−−=
−−=
+=
+=

36. 36
Example 3
Calculate the phasor currents I1 and I2 in the circuit
below.
j6Ωj5Ω
j3Ω
+
−
I1 I2
-j4Ω
12ΩV
012∠

37. 37
Solution
For coil 1, KVL gives
-12 + (-j4+j5)I1 – j3I2 = 0
Or
jI1 – j3I2 = 12
For coil 2, KVL gives
-j3I1 + (12 + j6)I2 = 0
Or
I1 = (12 + j6)I2 = (2 – j4)I2
j3
1
2

38. 38
Substituting into :
(j2 + 4 – j3)I2 = (4 – j)I2 = 12
Or
From eqn. and ,
A14.042.91
j-4
12
I2

∠==
2 1
2
3
3
A49.39-13.01
)14.04(2.91)63.43-(4.472j4)I-(2I 21


∠=
∠∠==