Transcript magnetically coupled circuit

CHAPTER 7:
MAGNETICALLY COUPLED
CIRCUIT
1
SUB - TOPICS
 SELF AND MUTUAL INDUCTANCE.
 COUPLING COEFFICIENT (K)
 DOT DETERMINATION
2
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.
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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)
Φ
V
_
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.
Hence;
d d di

.......(2)
dt
di dt
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Thus, (2) into (1) yields;
d di
vN
.......(3)
di dt
or
di
v  L .......(4)
dt
From equation (3) and (4) the self inductance L is
define as;
d
H ........(5)
LN
di
The unit is in Henrys (H)
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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.
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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
V1
_
L2
+
V2
Φ11
_
N1 turns
N2 turns
Fig. 2
9
 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
d11 di1
di1
v1  N1
 L1
.......(7)
di1 dt
dt
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Voltage induces in coil 2
d12 di1
di1
v2  N 2
 M 21 .......(8)
di1 dt
dt
Subscript 21 in M21
means the mutual
inductance on coil 2
due to coil 1
11
 Case 2:
Same circuit but let current i2 flow in coil 2.
+
Φ21
L1
L2 +
V1
_
N1 turns
V2
Φ22
i2(t)
_
N2 turns
Fig. 3
 The magnetic flux Φ2 from coil 2 has two
components:
* Φ22 links only coil 2.
* Φ21 links both coils.
Hence; Φ2 = Φ21 + Φ22 ……. (9)
12
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
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 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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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.
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 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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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.
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Φ12
Φ21
Φ11
Coil 1
Φ22
Coil 2
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 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.
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
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.
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 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
M
i1(t)
+
V1
_
+
L1
L2
V2 (t) = M di1/dt
_
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
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 Example 2
M
i1(t)
+
V1
_
+
L1
L2
V2 (t) = -M di1/dt
_
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
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 Same reasoning applies to the coil in example 3
and example 4.
 Example 3
M
i2(t)
+
+
V1= -M di2/dt
L1
L2
_
V2 (t)
_
 Example 4
M
i2(t)
+
V1= M di2/dt
_
+
L1
L2
V2 (t)
_
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Dot convention for coils in series
M
i
i
L1
(+)
Series –
aiding
connection
L2
L  L1  L2  2M
M
Series –
opposing
connection
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)
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 Circuit 3
R1
jb
M
Vs
+

I1
R2
ja
I2
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
ja
I2
R2
jb
M
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
jb
jc
Vs
+

M2
M1
-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
5Ω
j6Ω
I2
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Solution
KVL I1 : (4  j 3) I1  j 6 I 2  j 2 I 2  100
(4  j 3) 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)
In matrix form;
4  j3  j8   I1  100
  j8 5  j18  I    0 

 2   
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The determinants are:
4  j3
 j8

 30  j87
 j8 5  j18
100
 j8
1 
 500  j1800
0 5  j18
4  j 3 100
2 
 j800
 j8
0
Hence:
1
 I1 
 20.33.5 A

2
 I2 
 8.719 A

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Example 2
Determine the voltage Vo in the circuit shown below.
5Ω
j3Ω
+
10V
+

j2Ω
I1
j6Ω
Vo
-j4Ω
I2
_
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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)
In matrix form;
5  j5  j 4  I1  10
 


  j4

j 2  I 2   0 

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Answer:
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
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Solution
For coil 1, KVL gives
-12 + (-j4+j5)I1 – j3I2 = 0
Or
1
jI1 – j3I2 = 12
For coil 2, KVL gives
-j3I1 + (12 + j6)I2 = 0
Or
I1 = (12 + j6)I2 = (2 – j4)I2
j3
2
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Substituting 2 into 1 :
(j2 + 4 – j3)I2 = (4 – j)I2 = 12
Or
12
I2 
 2.91 14.04 A
4- j
From eqn.
2
and
3
3
,
I1  (2 - j4)I 2  (4.472  - 63.43 ) (2.9114.04 )


 13.01  - 49.39 A

38