EE 205 Dr. A. Zidouri
Electric Circuits II
More about Mutual Inductance
Lecture #22
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o
o
o
o
Mutual Inductance in Terms of Self Inductance
Mesh Current Analysis of magnetically coupled Circuits
Procedure for Determining Dot Markings
Energy Calculation
After finishing this lecture you should be able to:
¾
¾
¾
¾
Analyze Circuits with Mutual Inductance
Determine the Coefficient of Coupling
Determine the Dot Markings of neighboring Coils
Perform Energy Calculations in Magnetically Coupled Circuits
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EE 205 Dr. A. Zidouri
Mutual Inductance in Terms of Self Inductance
We have seen that
L1 = N12p1
L2 = N22p2
L1L 2
Therefore
But for linear system,
= N12N22p1p 2 = N12N22 (p11p 21 )(p 22p12 )
p21 = p12 ⇒
⎛ p ⎞⎛ p ⎞
L1L 2 = N12N22p12 ⎜ 1+ 11 ⎟⎜ 1+ 22 ⎟
⎝ p12 ⎠⎝ p12 ⎠
⎛ p ⎞⎛ p ⎞
= M2 ⎜ 1+ 11 ⎟⎜ 1+ 22 ⎟
⎝ p12 ⎠⎝ p12 ⎠
⎛ p11 ⎞⎛ p22 ⎞
1
1+
1+
⎜
⎟⎜
⎟
Replacing the positive quantity
by 2 gives a more meaningful expression
k
⎝ p12 ⎠⎝ p12 ⎠
M2 = k 2L1L 2
or
M = k L1L 2
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EE 205 Dr. A. Zidouri
k is called the Coefficient of Coupling and must lie between 0 and 1:
0 ≤k ≤1
When
Φ12 = Φ12 = 0 ⇒ p12 = 0
1
⇒ 2 = ∞,or k = 0
k
If no flux linkage between the coils, obviously
which presents the Ideal State.
M = 0 and k = 1, when Φ11 = Φ22 = 0 ,
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EE 205 Dr. A. Zidouri
Mesh Current Analysis
• To solve the following example 32-1, Let us reproduce the Mesh Current Equations that describe the
following Circuit seen in Lec31:
−
+
di
di
L
dt
dt
+
−
M
+
L2
−
di2
di
M 1
dt
dt
−
+
di1
di
− M 2 = 0 (Mesh 1)
dt
dt
di
di
R2i2 + L2 2 − M 1 = 0
(Mesh 2)
dt
dt
−vg + R1i1 + L1
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EE 205 Dr. A. Zidouri
Example 32-1
¾ Write a Set of Mesh-Current Equations that describe the Circuit in the following Figure:
¾ What is the Coefficient of Coupling of the Magnetically Coupled Coils?
d (i1 + i3 )
di
− 4.5 2 = 0
dt
dt
d (i + i )
di
6(i2 + i3 ) + 8(i2 − i1 ) + 4 2 − 4.5 2 3 = 0
dt
dt
−vg + 8(i1 − i2 ) + 9
− 4.5
−
di2
dt
+
d (i1 + i2 )
dt
+
20 i3 + 6( i2 + i3 ) + 4
4.5
k=
d ( i1 + i3 )
di2
di
+9
− 4.5 2 = 0
dt
dt
dt
(Mesh 1)
(Mesh 2)
(Mesh 3)
M
4.5
4.5
=
=
= 0.75
6
L1L 2
4×9
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EE 205 Dr. A. Zidouri
Procedure for Determining Dot Markings
The following six steps applied to the following figure determine a set of Dot Markings.
1) Arbitrarily Select one Terminal, (Say Terminal
D) of one coil and mark it with a Dot
2) Assign a current into the dotted terminal and
label it iD
3) Use the Right–Hand rule to determine the
direction of the magnetic field established by iD
inside the coupled coils and label this as ΦD
4) Arbitrarily pick one terminal of the second coil
(Say Terminal A) and Assign a current into it
(Show this as iA)
5) Use the Right–Hand rule to determine the
direction of the magnetic field established by iD
inside the coupled coils and label this as ΦA
6) Compare the directions of the two fluxes ΦD
and ΦA. If the fluxes have the same reference
direction, place a Dot on the terminal of the
second coil where the test current (iA) enters.
A
(Step 6)
C
(Step 5)
ΦA
(Step 4)
ΦD
B
(Step 3)
(Step 2)
D
Arbitrarily Dotted Terminal
(Step 1)
Fig.32-2 Set of Coils Showing Steps for Determining
a Set of Dot Markings
In this example the fluxes ΦD and ΦA have the
same reference direction, therefore a Dot goes on
terminal A.
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EE 205 Dr. A. Zidouri
Example 32-2
¾ The polarity markings on two coils are to be
determined experimentally. The experimental
setup is shown in Fig.32-3. When the switch
is opened, the dc Voltmeter kicks upscale.
According to the polarity and dot marking
shown, where should we place the other dot?
Answer:
9 The lower terminal of the unmarked coil
has the same instantaneous polarity as the
dotted terminal.
9 Therefore, place a dot on the lower
terminal of the unmarked coil.
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EE 205 Dr. A. Zidouri
Energy Calculations
For linear magnetic coupling
1 M12 = M21
2 M = k L1L 2
Consider the following Circuit
Case 1:
Increase i1 from 0 to I1, hold i1 = I1 and i2 = 0
∫
w1
0
I1
dw = L1 ∫ di1
W1 =
0
1 2
L1I1
2
Case 2:
Now we hold i1 = I1 and Increase i2
from 0 to I2
p = I1M12
di 2
+ i 2v 2
dt
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EE 205 Dr. A. Zidouri
Energy Calculations (cont)
The total energy stored in the pair of coils when i2 = I2
∫
w
w1
I2
I2
0
0
dw = ∫ I1M12 di 2 + ∫ L2 i 2 di 2
1 2
L2 I2 + M12 I1I2
2
1
1
W = L1I12 + L2 I22 + M12 I1I2
2
2
W = W1 +
If we reverse the procedure – That is, if we first increase i2 from zero to I2 and increase i1 from zero to
I1- The total energy stored is:
W=
1 2 1 2
L1I1 + L2 I2 + M21I1I2
2
2
At any instant of time, the total energy stored in the coupled coils is:
w(t) =
1 2 1 2
L1i1 + L2 i 2 + Mi1i 2 Assuming that both coil currents entered polarity marked terminals
2
2
In general:
w(t) =
1 2 1 2
L1i1 + L2 i 2 ± Mi1i 2
2
2
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EE 205 Dr. A. Zidouri
Example 32-3
Consider the circuit in Fig.32-5. Determine the coupling coefficient. Calculate the energy stored in the
o
coupled inductors at time t = 1s if v = 60 cos 4t + 30 V .
(
)
1
F
16
Solution:
The coupling coefficient is
k=
M
2.5
=
= 0.56
L1L 2
20
¾ Means that the inductors are tightly
coupled.
¾ To find the energy stored, we need to
obtain the frequency domain equivalent of
the circuit
(
)
60 cos 4t + 30o V ⇒ 60∠30o , ω = 4 rad
5H ⇒ j ω L1 = j 20Ω
2.5H ⇒ j ω M = j 10Ω
4H ⇒ j ω L 2 = j 16Ω
1
1
= − j 4Ω
F⇒
16
j ωC
s
The frequency domain equivalent is
shown in Fig.32-6 below:
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EE 205 Dr. A. Zidouri
Solution (Cont):
We now apply mesh analysis
(10 + j 20) I1 + j10 I 2 = 60∠30o
j10 I1 + ( j16 − j 4) I 2 = 0
Or
I1 = −1.2 I 2
This yields
I 2 ( −12 − j14) = 60∠30o ⇒ I 2 = 3.254∠160.6o A
I1 = −1.2 I 2 = 3.905∠ − 19.4o A
In the time domain
(
)
= 3.254 cos ( 4t + 160.6 )
i1 = 3.905 cos 4t − 19.4o ,
i2
o
At time t=1s, 4t= 4rad=229.2º,
(
= 3.254 cos ( 229.2
)
i1 = 3.905 cos 229.2o − 19.4o = −3.389 A,
i2
o
)
+ 160.6o = 2.824 A
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EE 205 Dr. A. Zidouri
Solution (Cont):
The total energy stored in the coupled inductors is
w=
1 2 1 2
L1i1 + L2i2 + Mi1i2
2
2
1 2 1 2
L1i1 + L2i2 + Mi1i2
2
2
1
1
= (5)( −3.389) 2 + (4)(2.824) 2 + 2.5( −3.389)(2.824) = 20.73 J
2
2
w=
60∠30o V
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EE 205 Dr. A. Zidouri
Self Test:
™ When k equals zero, two inductors in series combine by
a) addition
b) subtraction
c) product-over-sum
d) multiplication
answer: a
™ Two parallel inductors with no mutual inductance combine by
a) addition
b) subtraction
c) product-over-sum
d) multiplication
answer: c
™ If N=200 and a rate of change of flux is 0.4 Wb/s. The voltage induced in the inductor is?
a) 50
b) 80
c) 200
d) 500
answer: b
™ Tow inductors are wound on a ferromagnetic core. The first Inductor has N1=200 and a rate of
change of flux is 0.4 Wb/s. If the second Inductor has N2=125 turns and the coefficient of
coupling is unity, how much voltage is induced in the second inductor?
a) 50
b) 80
c) 200
d) 500
answer: a
™ The self-inductances of the coils in Fig.32-4 are L1= 18 mH and L2 = 32 mH. If the coefficient of
coupling is 0.85, calculate the energy stored in the system in millijoules when
a) i1 = 6 A, i2 = 9 A; b) i1 = -6 A, i2 = -9 A; c) i1 = -6 A, i2 = 9 A;
d) i1 = 6 A, i2 = -9 A.
What conclusion do you draw?
answer:
a) 2721.60 mJ;
b) 2721.60 mJ; c) 518.40 mJ;
d) 518.40 mJ.
Same answer when both current are entering or both leaving, and same answer when one is
leaving and one is entering.
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