12/25/2012
Overview
• Introduction
• Self Inductance
• Mutual Inductance
13-Self and Mutual Inductance
ECEGR 450
Electromechanical Energy Conversion
Dr. Louie
Inductance
2
Induced vs Applied Voltage
• Transformers and other machines have coils of
wire wrapped around permeable material
• Transformers are made of one or more inductors
on a common core
• We will start with a qualitative description of
inductance
• From circuit theory: vL  L
di
dt
• Previously showed that for steady-state AC
circuits: VL = IjXL
I
+ VL VL
I
Dr. Louie
3
Induced vs Applied Voltage
Dr. Louie
4
Induced vs Applied Voltage
• Let the current be: i(t)  1cos(t)
• The flux will be in phase with the current due
Ampere’s Law Ni  
(t) 
current lags voltage
• Why are induced and applied voltages opposite?
• It all has to do with sign convention
 V = iR + Ldi/dt (Ldi/dt is on the right hand side)
 V + e = iR (e is on the left hand side)
N
cos(t)

• Applying Faraday’s
Law:
2
2
• Therefore e = -Ldi/dt
d N
N

 sin(t) 
 cos(t  90) or
dt


d
d di
di
e  N
 N
 L
dt
di dt
dt
e  N
i
I
e
V
R
L
current leads voltage
Dr. Louie
5
Dr. Louie
6
1
12/25/2012
Inductance
Inductance
• Inductive reactance XL exists due to Faraday’s
Law
• Recall that
d
(note the polarity in the figure)
eN
dt
 jXL = jωL
 N is also known as the flux linkages (λ)
• The j operator accounts for the 90 degree phase
shift between current and induced voltage
• ω accounts for the dependency on frequency
• L is a description of how strong the current links
the flux through the coil
• Next we examine inductance
i
+
e
-
-
7
Dr. Louie
• Inductance depends on the physical characteristics of the
magnetic circuit
• Recall that
Ni


d
L N
therefore
di
2
N
L 

• Inductance is constant if the permeability of the magnetic
circuit is itself constant (not the case in ferromagnetic
materials)
• We will assume that we are operating in the linear region of
B-H curve
d
N
di
• Large inductance: great sensitivity of flux wrt
current
i

+
V
e
-
-
Inductance describes how the
flux linking a coil changes with
the applied current
N
9
Dr. Louie
Question
Which circuit has greater inductance?
A.
10
Dr. Louie
Question
Which circuit has greater inductance?
B.
i
8
Self Inductance
• Self-inductance (inductance) is defined as:
+
N
Dr. Louie
Inductance
L

+
V
A.
i
B.
i
i
Smaller reluctance. Current gives rise to greater flux
Dr. Louie
11
Dr. Louie
12
2
12/25/2012
Self Inductance
Mutual Inductance
• Consider two coils wrapped around a common core
• Let 1 be the flux that links coil 1
• Inductance is related to emf by:
d
d di
di
eN
N
L
dt
di dt
dt
• A coil with 1 H of inductance will have 1 volt induced in it if
the current changes at a rate of 1 A/s
• If we know the inductance, we do not need to compute the
flux
 Includes leakage flux (1l) and flux through the
core that links coil 2 (21)
 With coil 2 open: 1 = 1l + 21
i1 1l
i

21
+
e1
-
N1
13
Dr. Louie
Mutual Inductance
• Induced voltage in coil 1 is:
• Induced voltage in coil 2 is:
d21
d21 di1
di
e2  N2
 N2
 M21 1
dt
di1 dt
dt
d1
di
 L1 1
dt
dt
d21
di1
 M21: mutual inductance from coil 1 to coil 2
• L1: self inductance of coil 1
M21
i1
+
e1
-
N2
i
N1
21
+
e1
-
open circuit
N2
N1
15
Dr. Louie
d 2
di
 L2 2
dt
dt
d
d di
di
e1  N2 12  N1 12 2  M12 2
dt
di2 dt
dt
e2  N1
open circuit
d12
di2
+
e1
-
2l
12
N1
N2
Dr. Louie
+
e2 open circuit
-
16
Mutual Inductance
• Similar expressions can be derived in coil 2 if it is
connected to the source and coil 1 is open
N1
N2
Dr. Louie
Mutual Inductance
M12
14
Dr. Louie
Mutual Inductance
e1  N1
open circuit
N2
• We can write:
M12M21  N2
d21 d12
N
di1 1 di2
• Let k1 be the fraction of the flux of coil 1 that
links coil 2 21  k11
• Let k2 be the fraction of the flux of coil 2 that
links coil 1 12  k22
• Since L  N d we can write: M12M21  k1k 2L1L 2
i
di
+
e2
-
17
Dr. Louie
18
3
12/25/2012
Mutual Inductance
Mutual Inductance
• If the system is linear (standard assumption)
then:
• The circuit equivalent is:
i2
i1
M12  M21  M
 M: mutual inductance of coil 1 and coil 2
R1
• And we can reduce
R2
L2
L1
M12M21  k1k 2L1L 2 to
M  k L1L 2
i1
k  k1k 2
+
e1
-
 k: coefficient of coupling
k = 1 (tightly coupled coils, no leakage)
k = 0 (magnetically isolated coils)
Dr. Louie
Polarity dots indicate which way
the coils are wound.
If current enters one dot, it
leaves the other if connected to
a passive circuit.
i2
N1
19
N2
Dr. Louie
+
e2
-
20
Example
Example
Two identical coils are wound on the same magnetic
core. A current changing at a rate of 2000 A/s in
coil 1 induces a voltage of 20 V in coil 2. What is
the mutual inductance?
Two identical coils are wound on the same magnetic
core. A current changing at a rate of 2000 A/s in
coil 1 induces a voltage of 20 V in coil 2. What is
the mutual inductance?
di1
dt
 M21  M
e2  M21
M12
M
Dr. Louie
e2
20

 0.01H
di1 2000
dt
21
Dr. Louie
22
Example
Example
Two identical coils are wound on the same magnetic
core. A current changing at a rate of 2000 A/s in
coil 1 induces a voltage of 20 V in coil 2. If L 1 =
25mH, what percentage of the flux set up by coil 1
links coil 2?
Two identical coils are wound on the same magnetic
core. A current changing at a rate of 2000 A/s in
coil 1 induces a voltage of 20 V in coil 2. If L 1 =
25mH, what percentage of the flux set up by coil 1
links coil 2?
L1  L 2  L  0.25mH
M  k L1L 2  kL
k
Dr. Louie
23
0.01
 100  40%
0.025
Dr. Louie
24
4
12/25/2012
Magnetically Coupled Coils
Magnetically Coupled Coils
• It is possible to connect the magnetically coupled
coils together
 series or parallel
• Depending on the polarity the coils can be aiding
or opposing
• See 2.6 of text for more details
L1
L2
L1
Series Aiding
L2
• For aiding circuits:
di
di
di
M

(L1  M)
M is added
dt
dt dt
because of aiding
di
di
di
polarity
VL2  L 2
M

(L 2  M)
dt
dt dt
di
di
VLeff  VL1  VL2 
(L1  L 2  2M) 
L eff
dt
dt
where
M
L eff  L1  L 2  2M
L2
L1
VL1  L1
I
Vs
Series Opposing
25
Dr. Louie
R
26
Dr. Louie
Example
Example
The circuit below operates at 50Hz. The mutual
inductance M between the coils is 0.6 H. Compute I.
The circuit below operates at 50Hz. The mutual
inductance M between the coils is 0.6 H. Compute I.
L eff  1  1.5  2  0.6  3.7 H
Mutual inductances add in aiding
jXeff  j3.7  2  50  j1162 W circuits
Z  0.5  j1162 W
I  0.238  89.98 A
1H
1.5H
I
Vs  2770
1H
1.5H
I
Vs  2770
0.5W
27
Dr. Louie
0.5W
28
Dr. Louie
Example
Example
Repeat the problem but with opposing polarity of the
inductors.
Repeat the problem but with opposing polarity of the
inductors.
L eff  1  1.5  2  0.6  1.3 H
jXeff  j1.3  2  50  j408.4 W
Z  0.5  j408.4 W
I  0.68  89.93 A
1H
1.5H
Vs  2770
I
Dr. Louie
1H
1.5H
Vs  2770
0.5W
29
I
Dr. Louie
0.5W
30
5