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Giao trinh bai tap ece 110 hw5

ECE430
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Fall 2009

http://www4.hcmut.edu.vn/~nqnam/lecture.php

Lecture 3

1

Introduction
 Electromagnetic theory: basis for explaining the operation of all
electrical and electromechanical systems.
 There are magnetic field and electric field systems, the discussion is
restricted to magnetic field systems.
 Integral form of Maxwell’s equations



C




C

H  dl 

J
S

E  dl   

 J  n da  0
 B  n da  0
f

S

S

S

 n da

Ampere’s law

B
 n da
t

Faraday’s law

f

Conservation of charge
Gauss’s law

Lecture 3

2


1


Static magnetic circuits
 There are no moving components in static magnetic circuits.
 Toroid: N uniformly wound turns. r0 and r1 are inner and outer radii.
Consider the contour corresponding to the mean radius r = (r0 + r1) / 2,
assuming magnetic field intensity Hc is uniform inside the core. Using
Ampere’s circuital law, it can be determined that Hc(2r) = Ni. Or,

H c l c  Ni
where lc = 2r is the mean length in core. Assuming
B is a linear function of H in the core, the flux
density in the core is

Bc  H c  

Ni
Wb /m 2
lc

Lecture 3

3

Static magnetic circuits (cont.)
Flux is given by

 c  Bc Ac 

Ni
Ni
Ac 
Wb
lc
l c Ac

where  is the magnetic permeability of the core material, Ac is the core
cross-sectional area.
Define Ni as magneto motive force (mmf), reluctance can be defined as

l
Ni mmf

 c  R (At/Wb)
c
flux Ac
P = 1/R is called permeance. Flux linkage is now defined as  = Nc =
PN2i. By definition, self inductance L of a coil is given by L 

Lecture 3


N2
 PN 2 
i
R
4

2


Static magnetic circuits (cont.)
 There are similarities between electrical and magnetic circuits
mmf
flux
reluctance
permeance






voltage
current
resistance
conductance

 Toroid with air gap (no fringing): There is magnetic field intensity H in
both the air gap and the iron portion. lg – length of the air gap, lc – mean
length of the iron portion. Applying ACL around the contour c

Ni  H g l g  H clc 

Bg

0

lg 

Bc
lc
r 0

where 0 = 4 x 107 H/m is the air’s permeability, and r is the relative
permeability of the core material.
Lecture 3

5

Static magnetic circuits (cont.)
Applying Gauss’s law on the closed surface s covering one magnetic
pole, BgAg = BcAc. For the case of no fringing, Ag = Ac. Hence, Bg = Bc.
Divide the mmf by the flux to calculate equivalent reluctance

lg
l
Ni

 c  R g  Rc

 0 Ag Ac
Where Rg and Rc are reluctances of the air gap and the core,
respectively. In the equivalent magnetic circuit, these are in series.
 Suppose there is “fringing”, i.e., not all the flux is confined to the area
between the two faces of iron portion. In this case, Ag > Ac, i.e., effective
air gap area increases. This can be accounted for empirically,

Ac  ab, Ag  a  l g b  l g 
Lecture 3

6

3


Class examples
 Ex. 3.1: Find the required mmf to produce a given flux density. Air
gap and core length and area are known.

0.06
 47.7  103 At/Wb
7
4
10 4  10 10
0.001
Rg 
 7.23  106 At/Wb
7
4
4  10 1.1 10
Rc 

 



4









  Bg Ag  0.51.1 10

4

  5.5 10

4

Wb

Hence,

Ni  R c  R g   47.7  7230  10 3  5.5  10 5  400 At

Lecture 3

7

Class examples (cont.)
 Ex. 3.2: Find the flux through the coils. All air gaps are the same in
length and area. Iron’s permeability is infinite and ignore fringing.

0.1  10 
4  10 4  10   1.989  10
2

R1  R 2  R 3  R 

7

4

6

At/Wb
2500

In the equivalent circuit, positive directions
for 1, 2, and 3 are shown. The algebraic
sum of the fluxes at node a must be zero.
Let mmf of node a wrt. b be F, then

500
b

1
R

a

2
1500

2500  F 500  F F  1500


0
R
R
R
Hence,

R

R
3

F  500, 1  10 3 Wb,  2  0,  3  10 3 Wb
Lecture 3

8

4


In class quiz
 Problem 1: A toroid has a mean length with a radius of 500 mm, the
working flux density in the air gap is 0.6 Wb/m2, creating by a coil of 100
turns. An air gap with the length of 2 mm is made. Given a = 20 mm.
Ignore the reluctance of the core.
a) Find the required excitation current
b) Determine the self inductance of the coil
 Special question: Suppose you were asked to build a linear variable
inductor. Describe your solution, considering fringing effect and
reluctance of the core (if exists)?

Lecture 3

9

Mutual inductance
 Mutual inductance: parameter related induced voltage in one coil with
time varying current in another coil.
 Consider two coils wound on the same magnetic core, coil 1 is
excited whilst coil 1 is open. The total flux linking coil 1 is

11  l1   21
where l1 (called leakage flux) links to coil 1 only; whereas, 21 is the
mutual flux linking to both the coils, also the flux linking coil 2 due to
current in coil 1. The order of subscripts is important.
 Since coil 2 is open circuited, the flux linkage of this coil is

2  N 221
Lecture 3

10

5


Mutual inductance (cont.)
 21 is linearly proportional to the current i1, hence

2  N 221  M 21i1

 The induced voltage v2 (due to the change of flux linkage) is given by

v2 

d 2
di
 M 21 1
dt
dt

M21 is called the mutual inductance between the coils. Similarly,
induced voltage v1 in coil 1 can also be determined as follows.

11 is proportional to i1, hence 1  N111  L1i1 , then

v1 

d1
di
 L1 1
dt
dt

with L1 is the self inductance of coil 1, as you may know.
Lecture 3

11

Mutual inductance (cont.)
 Consider now the case where coil 1 is open and coil 2 is excited. The
same procedure can be used to calculate induced voltages.

22  l 2  12

1  N112  M 12 i2
v2 

2  N 222  L2i2

v1 

d1
di
 M 12 2
dt
dt

d2
di
 L2 2
dt
dt

where L2 is the self inductance of coil 2, as you may know.
 From energy considerations, it can be shown that M21 = M12 = M.
 Finally, consider now the case where both the coils are excited.

1   l1   21  12  11  12

 2   21   l 2  12   21   22
Lecture 3

12

6


Mutual inductance (cont.)
 Noting that M21 = M12 = M

1  N111  N112  L1i1  Mi2

2  N 2 21  N 2 22  Mi1  L2 i2

 By differentiating those, induced voltages can be calculated

v1  L1

di1
di
M 2
dt
dt

v2  M

di1
di
 L2 2
dt
dt

 Coefficient of coupling between the two coils is defined by k 
 It can be shown that 0  k  1, or equivalently, 0  M 

M
L1 L2

L1 L2

 Most air core transformers are loosely coupled (k < 0.5), whilst iron
core transformers are tightly coupled (k > 0.5, can approach 1).
Lecture 3

13

Example
 Ex. 3.4: Given reluctances of three air gaps in the magnetic circuit.
Draw equivalent circuit and compute flux linkages and inductances.

N 1i1  R3 1   2   R11

N 2 i 2  R 2 2  R 3 1   2 

100i1  51  2 2   10

100i 2   21  4 2   10

6

1

6

Solving these equations for 1 and 2

N1i1
R1
R3

6
1  25i1  12.5i 2   10 6  2  12.5i1  31.25i2   10

From

R2

1  N11  25i1  12.5i2  10 4
2  N 22  12.5i1  31.25i2  10 4

It can be seen that

L1  25  10

4

L2  31.25  10 4 H  3.125 mH

N2i2
2

H  2.5 mH

M  12.5  10 4 H  1.25 mH
Lecture 3

14

7


Polarity markings (dot convention)
 Lenz’s law: the voltage induced is in such a direction that the current
due to it opposes the flux causing the voltage.
 Signs of mutually induced voltages are monitored by a dot marking
convention. A current i entering a dotted (undotted) terminal in one
winding induces a voltage Mdi/dt with positive polarity at the dotted
(undotted) terminal of the other winding.
 Two problems: (1) given the coil configuration, determine the dot
markings. (2) given the dot markings, how they are used in writing
equations.

Lecture 3

15

Determining polarity markings
 Steps:
 Arbitrarily select one terminal of a coil and assign a dot in one coil.
 Assume a current flowing into the selected dotted terminal and
determine the flux flowing in the core.
 Select an arbitrary terminal of the second coil and assign a positive
test current to it.
 Determine flux direction due to this current.
 Compare directions of two fluxes. If both is additive, then a dot is
placed in the second coil where the test current enters.
 If the fluxes are in opposite directions, then a dot is placed in the
second coil at the terminal where the current leaves.
Lecture 3

16

8


Practical ways of determining dot markings
 For a device such as a transformer, there is no way of knowing how
the coils are wound, therefore a practical way is adopted:
A DC source is used to excite one
coil of the transformer.

+

Place the dot on the terminal to

_

which the + side of DC source is
connected.
Close the switch: up-scale kick in voltmeter => the dot on the other coil is
on the + side of the voltmeter. Down-scale momentary deflection in
voltmeter => the dot is placed the – side of the voltmeter.
Lecture 3

17

Writing equations with mutually coupled coils
 Given 2 mutually coupled coils and dot markings, write loop equations.
Choose arbitrary direction for currents.
Rule: Reference current enters a dotted (undotted) terminal, induced
voltage in the other coil is positive at the dotted (undotted) terminal.
Reference current leaves a dotted (undotted) terminal, induced voltage at
the dotted (undotted) terminal of the other coil is negative.

di1
di
M 2
dt
dt
di
di
v 2  i 2 R 2  L2 2  M 1
dt
dt
v1  i1 R1  L1

Lecture 3

R1
i1
v1

R2

M
i2

v2

18

9


Example
 Ex 3.6: Write loop equations for a circuit with mutually coupled coils.
Assuming zero initial voltage on capacitor

 L1

0

R1

i1

v1  i1 R1  i1  i2 R2

L2

R2

v1

d
i1  i2   M di2
dt
dt

(i1 – i2)

C
M

i2

L1

di
1 t
d
d
i2 dt  L2 2  M i1  i 2   L1 i2  i1 

C 0
dt
dt
dt
di
 M 2  i2  i1 R2
dt
Lecture 3

19

Lecture 3

20

In-class quiz
 Problem 3.15.

10



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