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(2r) = Ni. Or,

H c l c Ni

where lc = 2r 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 = Nc =

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 107 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 0.51.1 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 221

Lecture 3

10

5

Mutual inductance (cont.)

21 is linearly proportional to the current i1, hence

2 N 221 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 N111 L1i1 , then

v1

d1

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 N112 M 12 i2

v2

2 N 222 L2i2

v1

d1

di

M 12 2

dt

dt

d2

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 N111 N112 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 R11

N 2 i 2 R 2 2 R 3 1 2

100i1 51 2 2 10

100i 2 21 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 N11 25i1 12.5i2 10 4

2 N 22 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

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(2r) = Ni. Or,

H c l c Ni

where lc = 2r 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 = Nc =

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 107 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 0.51.1 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 221

Lecture 3

10

5

Mutual inductance (cont.)

21 is linearly proportional to the current i1, hence

2 N 221 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 N111 L1i1 , then

v1

d1

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 N112 M 12 i2

v2

2 N 222 L2i2

v1

d1

di

M 12 2

dt

dt

d2

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 N111 N112 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 R11

N 2 i 2 R 2 2 R 3 1 2

100i1 51 2 2 10

100i 2 21 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 N11 25i1 12.5i2 10 4

2 N 22 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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