Lecture 22Lecture 22 Mutual Inductance & Transformers 22.pdf · Mutual Inductance &...

Lecture 22Lecture 22Mutual Inductance &

Transformers

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Flux Linkages and Faraday’s LLaw

Magnetic flux passing through a surface area A:

φ d⋅= ∫ ABA∫

For a constant magnetic flux density perpendicular

BA=to the surface:

The flux linking a coil with N turns:


φλ N=

Faraday’s Lawy

Faraday’s law of magnetic induction:

dtde λ

=dt

The voltage induced in a coil whenever its gflux linkages are changing. Changes occur from:

• Magnetic field changing in time


• Coil moving relative to magnetic field

Lenz’s Law

Lenz’s law states that the polarity of the induced voltage is such that the voltagewould produce a current (through an external resistance) that opposes the original change in flux linkages.


Mutual Inductance & TransformersMutual Inductance & Transformers

1 Determine the inductance and mutual1. Determine the inductance and mutual inductance of coils given their physical parametersparameters.

2. Understand ideal transformers and solve circuits that include transformerscircuits that include transformers.

3. Use the equivalent circuits of real transformers to determine their regulationstransformers to determine their regulations and power efficiencies.


Inductance and Mutual InductanceDefinition of inductance L:

linkagesFluxL λ==

icurrentL ==

φλ N=Substitute for the flux linkages using φλ N=Substitute for the flux linkages using

Nφi

NL φ=


Inductance and Mutual Inductance

NiSubstituting

RNi

=φ

2NL =R


Faraday’s Lawy

Voltage is induced in a coil when its fluxVoltage is induced in a coil when its flux linkages change:

diLLidde )(λdt

Ldtdt

e ===


Mutual InductanceSelf inductance

Self inductance

11Lλ

= ← 22Lλ

= ←

inductance for coil 1

inductance for coil 2

11

11 i

L

λ=

=

22

22 i

L

λ

=

1i=

2i=

Mutual inductance between coils 1 and 2:

12212112M λλλλ==== ←←

Mutual inductance between coils 1 and 2:


2211 iiiiM ====

Mutual Inductance

Total fluxes linking the coils:

12111 λλλ ±=

21222

12111

λλλ ±= 21222 λλλ ±=


Mutual Inductance

Currents entering the dotted terminals produce aiding fluxes


Circuit Equations for Mutual Inductance

iLMiMiiL 2111

+±=±=

λλ

diMdiLde

iLMi

211

2212

±

+±=λ

λ

dididdt

Mdt

Ldt

e 211

11 ±==

λdtdiL

dtdiM

dtde 2

212

2 +±==λ


Example

WeberturnsampereR /)(107 −=

mHNL 11007

221

1 ===R

p )(

Self inductance:

mHNL 420010

7

222

2

71

===

R


1072 R

Example

15

7111

1 1010

100 iiiN −===R

φ iN 10200 15

1221 ×== −φλ

mHi

M 21

21 ==λ

Mutual inductance:


i1

Example

Does the flux produced by i aid or opposeDoes the flux produced by i2 aid or oppose the flux produced by i1?

didi 21

dididtdiM

dtdiLe

12

2111 −=


dtdiM

dtdiLe 12

22 −=

Exercise 15.13

• •• •

Mark the dot for coil 2


Fluxes from 1 and 2 aid each other in paths 1&2, oppose in path 3

Transformers

Can be used to step up or step down ac voltages


Can be used to step up or step down ac voltages

Electrical Power Distribution

)cos(d l d IVP = θ )cos(rmsrmsdelivered IVP = θ

2rmslineloss IRP = rmslineloss IRP

Minimize loss by increasing voltage and decreasing current. Modern transmission grids


use ac voltages up to 765,000 Volts

War of the Currents

In the "War of Currents" era in the late 1880s, Nikola Tesla and Thomas Edison became adversaries due to Edison's promotion of direct current (DC) for electric power distribution over the more efficient alternating current (AC) advocated by Tesla.efficient alternating current (AC) advocated by Tesla.

During the initial years of electricity distribution, Edison's direct current was the standard for the United States and Edison was not disposed to lose all his patent royalties. Direct current worked well for the incandescent lamps that were the principal load of the day From his work with rotary magnetic fields Teslaload of the day. From his work with rotary magnetic fields, Tesla devised a system for generation, transmission, and use of AC power. He partnered with George Westinghouse to

i li hi


commercialize this system.

War of the Currents

Edison's propagandaEdison s propaganda

Edison went on to carry out a campaign to discourage the use of alternating current. Edison personally presided over several g p y pexecutions of animals, primarily stray cats and dogs, to demonstrate to the press that his system of direct current was safer than that of alternating current Edison's series of animalsafer than that of alternating current. Edison s series of animal executions peaked with the electrocution of Topsy the Elephant. He also tried to popularize the term for being electrocuted as being "Westinghoused".


War of the CurrentsTopsy the Elephant (circa 1875 - January 4, 1903) was a member of a domesticated herd at Coney Island's Luna Park. She had been a part of the Forepaugh Circus Topsy was deemed an ill tempered and dangerousForepaugh Circus. Topsy was deemed an ill-tempered and dangerous animal since she had killed three men in as many years, including an abusive trainer who tried to feed her a lit cigarette.

i l h d id d h d hBecause Topsy was so violent, her owners decided to put her to death. A proposal of hanging was abandoned after the American Society for the Prevention of Cruelty to Animals protested. The elephant was offered a carrot poisoned with cyanide but did not eat it Later Thomas Edisoncarrot poisoned with cyanide, but did not eat it. Later, Thomas Edison suggested electrocution, using the Westinghouse alternating current system of electricity transmission, which Edison, a backer of direct current, argued was more dangerous than DC The ASPCA found this suggestionwas more dangerous than DC. The ASPCA found this suggestion acceptable, viewing electrocution as a more humane form of killing. Electrocution killed Topsy quickly. Edison recorded the execution with a motion picture camera, and showed his film to audiences around the


p ,country as part of his unsuccessful attempt to discredit AC.

Ideal Transformers

Law sFaraday'by )cos()( 111 dtdNtVtv mφω ==

)sin()cos()(1

1

1

1

101

1t

NV

dttN

Vdttv

Nφ(t) m

t tm ω

ωωφ∫ ∫ ==+=


10 011 NNN ω

Ideal Transformers

[ ])sin()(1

1222 t

dtd

NV

NdtdNtv m== ω

ωφ

)()cos( 12

12

1

tvNNtV

NN

dtNdt

m == ω

ω


11 NN

Ideal Transformers

( ) ( )tNt 2( ) ( )tvN

tv 11

22 =


Ideal Transformers

0 since02211 iNiN ≈≈=−= RRF φThe magneto-motive force (mmf) applied to the core:

)()( 1

2211

tiNti

iNiN

=

=

If the voltage is stepped up


)()( 12

2 tiN

ti = g pp pthe current is stepped down

Ideal Transformers

)()()()()( 12 tiNtvNtitvtp ==

)()()(

)()()()()( 12

11

222

ttit

tiN

tvN

titvtp ==

)()()( 111 tptitv ==

( ) ( )( ) ( )tptp 12 =

Net power is neither generated nor consumed by an ideal transformer


an ideal transformer

Ideal Transformers

( ) ( )tvNtv 2=( ) ( )tvN

tv 11

2 =

)()( 11

2 tiNti = )()( 12

2 N

( ) ( )tptp 12 =


Mechanical Analogg

d11

d2

( ) ( ) 12

212

2 dlldtv

NNtv ==

11

211

2

11

)()( FllFti

NNti

lN

==


22 lN

Transformer SummaryTransformer Summary

1. We assumed that all of the flux links all of the windings of both coils and that the resistance of the coils is zero. Thus, the voltage across each coil is proportional to the number of turns on the coil.

( ) ( )N 2( ) ( )tvNNtv 1

1

22 =


2 We assumed that the reluctance of the core is2. We assumed that the reluctance of the core is negligible, so the total mmf of both coils is zerozero.

( ) ( )tiNNti 2

12 =

N 2

3. A consequence of the voltage and current q grelationships is that all of the power delivered to an ideal transformer by the ysource is transferred to the load.


Analysis of a Circuit Containing anAnalysis of a Circuit Containing an Ideal Transformer


Example

Find the rms values of the currents and voltages with the switch open and closed. With the switch open:

2211011

22 === VVVNV rmsrms

sourcethefromtakenispowerNo0

5

22

1

11

2

→==

rmsrms

INI

N rmsrms


sourcethefrom taken ispower No021

1 →==rmsrms

IN

I

Example

With the switch closed:V

AVR

VI rms

rmsL

2.210222

2 =Ω

==

AAINI

L

44.02.210

22

1 =×==


N rmsrms 521

1

Impedance Transformations

LZ2 =IV

NNNN 112

2

)/()/(

=IV

I

LLLL ZNZZNZNN

NN2

1''2

2'12

121

)/(

)/(

⎟⎟⎞

⎜⎜⎛

=→⎟⎟⎞

⎜⎜⎛

==

I


LLLL ZN

ZZN

ZNN 2121 )/( ⎟⎟

⎠⎜⎜⎝

=→⎟⎟⎠

⎜⎜⎝

==

Impedance Transformations

ZNZ2

11⎟⎟⎞

⎜⎜⎛

==′V

LL ZN

Z21⎟⎟⎠

⎜⎜⎝

==I


Example 15.11

Find the phasor voltages and currents. First, transform ZL to the primary side of the transformer:

2010

22

⎟⎞

⎜⎛

+=

N

jZ L

200010001000

20001000)2010()10('2

2

1

+++

+=+=⎟⎟⎠

⎞⎜⎜⎝

⎛=

jZRZ

jjZNNZ LL

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.452828452200020002000

200010001000'1

∠=∠=+=

++=+=

j

jZRZ Ls

Example 15.11

oo

o

453536.045282801000

1 −∠=∠∠

== AZ

sVI

o

o

)20001000)(453536.0(

452828

'11

1

+−∠==

∠

jZ

Z

L

s

IVoo )43.632236)(453536.0(

))((11

∠−∠=

jL


o43.186.790 ∠=

Example 15.11

We can now calculate the current andWe can now calculate the current and voltage phasors on the secondary side using the turns ratio:

oo 45536.3)453536.0(1011

2 −∠=−∠== II N

t e tu s at o:

oo 43186079)43186790(1

45536.3)453536.0(10

2

12

2

∠∠

∠∠

VV

II

NN

oo 43.1860.79)43.186.790(101

1

22 ∠=∠== VV

N


Example 15.12

Th lt d R l b t f d t th dThe voltage source and R1 can also be transformed to the secondary side:

1N∠== 01000

101

1

2' VNNV ss

o

Ω=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 10)1000(

101 2

1

2

1

2'1 R

NNR


⎠⎝⎠⎝ 101N

Example 15.12

o )0100(20202010

2010102010 '

2 ∠++

=++

+=

jjV

jjV s

o

o)0100(

45220)10/20(400100

000001

∠∠

∠+=

−Tan

jj

ooo

o

43.1806.79)0100(43.6351045220

∠=∠∠

=

∠


o43.1806.79)0100(

45220∠∠

∠

Exercise 15.17

Find the values of V2 and the power delivered to the load. To find I2, transform Vs and Rs from the primary to the secondary idside:

∠=∠== 0400)0100(14

1

2'ss N

N ooVV

Ω==⎟⎟⎞

⎜⎜⎛

= 64016

12

2'

1

RRNR

N


Ω==⎟⎟⎠

⎜⎜⎝

= 640161

SSS RRN

R

Exercise 15.17

RVs 0385.0

4006400400

400'

'

2 ∠=+∠

=Ω+

= oo

I

R

R

L

S

08.153)400)(0385.0(

400640400

22 ∠=Ω∠==

+Ω+ooIV

WRIP Lrms6.29)400(

2385.0)(

22

2 =Ω⎟⎟⎠

⎞⎜⎜⎝

⎛==


rms 2 ⎟⎠

⎜⎝

Exercise 15.18

Find the turns ratio that maximizes the power transfer to the load. What turns ratio transforms the 400Ω load resistance into a

i t f 40Ω th i id ?resistance of 40Ω on the primary side?

40)400(2

12

1' Ω=Ω⎟⎟⎞

⎜⎜⎛

=⎟⎟⎞

⎜⎜⎛

=NRNR LL

1140

40)400(

12

1

22

Ω⎟⎞

⎜⎛

ΩΩ⎟⎟⎠

⎜⎜⎝

⎟⎟⎠

⎜⎜⎝

NN

NR

NR LL


101

101

40040

2

1

2

1 =→=ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛NN

NN

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