ECE 307 Master.pptx

8/9/2019 ECE 307 Master.pptx

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1ECE 307 Basic Electrical Engineering

ECE 307 Lecture 1DC Circuit Components, Connections,and KCL

Department of Electrical and ComputerEngineering

Clemson University


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Electric Circuit erforms a

function!o rocess

"nformationo #ransfer o$er C%aracteri&ed 'y!o (oltageso

Currentso o$er

Circuit Components )esistor (oltage *ource Current *ource *$itc%

Connections #erminal +ode ranc% Loop -es%

*imilar Electric Circuit

)eductions *ource #ransformation arallel *eries #%evenin E.uivalent +orton E.uivalent

/nalysis #ools

Kirc%off s Current La$ +ode (oltage -et%od

Kirc%off s (oltage La$ -es% Current -et%od

*uperposition

vervie$ of DC Electric Circuits

DC Lecture 1 2DC Circuit Components,Connections, and KCL


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vervie$ of Lecture

"ntroduce *i Elements Used in a DC Circuit Constant (oltage *ource Constant Current *ource )esistor pen Circuit, *%ort Circuit, *$itc%

Connect #%e Elements to uild a Circuit Define Connection #erminology! #erminal, +ode,

ranc%, Loop, -es% Define *eries and arallel Connections

Kirc%%off s Current La$ 4KCL5

Descri'es t%e currents at a connection


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DC vs6 /C

#%omas Edison4 e'ruary 11, 1897 cto'er 18, 1:315

+icola #esla10 ;uly 18


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(oltage, Current, and )esistance


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Constant (oltage *ource

Examples of Constant Voltage Sources:

Car Battery suppliesconstant 1>( to po$er t%elig%ts and accessories

AA Battery used in aflas%lig%t to supply aconstant 16


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Mathematical Model: (oltage is Constant Current

o Current is determined 'yconnections to ot%ercomponents

o Can supply any current

Symbol:

or

#ime ( o l

t a g e Constant 2? no c%ange over time

Behavior:

roduces a constant voltage t%atis not affected 'y connections toot%er components

@2

@

2


S v S v


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hat does it do !hen connected to other components"


@2 Electric

Circuit

Case #:

#%e electric circuit does $orA toc%arge t%e 'attery

2

@2

atteryC%argingCircuit

2

Current results from t%is connection Current results from t%is connection

Case $:

Does $orA to move electrons andsupply energy to t%e rest of t%eelectric circuit

+ote! Be $ill %ave to solve t%e entire circuit in orderto Ano$ t%e amount of current6

i i

S v S v


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Constant Current *ource

Examples of Constant Current Sources:

elder supplies constantcurrent to create t%e %eatneeded to fuse metal pieces6

Cell %hone chargersupplies constant current att%e 'eginning of a full2c%arge cycle6


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Time C u r r e n t Constant - no c!ange o"er time

Behavior:roduces a constant current t%at is

not affected 'y connections to ot%ercomponents

Mathematical Model: Current is Constant (oltage

o (oltage is determined 'yconnections to ot%ercomponents

o Can supply any voltage

Symbol:

si


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Does $orA to move electrons and supply energy to t%e rest of t%e electriccircuit

ElectricCircuit

2

/ voltage can 'emeasured acrosst%e currentsource as a resultof t%isconnection

@

2v

+ote! Be $ill %ave to solve t%e entire circuit inorder to Ano$ t%e amount of voltage6

si

ECE 307 B i El i l E i i


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)esistor

Examples of Resistors:

&he 'eating Element of atoaster 4t%e part t%at gets %ot5resists t%e flo$ of electrons andgets %ot6

&he (ilament in a lig%t 'ul'4t%e part t%at produces lig%t5resists t%e flo$ of electrons andgets very %ot to produce lig%t6

/n extension cord is usedto connect appliances and%as very lo$ resistance6

#%e resistance of t%e extension cord is not &ero and can 'esignificant depending on t%e application6

ECE 307 B i El i l E i i


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)esistor

(oltage

C u r r e n

tBehavior:

Constant relations%ip 'et$eenvoltage and current6

Mathematical Model:

Constant ) summari&es material properties, temperature, and si&e6

)hm*s +a! relates t%e voltageand current using )!

v = iR

Symbol:

R

1slope

is constant6 R

ECE 307 B i El t i l E i i


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)esistor


)esists t%e flo$ of electrons6 / resistor removes energy from t%e circuitand dissipates it as %eat6

ElectricCircuit

/ voltage can 'emeasured acrosst%e resistor as aresult of t%isconnection

@

2

v

Current results from t%is connection

2

+ote! Be $ill %ave to solve t%e entire circuit in order toAno$ eit%er t%e voltage or t%e current6 Be do Ano$ t%atvoltage and current are related 'y %m s La$ as v = iR

i

R

ECE 307 B i El t i l E gi i g


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/pplication of %m s La$

) is a circuit model of a p%ysical material t%at%as lengt%, $idt%, %eig%t, and property ofconductivity t%at constitute t%e resistance6E amples 4given same si&e5!

Copper $ire 4lo$ resistance5 +ic%rome $ire 4medium resistance5

Fold $ire 4lo$ resistance5

"nsulation on $ire 4%ig% resistance5

or t%e current referenced into a voltage drop as s%o$n v=iR

@

2

i

Rv

ECE 307 Basic Electrical Engineering


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/pplication of %m s La$

-ust adGust t%e sign 4@H25 in %m s la$ for ot%er referenced directions,

#%is is t%e standardconvention

v= - iR v= iR

@

2@

2

i i

Rv Rv



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E ample 1! /pplying %m s La$ given fi ed voltage andcurrent references

v= - iR 242>/54105 >0(

v= iR >/410 5 >0(

v= iR 42>/54105 2>0(

#%ese are all t%e same circuit6 >/ goes into t%e resistor from t%e top and t%ere is a>0 ( drop from top to 'ottom6

@

2

@

2 @

210

> A

v10

> A

v10

> A

v



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pen and *%ort Circuits

ElectricCircuit

+ voltage can 'emeasured acrosst%e opening

@

2

v

+o Current

2

Short Circuit:

+o resistance to current flo$, I &ero )

)pen Circuit:

+o current can flo$, I infinite )

ElectricCircuit

+ voltage can 'emeasured acrosst%e s%ort

@

2

v

Current

2

i i

)esistance is )esistanceis zero



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*$itc%

Examples of S!itches:

S!itch on t%e $all turnst%e lig%ts on or off 6

Eac% 'utton on t%e p%oneor Aey on t%e Aey'oard isa S!itch 6

S!itch on t%e coffeemaAer turns t%e appliance

on or off 6



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*$itc%

Behavior:

Can stop or allo$ t%e flo$ of current6

Mathematical Model: +o current flo$s $%en t%e s$itc%

is open 6 Current flo$s freely $%en s$itc%

is closed 6

Symbol:



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*$itc%


Connects or disconnects part of t%e circuit

ElectricCircuit

+ voltage can 'emeasured acrosst%e opening

@

2

v

+o Current

2

Closed:

/cts liAe a s%ort circuit

)pen:

/cts liAe an open circuit

ElectricCircuit

+ voltage can 'emeasured acrosst%e s$itc%

@

2

v=0V

Current

2

i i

R R



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Connect Components to Create an Electric Circuit

E ample! Connect a 'attery and a resistor to 'uild a porta'le lig%t6

#e"ice$ %las!lig!t

1&5' 11&5'

Circuit (o)el

1&5*

1&5*



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Electric Circuit 4Electric +et$orA5

Connection of Components#erms to descri'e t%e connection of components #erminal +ode ranc% Loop -es%

+earning a ne,"oca ular. /



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Electric Circuit Connections

&erminal point $%ere a component or part of t%ecircuit connects to ot%er components or ot%er parts oft%e circuit

#erminalsv1@2 )

v1@2

) 1

) >

) 3Eac% element alone%as t$o terminals

/ connection ofcomponents %as terminals

v1@2

) 1

) >

) 3

) 9

#%ese internalconnections arenot t%e terminalsof t%econnectedcomponents.



25/504

25g g

Electrically any$%erein %ere is t%e

connection point

icA a point and call it t%e node

) 3

) 1

) >

) 3

) 1

) 9

,ode connection point




26/504

26g g

E ample! "dentify +odes

Electrically,

any$%ere alongt%e $ires in t%isarea is t%econnection point

icA a point and

call it t%e node

+ote t%at t%is connectionis an electrical circuit andcould 'e a model for asystem suc% as t%eelectrical system in a car

) 1@2 ) > ) 3v1

La'el t%e nodes in t%e circuit6



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27g g


Branch portion of a circuit $it% only t$o e ternalterminals

terminals

Eac% element is a 'ranc%6#%e connection 'et$een and is also a 'ranc%

Eac% element is a 'ranc%6 #%econnection is + # a 'ranc%

'ecause it %as 3 e ternal terminalst1 , t3, and t 9 and t%ree

components connect at t > 6

) 1

v1@2 )

1

) >

v1@2

1t 3t

3t

>t

1t

9t

1t

>t

3t



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28g g

E ample 9! "dentify ranc%es

ind 'ranc%es in t%e circuit6

>8/@2

90(

Eac% of t%e components are individually a 'ranc%, 8

components means 8 'ranc%es6

irst identify t%e nodes6

#%ere is a branch containing t%e 90( source and t%e 3 and 1 resistors t%e connection %as t$o terminals6

#%e branch at t%e rig%t %as t$o terminals t%at connect t%is

su'2circuit to t%e rest of t%e circuit6

3 >

9=

1 >



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29g g

arallel Connection

Elements of a circuit $%ic% s%are t%e same t$o nodes

-1 32i1

Electricall.an.,!ere along

t!e ,ires in t!isarea is t!econnection oint

ic a oint an)call it t!e no)e

" 1

*ll com onents o t!e circuit connect to t!e same t,ono)es an) are t!ere ore in arallel&

e o ten use t!e s.m ol : to in)icate t!at elementsare in arallel !ere v 1 i 1 R 1 R 2 R 3



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30

>8/-

E ample

>3

9=1 >



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31

1

23

" 1- 54


+oop a closed connection of 'ranc%esMesh a loop t%at does not contain ot%er loops

Jo$ many nodes

Jo$ many mes%es

Jo$ many loops

#%e t$o mes%es plus t%et%ird e terior loop

>@1 3

orm a closed connection of 'ranc%es 'y starting at a node andtraversing t%e circuit until $e get 'acA to t%e starting node6Cannot use t%e same node t$ice6



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32

-

1

-

2

3 4" 2i1" 1

E ample =! "dentify nodes, loops, mes%es

=o, man. no)es;

=o, man. mes!es;=o, man. loo s;

!at is not a loo ;

5

4

5

* at! t!at crosses t!e same no)e t,ice

5

1 N 3 N > N

9 N



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33

Kirc%%off s Current La$ 4KCL5

E.ual amounts of c%arge enter and e it a node6 /lge'raic sum of currents into and out of a node is

&ero!

Convention 1! Current la'eled as pointing into anode is given a negative sign in t%e summation andcurrent la'eled as pointing out of t%e node is

positive6 Convention >! Current la'eled as pointing into a

node is given a positive sign in t%e summation andcurrent la'eled as pointing out of t%e node is

negative6

1

0 N

n

i=

=



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34

E ample 7! /pplying KCL

ind i3 in terms of t%e ot%er currents6

+ode

Using Convention 1!

Using Convention >!ot% conventions yield t%e

same result6 Be $illgenerally use Convention 1

*olve for i3!

*olve for i3!

1i>i

3i

9i

9

1 > 3 91

4i out of node @5

0 0n

i i i i i=

= + + =3 1 > 9i i i i= +

9

1 > 3 91

4i out of node 250 0n i i i i i= = + =

3 1 > 9i i i i= +



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35

E ample 7 4cont5! /pplying KCL

Fiven!

ind !

+ote! i2 = -3A meanst%at t%e current actuallyflo$s in t%e directionopposite to t%e arro$

+ode

1i>i

3i

9i

1 3/i = 9 >/i = +

3i 91 > 3 9

14i out of node @5

0 0n

i i i i i=

= + + =( )3 1 > 9 4 < 5 3 4 > 5

0/

i i i i A A A= + = + + + +=



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36

E ample 8! /pplying KCL

*ame pro'lem $it% some of t%e current directionsc%anged6 ind i3 in terms of t%e ot%er currents6

+ode

Using Convention 1!

Using Convention >! ot% conventions yield t%e

same result6 Be $illgenerally use Convention 1

*olve for i3!

*olve for i3!

1i

>i

3i

9i

9

1 > 3 91

4i out of node @5

0 0n

i i i i i=

= + + + =

3 1 > 9i i i i=

9

1 > 3 91

4i out of node 25

0 0n

i i i i i=

= =3 1 > 9i i i i=



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37

) 1@2 ) > ) 3

) 9

v1

E ample :! /pplying KCL in a Circuit

KCL at +ode 1 !

at +ode >!

i1 i2 i3

i4

i s

/pply KCL at eac% node6

"dentify t%e nodes in t%e circuit and la'el6

1 N

> N

3 91

4i out of node @5

0 0S n

i i i i i i=

= + + + = 3 91

4i out of node @5

0 0S n

i i i i i i=

= + =



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38

E ample : 4cont5! /pplying KCL in a Circuit

,e ne, some actual "alues$

1 > 3


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39

E ample 10! *olving a Circuit Using KCL

B%at is t%e current t%roug% t%e


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40

*eries Connections

Elements of a circuit connected so t%at t%e current outof one component goes into t%e ne t6

>8/@290(

Be say t%at t%e 90( source, t%e 3 resistor, and t%e 1 resistor are connected in series-

3 >

9=1 >



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41

E ample 11! *%o$ t%at series components %ave t%e same current6

>8/@290(

ia

ib

ic

+ 1 + > + 3

+ 9 + < + =

Brite KCL e.uations at nodes + 1 and + 9!

9

a b ci i i= =

3 >

9=

1 >

9

11

4i out of node @5

! 0 0a bn

N i i i=

= =9

>1

4i out of node @5

! 0 0b cn

N i i i=

= =



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42


function!o rocess

"nformationo #ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er

Circuit Components Resistor Voltage Source Current Source S!itch

Connections &erminal ,ode Branch +oop Mesh


)eductions *ource #ransformation %arallel . same voltage Series . same current #%evenin

+orton

/nalysis #ools

/irchoff*s Current +a! +ode (oltage -et%odKirc%off s (oltage La$

-es% Current -et%od*uperposition

*ummary

DC Lecture 1 2DC Circuit Components,Connections, and KCL



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43

Cypria

*n)re, (ount?os!ua (ount


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44

ECE 307 Lecture >KCL and K(L


Clemson University



45/504


function!o rocess

"nformationo

#ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er o Energy




)eductions *ource #ransformation arallel *eries #%evenin +orton

/nalysis #oolsKirc%off s Current La$

+ode (oltage -et%odKirc%off s (oltage La$



DC Lecture > KCL andK(L



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B%y Do Be erform Circuit /nalysis

Be analy&e a circuit or solve a circuit in order toidentify t%e currents, voltages, and po$erconsumption in t%e circuit6

Circuits are analy&ed 'y application of t%e t%ree la$s!

%m s la$ Kirc%off s current la$ Kirc%off s voltage la$



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*ense c%anges in lengt% 4temp, force, 665

B%y Do Be erform Circuit /nalysis#o Understand )eal Devices *uc% as a *train Fauge

cvcv

!tt $ ,,,&societ.o ro ots&com images sensorsF tstraingauge&@ g



48/504

B%y Do Be erform Circuit /nalysis#o Understand )eal Devices *uc% as a *train Fauge

!eatstone Bri)ge

(easurement Circuit

*t start$ 1 4 G 2 3 * ter sur ace strain$ 1 4 an) 2 3 " a an) "

Knee Implant Sensor http://www.medgadget.com/ar chives/200 /!!/wireless"sensor.html



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)eminder! KCL

+et flo$ of current out of a node is &ero6

/lge'raic sum

/ negative sign isused to account for acurrent t%at %as a

reference direction pointing into t%enode6

+a

i1

i2

)eference directionout of node @

)eference directioninto node 2

( )currents out of t%enode are positive

0

N

ni =



50/504

E ample 1! Use KCL to *olve Circuit

Brite KCL at node c

Fiven! v> 9(ind i3 2@ @ 2

i1 i3 i4

+ a

+ c

+ 'i1 i4

i3

v2v5

i2

+ote! B%en talAing a'out a voltage orcurrent a reference direction must 'egiven6

t%er voltages and currents are assumed4ar'itrary5 in order to analy&e t%e circuit6

La'el all nodes to see t%at $e could apply KCL at node c in order to find i36

+ c

i5 + d

10(1/

>

( ) ( ) ( )3

1 3 91

4i out of node @5

0 0n

i i i i=

= = + + + =

3 1 9i i i

= +



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E ample 1 4cont5! Use KCL to *olve Circuit

2@ @ 2i2

i4

+a

+ c

resistor

i1 is t%e currentin t%e currentsource

*u'stituteand solve!

2@ @ 2 +

a

+ c

+ 'i1 i4

i3

v2v5

i2

1 e.uation $it% 3 unAno$ns, use giveninformation to reduce t%e unAno$ns!

+ di5

10(1/

>

>> > > >

>

9() >/

) >v

v i i= = = =

1 1/i =

3 21/@>/ 1/i =

3 1 >i i i= +> 9i i=



53/504

E ample >! Do2over of E ample 1

*olution! Brite KCL at node a

Fiven! v2 9(ind i3 2@ @ 2

+ a

+ c

out of +ode a are

positive

0i i i i= + =

3 < >i i i= +

10(1/

>

3 1 >i i i= +



54/504

E ample > 4cont5! Do2over of E ample 1


4same5

4same5

*ame result M

2@ @ 2 +

a

+ c

resistor

i1 is t%e currentin t%e currentsource

+ di5

3 1 >i i i= +10(1/

>

>> > > >

>

9( ) >/

) >v

v i i= = = =

1 1/i =

3 21/@>/ 1/i =



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(oltage "s a )elative -easure

(oltage descri'es $orA done to move c%arge 4'attery5or t%e $orA done 'y moving c%arge 4lig%t 'ul'56

"n a circuit it is t%e $orA done to move c%arge 'et$een > points

-ust la'el > points in t%e circuit to descri'e a voltage ne side $it% 2 ne side $it% @

v4 + + 1

@2

@

2

v1

@ @@

v3v2



56/504

Kirc%%off s (oltage La$ 4K(L5

*um of t%e voltage drops around a closed pat% is &ero

/lge'raic sum

/ negative sign isused to account for avoltage rise6

/lternatively, $e couldconsider voltage rises as

positive6 Eit%er $ay $orAs 'ut$e must 'e consistent

"n t%is second case, anegative sign is assigned tovoltage drops

/lternatively, $e could %ave usedt%e counter clocA$ise direction6-ust 'e consistent

( )voltage drops in t%e

clocA2$ise directionare positive

0 N

vn =



57/504

Kirc%%off s (oltage La$ 4K(L5

/lge'raic sum

/ negative sign isused to account for avoltage rise6

"n t%e CB direction, enter t%e @ terminaland leave from 2 2? voltage drop 2? @@ 2v1

2v2@

CB Direction

"n t%e CB direction, enter t%e 2 terminal andleave from @ 2?voltage rise 2? 2

/ voltage rise means you enter from t%e 2 and leave from t%e @ in t%edirection of t%e loop6 / voltage drop means you enter from t%e @ andleave from t%e 2 6

( )voltage drops in t%eclocA2$ise direction

are positive

0

N

vn =

1 D

> D



58/504

E ample 3! /pplication of K(L 4E 6 >6=5

D i device could 'e aresistor or somet%ing else

vi are t%e component

voltages

16 )ecogni&e t%at D> and D 3 %ave t%e same t$o nodes parallel

>6 Jo$ many loops in t%e circuit t%ree36 Jo$ many loops include v2 #$o, Loop 1 and Loop >

K(L Loop 1 @ 2v2

@

2

v 4

+ 3 + > + 1

+ 9

2v3@

2v1@

K(L Loop 3

K(L Loop >

Fiven v1 =(,

v4 1(, v s 1>(ind v2

S v

1 D

> D

3 D

9 D



59/504

E ample 3 4cont5! /pplication of K(L 4E 6 >6=5

K(L Loop 1 @ 2v2

@

2

v 4

+ 3 + > + 1

+ 9

2v3@

2v1

@

#raverse Loop 1 in t%e clocA2$ise 4CB5 direction

Use t%e convention t%at voltage rises 42 to @5 $ill 'e negative$%ile voltage drops 4@ to 25 $ill 'e positive

/pply K(L to Loop 1

S v

1 D

> D

3 D

9 D

( ) ( ) ( ) ( )9

1 > 9i 1

0i S v v v v v=

= + + + =



60/504

E ample 3 4cont5! /pplication of K(L 4E 6 >6=5

K(L Loop 1 @ 2v2

@

2

v 4

+ 3 + > + 1

+ 9

2v3@

2v1

@ K(L Loop >


'o! !ould 0 apply /V+ to +oop #" Could 0 "

%arallel elements have thesame voltage

S v

1 D

> D

3 D

9 D

> 1 92 2 1>(2 =(21(

> 3i 1

> 3

2 0iv v v

v v=

= + =

=



61/504

arallel Connections

Elements of a circuit $%ic% s%are t%e same nodes also%ave t%e same voltage6

Use K(L to s%o$ t%at t%e voltage across all parallel elements is t%e same6

K(L Loop 1 K(L

Loop >

K(L Loop 3

@

2v1

@

2v2

@

2

v3

%arallel elements have thesame voltage

+ 1

+ >

1 > > 1drops, CB1! 0 v v v v v= + = =

> 3 > 3drops, CB

>! 0 v v v v v= + = =1 3 1 3

drops, CB

3! 0 v v v v v= + = =

1 > 3v v v= =



62/504

E ample 9! Clarification of arallel Elements

E ample of arallel Elements! 9 and >8/ source are in parallel6

+e$ Nuestion! "f v4! = v 4 does t%at mean t%at t%e 9 resistor and t%e 90 arein parallel

/ns$er! + , t%ey must also s%are t%e same t$o nodes to 'e in parallel6

+ 9

+ > + 3 + 1

+ < + =

@290( >8/90

@

2v4!

@

2v4

9


63/504

E ample

Oou are given t%is !

B%at is v

K(L

K(L loop 1

" -v R

"- v

drops, CB0

N

nv =

1>(

(24


64/504

E ample =! Use of K(L to *olve / Circuit 4a little morecomplicated5

2 =( @

"f v2 9(, $%at is vcK(L loop 1

+ a

+ '

+ c

+ d

@2 vc/pply K(L to Loop a'c

10(1/

>

>v+

>drops, CB

>

10( 0

10( 2 49(5 @10( =(

N

n #

#

v v v

v v

= + + =

= +

C)



65/504

E ample 7! Use of K(L P KCL #oget%er to *olve a Circuit

*ince $e are given t%e sourcevoltages and resistance values, $e$ant t%e current in eac% 'ranc% andt%e voltage across eac% element6

*teps!16 La'el nodes>6 La'el and assign directions for t%e current in eac% 'ranc%

4ar'itrary536 /ssign unAno$n element voltages in terms of t%e ar'itrary

assignments made for currents

< a < < c

< )

i1 i2

i3

3>(

9

>

8 >0(1v

+

>v +

+

3v


E ample 7 4cont5! Use of K(L P KCL #oget%er to *olve a


66/504

E ample 7 4cont5! Use of K(L P KCL #oget%er to olve aCircuit

15 KCL at node '!

>5 K(L around loop 1!

35 rom %m s la$ $e see t%at!

95 Com'ine >5 and 35 to get!

K(L loop 1K(L loop >

+ a + ' + c

+ d

i1 i2

i3

@2 v>@

2v3

" -v11 > 3

out of node '

0i i i i= + = 3>(

9>

8 >0(

1 3drops, CB

3>( 0nv v v= + + =

1 1 3 3> i and 8 iv v= = 1 > 3> 40 5 8 3>(i i i + + =


E ample 7 4cont5! Use of K(L P KCL #oget%er to *olve a


67/504

E ample 7 4cont5! Use of K(L P KCL #oget%er to olve aCircuit

!

=5 rom circuit $e see t%at!

75 Com'ine

"- v2"

-v3

" -v1

+ a + ' + c

+ d

3>(

9>

8 >0(

3i

>i1i

> 3drops, CB

>0( 0 N

nv v v= + => > 3 39 i and 8 iv v= =

1 > 340 5 9 8 >0(i i i+ + =



68/504

Jo$ to Fet t%e *olution

#%ree e.uations $it% t%ree unAno$ns gives us a system of simultaneous e.uations

$%ic% are!

#%is system of e.uations can 'e solved 'y!16 *u'stitution>6 Cramer s rule36 / calculator suc% as #" 8:96 -/#L/ , EQCEL, -/ LE

1 > 3

1 > 3

1 > 3

0

> 40 5 8 3>(

40 5 9 8 >0(

i i i

i i i

i i i

+ = + + =

+ + =



69/504

*olution 'y Cramer s )ule 4see pp10832108< of 'ooA5

*: (atriH

*: (atriH ,it! :re lacing irstcolumn

*: (atriH ,it! :re lacing secon)

column*: (atriH ,it! :

re lacing t!ir)column

Determinant

4215

1 1 1

> > >

1 1 1 1 1 1 1 1415405485 415485405 4 154>5495

> 0 8 > 0 8 > 0 5415 405405415

0 9 8 0 9 8 0 9

0 1 1 0 1 1 0 13> 0 8 3> 0 8 3> 0 >>9 9/

>0 9 8 >0 9 8 >0 9

1 0 1 1 0 1 1 0

> 3> 8 > 3> 8 > 3> 0 8 0 >0 8 0 >0

D D

D D i

D D i

+ + = = = =

= = = =

= = = =

3 3 3

1/

1 1 0 1 1 0 1 1

> 0 3> > 0 3> > 0 1=8 3/

0 9 >0 0 9 >0 0 9

D D i= = = =

31 >1 > 3R R D D D

i i i D D D= = =



70/504

*olution Using #" 8: Calculator

#"28: #itanium

ans 960000 2160000 360000

{ } { }( )&eros , > 8 3>, 9 8 >0 , , , $ y z $ z y z $ y z + + +

1 > 3

1 > 3

1 > 3

0

> 40 5 8 3>(

40 5 9 8 >0(

i i i

i i i

i i i

+ = + + =

+ + =

1

>

3

i

i

i


l


71/504

*olution Using a -atri /pproac%

#%ree e.uations $it% t%ree unAno$ns gives us a system of simultaneous e.uations

$%ic% are!

T!ese can e ,ritten inmatriH orm *iG as$

1 > 3

1 > 3

1 > 3

0

> 40 5 8 3>(

40 5 9 8 >0(

i i i

i i i

i i i

+ = + + =

+ + =

0

3>

>0

1 > 3

1 > 3

1 > 3

1 1 1

> 0 80 9 8

i i i

i i ii i i

1

>

3

1 1 1

> 0 80 9 8

i

i

i

=


* l i U i i / %


72/504

*olution Using a -atri /pproac%

y %m s la$!

i1 9 /i2 21 /

i3 3 /

v1 = % (

v2 = -4 (v3 = 24 (

1 1

> >

3 3

> 8

9 9 (olts8 >9

v i

v iv i

= =

1

>

3

91 /

3

ii

i

=


* l i U i i / % 2 /#L/ * f $


73/504

*olution Using a -atri /pproac% 2 -/#L/ *oft$are

Feneral ro'lem! / ' Feneral *olution! / 21 ' $%ere / 21 is matri inverse6

4-/#L/ is availa'le for all students to install, license $orAs oncampus or 'y ( +5


* l i U i i / % 2 /#L/ * f $


74/504

*olution Using a -atri /pproac% 2 -/#L/ *oft$are

*GI1 1 -1J 2 0 8J 0 4 8K 1 1 -1 * G 2 0 8 0 4 8

GI0J32J20K 0 G 32 20

in" * Lans G 4&0000 -1&0000 3&0000

Mser >n ut

Mser (*T+*B

out ut 1

>

3

ii

i


* l i U i i / % 2 * E l


75/504

*olution Using a -atri /pproac% 2 -* E cel

Result

4

-1

3Enter t!e coe icientsan) constants

Nelect an area t!e same siOe as t!e constants 3 ro,s 1 colClic on t!e ormula ar an) enter G mmult min"erse *2$C4 E2$E4T!en simultaneousl. ress Control s!i t an) enter&

1

>

3

i

i

i


*


76/504

*ummary

%m s La$! v i)6 #%e current flo$ is proportionalt%e voltage

Kirc%off s Current La$ 4KCL5! Si n 0

#%e alge'raic sum of currents at a node must e.ual&ero6

Use! *um of t%e currents out of a node 0

Kirc%off s (oltage La$ 4K(L5!

Tvi 0 #%e alge'raic sum of t%e voltage drops around a closed

loop must e.ual &ero6 Use! *um of t%e voltage drops around a CB loop 0

Be use t%ese 'asic la$s to solve a circuit


Science As Art at Clemson


77/504

Calc#lated ChaosEric %enimore

#escri tion$T!e creation o ire,or sin"ol"es no,le)ge oc!emistr. ,!at materials toinclu)e to get t!e )esire)colors !.sics ).namicsan) artistr. ,!at colorss!a es atterns an) soun)s

s!oul) t!e ire,or emit suc!t!at it is [email protected] le to ,atc! &T!is icture is an eHtremel.clear ocuse) close-u o t!einstant ,!en a ire,or is

)etonating&

4%ttp!HHgeo6ces6clemson6eduHgalleryHmain6p%p5



78/504

ECE 307 Lecture 3o$er and Energy and E.uivalent )esistance

Department of Electrical and ComputerEngineering Clemson University


i $ f DC El t i Ci it


79/504

Electric Circuit

erforms afunction!

o rocess"nformation

o

#ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er o Energy

Circuit Components )esistor

(oltage *ource Current *ource *$itc%

Connections

#erminal +ode ranc% Loop -es%




+ode (oltage -et%od

Kirc%off s (oltage La$ -es% Current -et%od

*uperposition


DC Lecture 3 o$er andEnergy and E.uivalent)esistance


o$er in Electric Circuits


80/504

o$er in Electric Circuits

Batts

Convention#%is is t%e po$er dissipated 'yt%e element $%en voltage andcurrent are dra$n as s%o$n6

genericcomponent

"nterpretation! "f p ? 0 t%en t%e device consumes po$er "f p 0 t%en t%e device supplies po$er

2

( )B p vi=

v

i


E ample 1! Using t%e *ign Convention in o$er Calculations


81/504

E ample 1! Using t%e *ign Convention in o$er Calculations

dissipates po$er

device supplies po$er, e6g6 'attery

dissipates po$er

ecause of la'eling

@

2v

@

2v

@

2v

i >///


82/504

/merican Bire Fauge 4/BF5


E ample >! o$er Loss in a Bire


83/504

E ample >! o$er Loss in a Bire

ne gauge $ire %as a resistance of 061>3:VH1000ft6 /ssume >,000 ft and 1/

current6 B%at is voltage drop along t%e $ire and po$er consumed 'y t%e $ire

9601=VH1000ft ) 8603> V p=vi and v=iR /lso p=vi and i=v&R

*implified t$osteps into one

@

2v

+ote! 1= gauge$ire is a smallerdiameter t%an 1gauge

#%is is t%e amount of po$ergiven off from t%e $ire as %eat

)epeat for 1= gauge $ire

'G> G 1* 0&2478 G 0&2478'G'>G 0&2478' 1* G 0&2478

G2 000 t 0&1239 1000 tG 0&2478

1/i =

>

>41/5 :V :B

p i R== =

>v p

R

=


E ample 3! o$er Calculation Components Can *upply or


84/504

Consume o$er

)4very small,ignore5

)4very small,ignore5

16


85/504

o$er

v> v1

i 10V

@ 2v)

1

>

v

> >

v

4 9(54 06=/5 >69B 4supplies po$er5

406=/5 410 5 36=B 4consumes po$er5

4>(54 06=/5 16>B 4supplies po$er5 R

p vi

p i R

p vi

= = = = = == = =

1

>

> >

406>/549(5 068B 4supplies po$er5

) 406>/5 410 5 069B 4consumes po$er5

406>/54>(5 069B 4consumes po$er5

v

R

v

p vi

p i

p vi

= = = = = =

= = =

1 >

ind po$er consumed 'y eac% component

if 9(, >( >(, 06>/ ! Rv v v i= = = =

1 >

ind po$er consumed 'y eac% component

if 9(, >( =(, 06=/

4+one of t%e reference directions %ave c%anged5 Rv v v i= = = =


Energy in Electric Circuits


86/504

Energy in Electric Circuits

Energy consumed

Units

;oules Bs (/s W (/% 2?/%

Fiven constant po$er

Constant voltage and current

B%en t%e voltage is "mplied Example: 1222 mAh for aremote control car !here $#V isimplied

s seconds% %ours

( ) ( ) ( )0

0E ( t

( t

p d p t p t vi t = = =


E ample


87/504

E ample

/ 1>( 'attery is rated at 1(41


88/504

)esistors in eries

)esult >! Using %m s la$ for eac% ), t%evoltage and current for t%e $%ole resistor

'ranc% is related to t%e sum of t%e individualresistances

v

"v1

-"v2-"v3

-

*umming voltage drops around mes% 1!

Neries esistors are on t!e Name Branc! - Name Current

)esult 1! #%e total voltage across all of t%e

resistors gets divided 4split5 'et$een t%eindividual resistors!

(oltage Divider

Circuit )eduction

1)

>)

3)

1m

i 1 > 3n

9

nE1

E 2 E 0 v v v v v+ ++

1 > 3

1 > 3

) ) )

4) @) @) 5

v i i i

i

= + +=

1 > 3v v v v+ +=


)esistors in *eries (oltage Divider


89/504

)esistors in eries (oltage Divider

Eac% resistor voltage isa scaled version of v

v is scaled or divided 'et$een t%e resistors

v

"v1

-"v2-"v3

-

Continuing from previous slide, solve for i

Brite eac% resistor voltage!

#%is result is true for any num'er 4+5 of resistors inseries6 )esistor voltage for t%e n t% resistor is!

(oltage Divider E.uation

1 > 3) ) ) vi = + +

>> >

1 > 3

) )

) @) @) v i v= =

11 1

1 > 3

) )

) ) ) v i v= = + +

1)

>)

3)

1m

i

33 3

1 > 3

) )

) ) ) v i v= = + +

1 > 3

) ) ) )

nnv v= + +

1

)

)

nn N

)

)

v v

=

=


)esistors in *eries Circuit )eduction


90/504

)esistors in eries Circuit )eduction

(rom the voltage source point of vie!3 t%ere is nodifference 'et$een t%e 3 individual resistors inseries and one large resistor

Be can redra$ t%e circuit using a singleresistance

Be %ave lost information a'out t%e individual

components, e6g6 v1, v>, v3

v

"v1-"v2-"v3-

v

rom earlier slide!

$%ere

#%is result is true for any num'er 4+5 of resistors inseries6 #otal resistance is!

E.uation for E.uivalent)esistance of *eries )esistors

1)

>)

3)

1m

i

) e*

i

1 > 3 1 > 3) ) ) 4) @) @) 5 ) e*v i i i i i= + + = =

1 > 3) ) @) @) e* =

1

) ) N

e* )

) =

=


E ample =! *eries Lig%t *tring


91/504

E ample ! eries Lig%t tring

*tep 1! ind e.uivalent resistance and solve fort%e current

*tep >! "ndividual 'ul' voltages

C%ecA!

4

"v1-"v2-

"

v1!!-

v1!!

1) 1=

100( >) 1=

100) 1=

i

) e*

i

1 1 > 3 100) 1/41 5 1( 666v i v v v= = =

( )100

1

) ) 1 1 6 1 100e* ) = = + + + = L100(

100(1/

100

i = =

100

1 > 3 1001

100( 666 0() )

v v v v v=

= + + + + + =


)esistors in arallel


92/504

)esistors in arallel

+ '

+ a

arallel esistors N!are t!e Name 2 31

out of a

0nn

i i i i i=

= + + + =

1 > 3

1 > 3

) ) )

1 1 1@ @

) ) )

v v vi

v

= + +

=

1) v

i

>) 3)

1i >i 3i

1 > 3i i i i= + +


)esistors in arallel 2 Current Divider


93/504

)esistors in arallel 2 Current Divider

+ '

+ a

Eac% resistor currentis a scaled version ofi

i is scaled ordivided 'et$een

t%e resistors

Continuing from previous slide, solve for v in terms of i

Brite eac% resistor current!

#%is result is true for any num'er4+5 of resistors in parallel6 )esistorcurrent for t%e n t% resistor is!

Current Divider E.uation

1

1 > 3

1 1 1@ @) ) )

v i

=

1) v

i

>) 3)

1i >i 3i

1

1 > 311 1

1 1 1@ @

) ) )

) ) vi i

= =

1

1

1)

)

N

) ) n

n

i i

= =

1

1 > 3>

> >

1 1 1@ @

) ) )

) ) v

i i

= =

1

1 > 33

3 3

1 1 1@ @

) ) )

) ) v

i i

= =

1

1 > 3

1 1 1@ @

) ) ) ) n n

i i

=


)esistors in arallel 2 Circuit )eduction


94/504

)esistors in arallel 2 Circuit )eduction

+ '

+ a

+ '

+ a

(rom the voltage source point of vie!3 t%ere isno difference 'et$een t%e 3 individual resistors in

parallel and one smaller resistor Be can redra$ t%e circuit using a single

resistance Be %ave lost information a'out t%e

individual components, e6g6 i 1, i>, i3

rom previous slide!

#%is result is true for any num'er 4+5 of resistors in parallel6 #otal resistance is!

E.uation for E.uivalent)esistance of arallel )esistors

$%ere

) e*v

) e*

vi =

i

1) v

i

>) 3)

1i >i 3i

1

1 > 3

1 1 1) @ @) ) ) e*

=

1

1 1) )

N

) e* ) ==


E ample 7! arallel Lig%t *tring


95/504

E ample 7! arallel Lig%t tring

4 "n eac% 'ranc%, $%at is t%e 'ul' resistance sot%at pi 1B

ind e.uivalent resistance of t%e string and solvefor t%e current i

@

2

v R

1)

100+

i

>) 100)

100i

>1 1

1 1 1 11 1

1B) ) v v

p i v v = = = =

> >1

14100(5

) 10A 1B 1Bv= = =

>11

1

100(10 10 / 10m/

) 10A v

i = = = =

100>

1

1 1 1100 10 H

) ) 10A

100() 100 1/

100

) e* )

e* i

=

= = =

= =

>i1i

) e*

i

100(


E ample 8! Compare *eries and arallel Lig%t *trings


96/504

p p g g

4 4

*tring failure modes! "f one of t%e series 'ul's 'reaAs 4open2circuit5, t%en i 0 for all 'ul's6 "f one of t%e parallel 'ul's 'reaAs 4s%ort2circuit5, t%en v 0 for all 'ul's6

@

2v1

@

2v2

@

2

v1!!*ame current and po$er 2?from t%e So'rce oint o( +ie ,t%ere is no difference 'et$eent%e series and parallel strings

arallel re.uires larger resistance

arallel resistor %as %ig%er voltage*eries resistor %as %ig%er current

/ll 'ul's consume t%e same po$er

100( 1) 100(

i>) 100)

1i

>i 100i1 1 R =

> 1 R =

100 1 R =

i

100() 100 1/ 100B

100e* tota i p= = =

) 1 ) 10A ) ) = =

1( 100() ) v v= =1/ 10 /) ) i i m= =1B 1B) ) p p= = =


)educing Circuits Using *eries and arallel E.uivalent) i t


97/504

)esistances

/ se.uence of series and parallel reductions can 'eused in a circuit6

Foal is to simplify t%e circuit $it%out losing t%eidentity of t%e voltage or current of interest6

rocedure16 erform a reduction 4series or parallel5>6 )edra$ circuit

36 / reduction may create t%e opportunity for a ne$reduction6

)epeat


E ample :! Use ot% *eries and arallel )eductions to ind )e.


98/504

p ) )

Step1/ R 2 and R3 are in series!

#%e circuit reduces to

1 > 3) ) ) ! ) and ) are in parallele*Step

1e* R

1) e*


E ample 10! Using )esistor )eductions to *implify Circuit*olution


99/504

*olution

) 1

) 3

1/ ) >

ind i1 and i2

Com'ine R2 and R3 in series!

Use Current Divider!

i1 )e. 1H4) 1@)e. 15 X i


100/504

ind v R2

P v R3

Circuit *olution

) >@

2v R2

@ 2v R3) 1

) 3

1/Use current i > and %m s La$

( ) ( )>

> >) 06= 30 18( Rv i= = =( ) ( )

3 > 3) 06= >0 1>( Rv i= = =

>i

1i

i


E ample 11! *e.uence of *eries and arallel )eductions to ind)e


101/504

)e.

rom inspection!

B%ere!

1>39 3 9 < =) ) @) ) @) ) @) e* =

$y $ y

$ y= +

ind e* R


E ample 11 4cont5! *e.uence of *eries and arallel )eductionsto ind )e


102/504

to ind )e.

Fiven!

< =) ) 3 9 < =) ) @) ) @) ) @) > >6= 96=e* = = +

1 > 3 9

< =

) > ) 9 ) < ) ) 3

= = = = = =


otentiometer


103/504

Circuit-odel

"dea C%ange resistance as

t%e s%aft rotates

Devicest%at useangle

sensing

Fiven/ngle

T!is is a'oltage#i"i)er


otentiometer


104/504

(easurementan)

*nalog-#igital

Con"ersion

#igitalCom uterController


*ummary


105/504

y

o$er and Energy in DC Circuits p vi

or current t%roug% a voltage drop, p ? 0 means po$er is dissipated in t%e component and t%e

circuit supplies po$er to t%e component or current t%roug% a voltage rise, p 0 means

po$er is supplied 'y t%e component andcomponent supplies po$er to t%e circuit6

"n a resistor

E pYt viYt

>> v p i R

R= =


*ummary 4cont5


106/504

E.uivalent Circuits #%e terminal 'e%avior, voltage and current, is

unc%anged *eries and parallel reductions can 'e used to create

simplified, electrically e.uivalent circuits6 *eries resistors com'ine into an e.uivalent resistance as

arallel resistors com'ine into an e.uivalent resistanceas

1

) ) N

e* ) ) =

=

1

1 1) )

N

) e* ) ==


*ummary 4cont5


107/504

)esistor )esults *eries resistors divide a voltage 6

E ! for > resistors in series, voltage accross resistor 1 is

arallel resistors divide a current E ! for > resistors in parallel, current in resistor 1 is

11

1 >

)

) ) v v=

+

1

1 >

1 > 1 >1

1 1

1 1 ) ) ) ) ) )

) ) i i i

+ + = =

108ECE 307 Basic Electrical EngineeringScience As Art at Clemson4%ttp!HHgeo6ces6clemson6eduHgalleryHmain6p%p5


108/504

Bamboo $ipewor%s

=oll. Tuten

#escri tion$

T!is !oto s!o,s amosPuito ree)ing !a itat ina Nout! Carolina Ooological

ar & !ile collecting

mosPuito lar"ae t!e cano .a o"e ,as !otogra !e) orlater anal.sis o cano .co"erage ,it! gra !ic artsso t,are&

4%ttp!HHgeo6ces6clemson6eduHgalleryHmain6p%p5



109/504

ECE 307 Lecture 9#%e +ode (oltage -et%od

Department of Electrical and ComputerEngineering Clemson University




110/504

Electric Circuit

erforms afunction!

o rocess"nformation

o #ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er o Energy

Circuit Components )esistor

(oltage *ource Current *ource *$itc%

Connections

#erminal +ode ranc% Loop -es%

*imilar

Electric Circuit

)eductions

*ource #ransformation arallel *eries #%evenin +orton


+ode (oltage -et%od

Kirc%off s (oltage La$ -es% Current -et%od*uperposition

DC Lecture 9 #%e +ode(oltage -et%od


E plosive Fas *ensor


111/504

Fas t%atmay 'e

e plosive

@2

i

Aas lo,st!roug! sensor


112/504

/ voltage must 'e measured 'et$een t$o points in acircuit, $e %ave 'een measuring across elements6

/not%er possi'ility! one node is c%osen as t%ereference point for all ot%er voltages6 #%e )eference +ode is marAed 'y #%e )eference +ode often 4'ut not re.uired5 %as a

p%ysical meaning in $%ic% case it is called a ground /ll node voltages are s%o$n relative to t%e

reference node6

- ) v


E ample 1! C%oosing a )eference +ode and La'eling +ode(oltages


113/504

( g

rame of t%eve%icle is aconductor Could consider t%e frame as a

reference node in analysis

'e!icle

+ote! #%e das%ed lines indicate t%at t%ere are several parallel elementsconnected 'et$een + 1 and + 9 and all %ave t%e same voltage6

La'el t%e +ode (oltages in t%e Circuit

Z of node voltages Z of nodes 21

1>( >v1v

3v

1 N

3 N

> N

1v1v1v

9 N


#%e +ode (oltage /pproac%


114/504

Solve theentire circuit

using the ,ode

Voltage Method

(indcomponent

voltages

(indother

electrical5uantities

such ascurrent orpo!er/ll

+ode

(oltages

*pecificComponent

(oltages

Current,

o$er, etc


+ode (oltage -et%od *ystematic /pplication of KCL


115/504

16 La'el all n of t%e nodes and *elect a )eferencenode

>6 Decide if t%e remaining n21 node voltages aredependent or independent6 / connected voltage

source $ill maAe a node dependent6 Count t%e mdependent nodes636 Brite KCL e.uations at eac% of t%e n212m

independent nodes6 Brite m e.uations to relate

t%e dependent node voltages to t%e sourcevoltages6

96 *olve n21 e.uations6


reliminary 2 Jo$ to Brite t%e Current at a +ode Using +ode(oltages


116/504

( g

)eference +ode

@ 2 v1 N b

N c

N d N a

vavb

vc

vd

@@@

2

) 1

) >

) 3

i1i2

i3

2

"mportant oint! #%e ) 1 component current $as $rittenin terms of t%e t$o node voltages 4v ' and v a5

1! 0a b 0+ v v v + =

11

1 1) ) b av v v

i

= =

1*olve to yield! b av v v=




117/504

( g

)eference +ode

)

@

2

+ode of"nterest

vnode o( interest vad acent

/dGacent +ode

K(L t%at includes )esistor of"nterest , +ode of "nterest ,and /dGacent +ode

@

Can no$ apply t%is to all nodes and resistors in t%ecircuit $it%out t%inAing a'out t%e sign convention

@2 v R

%m s La$ for )esistor of "nterest

+o$ e tend t%e result of t%e previous slide to any resistor in t%e circuit

adGacent ) node of "nterest,

) node of interest adGacent

0) drops #2

v v v v

v v v

= + ==

node of interest adGacent) out of node of interest ) )

v vvi = =

ut of node of interesti




118/504

( g

)eference +ode

)

2

+ode of"nterest

vnode o( interest

/dGacent +ode

@

%m s La$ for )esistor of "nterest#%ere is no adGacentnode in t%is case

@ 2 v)

N ecial Case ,!ere a)@acent no)e is t!e e erence


119/504

*olve simultaneous e.uations to find t%e +ode (oltages v a and v '

Z nodes 1 2 Z voltage sources n212 m 32120 > KCL e.uations

)edra$ to

emp%asi&enodes

va

@

2

v '

@

+ a + '

+ c

) >

) 1 ) 3i s va

2

vb

@

i s) 1

) >

) 3

@ i2i3

i1

i s + '

+ c

+ a

out of 1 >node a

0) ) a a b

s

v v vi i

= + + = 1 > >

> > 3

1 1 1) ) )

1 1 1 0) ) )

a s

b

v i

v

+ = +

out of 3 >node '

2@ 0

) ) b b av v vi = =


+ode (oltage -et%od


120/504

Be no$ Ano$ %o$ do all of t%e steps in t%e +ode(oltage -et%od

16 La'el all nodes and select )eference +ode Z nodes n

>6 "dentify dependent nodes 4voltage sources5 Z dependent nodes m

36 Brite n212m KCL e.ns6 @ m e.ns6 to descri'edependent nodes 4al$ays need a total of n21

e.ns596 *olve


E ample >! *traig%tfor$ard +(-, 3 +odes


121/504

16 La'el all nodes andselect )eference +ode

>6 "dentify dependent nodes 4voltage sources5

)eference +ode

+ a + '

+ c2

vb

@ @

va

36 Brite n 21 m > KCLe.ns6

n 3

m 0

96 *olve

i2

) 1

) >

) 3

i11

out of node a 3 > 1) ) ) a b a b av v v v vi i = + +

1

>

1 > 3 > 3

> 3 > 3 9

1 1 1 1 1) ) ) ) )

1 1 1 1 1) ) ) ) )

a

b

v iv i

+ + + = + + +

>

out of node ' > 3 9) ) )

b a b a bv v v v vi i = + + +

out of node of interesti





E ample 3! our +odes and a Dependent *ource


122/504

16 La'el all nodes and

select )eference +ode>6 "dentify dependent nodes 4+o KCL at node a5

36 Brite n212m > KCL e.ns @1 e.uation to descri'e dependent node

n 9

m 1

)eference +ode

96 *olve

@2

vcvbva

out of node ' 1 3 >

0) ) )

b a b c bv v v v vi = + + =

1 1 > 3 3

3 3 9

1 1 1 1 1) ) ) ) )

01 1 1

0) ) )

1 0 0

a

b s

c s

v

v i

v v

+ + + =

out of node c 3 9

0

) )

c b c s

v v vi i

= + =a sv v=


E ample 9! +(- Complete Circuit *olution


123/504

+ 1 + >

/t node 1/t node >


>6 "dentify dependent nodes

4voltage sources5

36 Brite n 21 2 m > KCL e.ns6

n 3

m 0

ind voltage at nodes 1 and >6

+ 3

@

2

v1

@

2

v2

1 1 >3/ 0> >v v v+ + =

> 1 >>/ 0> 9

v v v + =


E ample 9 4cont5! +(- Complete *olution


124/504

/t node 1

/t node >

/dd

96 *olve

1 >> =v v = 1 >> 3 8v v + =

>> >v =

> 1(v =

1 >6


125/504

/ny component voltage can 'e $ritten using one ort$o node voltages6


using the ,odeVoltage Method

(indcomponent

voltages

(ind

otherelectrical5uantities

such ascurrent or

po!er

/ll +ode

(oltagesComponent

(oltagesCurrent,

o$er


Component (oltage from +ode (oltage


126/504

Use K(L and t%e la'eled node voltages to find t%ecomponent voltages

901

ind v R1!! !K(L! 2 v1!! "v R1!! " v 11! =! v R1!! =v 1!! - v11!

ind v R4!! !K(L! 2 v4!! -v R4!! "v 4!1 =! v R4!! = 2v4!! "v 4!1

ind v R3!! !K(L! 2 v3!! 2v R3!! 0 v R3!! 2v3!!

nly need one node voltage $%en

component is attac%ed to reference node6

#%is is avery 'igcircuit

2

100

>00

110

1>>

300 - - - - - -

900

v)100@ 2

v R4!!@

2v R3!!

@

2

@

2

v1!!

@

v110

v4!1

@

@

v4!!

v3!!


Use Component (oltages to ind /ny t%er Electrical Nuantities


127/504

%m s la$, po$er e.uation, energy, etc6



(indcomponent

voltages

(ind

otherelectrical5uantities

such ascurrent or

po!er

/ll +ode

(oltagesComponent

(oltagesCurrent,

o$er


E ample


128/504



36 Brite n212m > KCL e.ns6 @ 1 e.uationto descri'e dependent node

n G 4

m G 1

ind i using +(- &

96 *olve

a a b a cv v v v vi = + + =

1=(, 18(a cv v= =

>/8

avi = =>0(bv =

out of node c

1/ 0>

c av vi = + =

i

bvcv


E ample =! +(- to ind Component Current, < +odes


129/504

ind current t%oug% t%e 10 resistor6

i10R


n 6 "dentify dependent nodes 4voltage sources5

36 Brite n212m 3 KCLe.ns6 @ 1 e.uation todescri'e dependent node

m 1i10V

96 *olve

90(av =

out of node '

0> 10 :

b a b b cv v v v vi = + + =

out of node c

0: 9 8

c b c c d v v v v vi = + + =

out of node d

1/ 08

d cv vi = =

av bv cv d v


E ample = 4cont5! +(- to ind Component Current, < +odes


131/504

96 *olve 4cont5


132/504

96 *olve 4alternative5 *olve 'y computer or calculator as asystem of 9 e.uations and 9 unAno$ns

#"28:*imultaneous E.uations

/pplication90(av =

out of node '

0> 10 :

b a b b cv v v v vi = + + =

out of node c

0: 9 8

c b c c d v v v v vi = + + =

out of node d

1/ 08

d cv vi = =

9< =9 10 0 00 8 3< : 0

0 0 1 1 8

1 0 0 0 90

a

b

c

d

vv

v

v

=

90301>>0

a

b

c

d

vv

+ v

v

=


E ample 7! -ore +(- /pplication ractice


133/504



36 Brite n 212 m 1 KCL e.ns6 @ 1e.uation to descri'e dependent node

n 3

m 1

ind i and vb

96 *olve(bv = >/9b av vi

= =

i


E ample 8! Even -ore +(- /pplication ractice


134/504

ind t%e voltage across t%e 90 V resistor usingt%e +(-6 ind t%e current t%roug% t%e 30 V resistor6

a '

c

KCL at node '!

16 La'el all nodes and select)eference +ode


36 Brite n 212 m 1 KCL e.ns6 @ > e.uations todescri'e dependent nodes

n 9

96 *olve

d

@

2v!

(oltages of Dependent nodes!

10(av =3>(cv =

ut of '

010 90 >0

b a b b cv v v v vi = + + = 109 1968 10 >>

0673/30 30 30

c av vi = = = =0 1968


135/504

C%ecA! Use +ode (oltages to /pply KCL at nodes

a '

c

d

30 0673/i =

9019680 >03> 19680068


136/504

/'sor's o$er Delivers o$er

C!ec $ o,er

*um all po$er ?0

*um all po$er 0

C%ecA

4delivered5 8Btota p =

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

( ) ( )

10(

3>>

30

>

>0

>90

>10

10 16>1: 1>61:B

3> 169B

063719 90


137/504

+ode (oltage -et%od

*olve circuit using Kirc%off s current la$ tofind all +ode (oltages

Use t%e +ode (oltages to find any voltage, current,

po$er etc6 in t%e solved circuit



(indcomponent

voltages

(indother

electrical5uantities

such ascurrent or

po!er


*ummary 4cont5


138/504

*teps of +ode (oltage -et%od16 La'el all nodes and select )eference +ode

Z nodes n

>6 "dentify dependent nodes 4voltage sources5 Z dependent nodes m

36 Brite n212m KCL e.ns6 @ m e.ns6 to descri'edependent nodes 4al$ays need a total of n21e.ns5

96 *olve


139/504

Close Encounter: Carbon,anotube Meets %rotein

*ung%o C%oi P Kurt FecAeler

Description!#%e computer2assistedillustration s%o$s t%e productof a car'on nanotu'einteracting $it% a protein for

t%e first time6 #%is novel classof nano'iocomposites isdesigned for 'iomedical andsensor applications6



140/504

ECE 307 2 Lecture K(L e.ns6

n >m 0

:orm (or resistors in s9ared branc9

:orm (or resistors in ;o'tside< branc9

Feneral orm in a -es% of "nterest!

4imes9 o( interest 2 iad acent 5)

@2

@2 @2

i2i1

1 1 1 > 1 > >drops inCB directionin mes% 1

> > 3 3 > 9 > 1 >drops inCB directionin mes% >

) 4 5) 0

) ) 4 5) 0

n

n

v + i + i i

v + i + i i i

= + + + =

= + + + + =


E ample 1 4cont5! *et up *olution 'y /pplying t%e *teps of t%e-C-


152/504

96 *olve

K(L E.uations from previous slide!

*olve 'y any means, e ample of matri solution!

i1 i>

@2

@2

@2

4$ould need actualvalues for t%evoltage sources andresistors to solve5

1 1 1 > 1 > >drops inCB directionin mes% 1

> > 3 3 > 9 > 1 >drops inCB directionin mes% >

) 4 5) 0

) ) 4 5) 0

n

n

v + i + i i

v + i + i i i

= + + + =

= + + + + =

1

>

i

i

1 > > 1 >1

> > 3 9 > 3>

) @) )

) ) @) @)

+ + i

+ + i

=


E ample >! *olve t%e circuit 'y applying t%e *teps of t%e -C-6

>V 3V


153/504

>(

1V

V

9V


154/504

9AV

16 La'el all mes%es>6 "dentify dependent mes%es 4current sources5

36 Brite n2m 1 K(L e.ns/+D m 1 e.uation in t%edependent loop

n >m 1

96 *olve

1

>drops inCB directionin mes% >

1/

16>< 10 /i

=1 1/i =


E ample 9! ne of t%e 'ranc%es %as a current source 4a littlemore difficult5


155/504

4 R

5'1*

i1 i2

16 La'el all mes%es>6 "dentify dependent mes%es 4current sources5

36 Brite n2m 1 K(L e.ns

/+D m 1 e.uation in t%edependent loop

n G 2m G 1

96 *olve

#i"i)e) . R in ottomePuation to ma e t!eunits consistent( )

1

> 1drops in

CB directionin mes% >

1/

>

1 0 1/ 160 /

9000 9000


156/504

9AV

6 Can t $rite t%at i1 or i2 is e.ual to t%evalue of t%e current source 4as in t%e

previous e ample5 since 'ot% go t%roug%it6

v $

@

2

9AV

>6 "dentify dependent mes%es 4current sources5

36 Brite n2m 1 K(L e.ns/+D m 1 e.uation fort%e dependent mes%

m 1

1/


E ample < 4cont5 ! #$o -es%es *%are a Current *ource 4mostdifficult5

9AV


157/504

9AV


158/504

9AV


159/504

/ny 'ranc% current can 'e $ritten using one or t$o

mes% currents6

'oltage

o,er

Solve theentire circ#it

#sing the MeshC#rrent Method

&indcomponent

c#rrents

&indother

electrical'#antities

s#ch asvoltage or

power (llMesh

C#rrents

Speci)icComponentC#rrents

160


Component 4 ranc%5 Current from -es% Current


160/504

ind i a!ia i2!1 - i1!2

ind ib!i

b= i

3!1- i

3!2

ind ic/ic= i 3!1 - i2!1

ind id /id = i 2!!nly one current

needed at t%e edge oft%e circuit

#%is is partof a very

'ig circuit

i1!2 i1!3i1!1

i2!!

i2!1 i2!3

i3!! i3!1 i3!2

ia

ic

i '

id

161


E ample =! /pply t%e -C- *olution


161/504

%in) t!e Branc! CurrentsN!o,n$i f

i c

i e

i d

i b

i a

Ai"en t!at .ou !a"e alrea). sol"e) t!e circuit$

i 1

i 3

i 2 - 0 -

1

>

3

06


162/504

%m s la$, po$er e.uation, energy, etc6

(oltage,o$er

Solve the

entire circuitusing the Mesh

Current Method

(indcomponent

currents

(indother

electrical5uantitiessuch as

voltage orpo!erAll

Mesh

Currents

SpecificComponent

Currents

163


E ample 7! /pply t%e -C- *olution


163/504

Fiven t%at you %ave already solved t%e circuit using -C-!

ind po$er consumed 'y ) 3 1=V!

Jo$ muc% po$er is supplied

'y + s2 110(!

@2

@2

i1

i3

i2

ia

i '

1

>

3

17611/136=/

ii

i

=

( ) ( )> >> 3 3 3) ) 116>=/ 1=>,0>8B >60>8 B

a p i i

)

= = = = =

( ) ( )( )( )

> > >

136


164/504

Use -C- to find t%e current t%roug% t%e 8 resistor6n G 2


36 Brite n2m > K(L e.ns

m G 016 La'el all mes%es

96 *olve 2368i

( )

( )

1 1 > drops

-es% 1

> 1 > drops

-es% >

10( > 8 0

8 9 >0( 0

v

v

i i i

i i i

+ + =

+ + =

165


E ample :! ull Use of -C- to ind a (oltage of "nterest

Use -C- to find t%e voltage across t%e 8


165/504

resistor6

#"28: #itanium

n 3


36 Brite n2m 3 K(L e.ns

m 0

16 La'el all mes%es

96 *olve

i

3i

( ) ( ) ( )

( )

1 1 > > 1 > > 3 drops drops

-es% 1 -es% >

3 > 3 drops

-es% 3

90 > 8 0 R 8 = = 0

= 9 >0 0

v v

v

i i i i i i i i

i i i

+ + = + + =

+ + =

( ) ( ) ( ) ( ){ } { }( )&eros 90 > 8 ,8 = = , = 9 >0 , , , $ $ y y $ y y z z y z $ y z + + + + + +1 > 3/ 068/i i i= = =

( ) ( )0 1 >8 8 >868(v i i A A= = =

166


E ample 10! ull Use of -C- to ind a Current of "nterest4more difficult5Use -C- to find t%e current t%roug%


166/504

t%e 10 resistor6

#"28: #itanium

n 3


36 Brite n2m > K(L e.ns /+D m 1 e.uation for t%e dep6 loop

m 1

16 La'el all mes%es

96 *olve

i 3i

3 1/i =

( ) ( ) ( )1 1 > > 1 > > 3 drops drops

-es% 1 -es% >

90 > 10 0 R 10 : 9 0v v

i i i i i i i i + + = + + =

1 >/i i= =( ) ( ) ( ){ } { }( )&eros 90 > 10 ,10 : 9 1 , , $ $ y y $ y y $ y + + + + +

10 1 > 3/i i i = =

167


" %ave to solve a pro'lem, do " use +ode (oltage or -es%Current

L A i i f $ill f $ i


167/504

*eries and parallelcom'inations can 'e used"n conGunction $it% -C-or +(-6

LooA at circuit, often one $ill use fe$er e.uations

> mes%es, 11 nodes Direct use of -C- $ill %ave fe$er e.uations

= mes%es, > nodesDirect use of +(- $ill %ave fe$er e.uations

) e.ie.

168


*ummary

% C t t% d


168/504

-es% Current -et%od

*olve circuit using Kirc%off s voltage la$ tofind all -es% Currents6

Use t%e -es% Currents to find any current,voltage, po$er etc6 in t%e solved circuit

Can solve any circuit 'y +ode (oltage -et%odor -es% Current -et%od 'ut one may 'e easier fora given circuit6


(indcomponent

values

(ind

other electrical5uantities such

as current3voltage3 or

po!er

169


Science As Art at Clemson4%ttp!HHgeo6ces6clemson6eduHgalleryHmain6p%p5


169/504

'istological %rocessing of Cellbased (iber !ith RegenerativeMedicine Application

Billie ;ones

#%is p%otograp% displays five2micron %istological sections of a%ollo$ fi'er $it% a greenfluorescent protein t%at la'elsepit%elial gland cells6 #%e

cellular fi'er $as sectioned inorder to determine t%e fi'ermorp%ology, as $ell as t%ecellular distri'ution t%roug%out6

170


ECE 307 2 L


170/504

ECE 307 2 Lecture =E.uivalent Circuits


Clemson University

171


C ti ) d



171/504


function!o rocess

"nformationo #ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er o

Energy



*imilar

Electric Circuit



+ode (oltage -et%odKirc%%off]s (oltage La$


DC Lecture =E.uivalent Circuits

172


#emperature *ensing 2 )esistance #emperature Detectors 4)#Ds5

"n a -etal increasing temperature 4#5 $ill increase t%e resistance 4) 5


172/504

%latinum: @06003:>H^C, ) 0 100 at # 0^C

n a etal, increasing temperature 4#5 $ill increase t%e resistance 4) # 5

-easure t%e resistance ) t%en $e Ano$ ##%is is a means of measuring temperature6

*everal $ays to measure )6

@

2v>

R R R > +

> R

R

0 0

0 0

4< 5

4< 5

> >

>

>

Rv + R R

R R > v +

R R R >

= ++= + +

173


#emperature *ensing 2 )esistance #emperature Detectors 4)#Ds5


173/504

(easure ' T

Use t%e resistortemperature curve tocalculate temperature #

*olve t%e voltage e.uation for #

( )0

0

4 < 5 >

>

v R R v + >

+ v R + =

174


#emperature *ensing 2 #%ermistor "n a +egative #emperature Coefficient 4+#C5 #%ermistor, increasing temperature $illd % i 6


174/504

decrease t%e resistance6/ ositive #emperature Coefficient 4 #C5 refers to materials t%at e perience an increase in electrical resistance $%en t%eir temperature is raised6 #%e material is a semiconductor 4later lecture5 and not a metal

-easure t%e resistance ) t%en $e Ano$ ##%is is a means of measuring temperature6

uild a circuit tomeasure resistance

4voltage divider is one possi'ility5

175


#%evenin and +orton E.uivalents

)eplace part of a circuit $it% a simpler circuit t%at is


175/504

)eplace part of a circuit $it% a simpler circuit t%at is

electrically e.uivalent6

176


Oou Jave /lready Created Electrically E.uivalent Circuits

*eries and arallel reductions create electrically


176/504

*eries and arallel reductions create electrically

e.uivalent circuits6

a a

*ome .uantities are e actly t%e same $%ile ot%ers are different6E.uivalent means t%e circuits are t%e same in some importantaspects 'ut are not identical6

-

va'

-

va' 1 R+

i

> R 3 R

1i >i 3i

e* R+

i

177


Electrically E.uivalent for ur BorA

Electrically E uivalent 2 from t%e load perspective


177/504

Electrically E.uivalent 2 from t%e load perspective,

t%e same (oltage and Current e ist6


178/504

/ny part of t%e circuit $it% t$o terminals 4a one2port

net$orA5 can 'e replaced 'y a single voltage sourceand a resistor in series6

rom t%e ) 9 perspective, v andi, it can t tell$%ic% circuit it isin6

179


rocedure to find #%evenin E.uivalent Circuit

/6 ind E uivalent )esistance


179/504

/6 ind E.uivalent )esistance

)emove Load 4resistor or su'2circuit5 )emove *ources

( source 2? s%ort circuit " source 2? open circuit

ind ) 6 ind pen Circuit (oltage

)emove Load *olve Circuit 4node voltage, or mes% current5 ind voltage at load terminals 4still $it%out t%e load5

180


E ample 1! ind #%evenin E.uivalent

%in) T!e"enin EPui"alent or circuit to t!e le t o


180/504

/6 ind E.uivalent )esistance)emove Load)emove *ources

( source 2? s%ort circuit

" source 2? open circuit

%in) T!e enin EPui alent or circuit to t!e le t o


181/504

/6 ind E.uivalent)esistance 4continued5

arallel Com'ination

*eries Com'ination

3>1

>13>1 ) ) )

) ) ) ) __) ++=+=

182


E ample 1 4cont5! ind #%evenin E.uivalent

0i =


182/504

6 ind pen Circuit (oltage)emove Load*olve Circuit 2 mes% current

ind voltage at load terminals 4still$it%out t%e load5 use i1 to findva'

0

+ote t%at t%is is not a-es% since it is not a

$indo$ in t%e circuit

16 La'el mes%es n 1>6 +o dep mes%es n 036 Brite n2m 1 K(L96 *olve

K(L at output loop!

i1abv

0i

3v

>v

3

n s 1 1 1 >n 14drops5

s1

1 >

i i 0

i

v v R R

v R R

== + + =

=+

3

n > 3 a'

n 14drops5

>a' > 1 > s

1 >

0v v v v

Rv v i R v

R R

=

= + + =

= = =+

183


E ample 1 4cont5! ind #%evenin E.uivalent

>1 )) )

) +=


183/504

lacing t%e #%evenin e.uivalent into t%e original circuit!

-

i and v $ill

'e t%e same as $%ent%e original circuit$as connected6

>s

1 >>

Rv v

R R=

+

3

>1

>1# ) ) )

))) +

+=

i

v

184


+orton E.uivalent Circuit



184/504


net$orA5 can 'e replaced 'y a single current sourceand a resistor in parallel6

rom t%e ) 9 perspective, v andi, it can t tell$%ic% circuit it isin6

i + 4cont5! ind t%e +orton E.uivalent Circuit

/6 ind E uivalent )esistance 4cont5


187/504

/6 ind E.uivalent )esistance 4cont5

( )( ) + 1 > 3 9 3 9

1 >3 9

1 >3 9

1 >

) ) __) ) __ ) )

) ) ) __ ) )

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188


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