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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
Constant (oltage *ource
S v S v
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hat does it do !hen connected to other components"
Constant (oltage *ource
@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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Constant Current *ource
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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Constant Current *ource
hat does it do !hen connected to other components"
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
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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
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)esistor
(oltage
C u r r e n
tBehavior:
Constant relations%ip 'et$eenvoltage and current6
Mathematical Model:
Constant ) summari&es material properties, temperature, and si&e6
)hm*s +a! relates t%e voltageand current using )!
v = iR
Symbol:
R
1slope
is constant6 R
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)esistor
hat does it do !hen connected to other components"
)esists t%e flo$ 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
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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
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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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*$itc%
hat does it do !hen connected to other components"
Connects or disconnects part of t%e circuit
ElectricCircuit
+ voltage can 'emeasured acrosst%e opening
@
2
v
+o Current
2
Closed:
/cts liAe a s%ort circuit
)pen:
/cts liAe an open circuit
ElectricCircuit
+ voltage can 'emeasured acrosst%e s$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.
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25g g
Electrically any$%erein %ere is t%e
connection point
icA a point and call it t%e node
) 3
) 1
) >
) 3
) 1
) 9
,ode connection point
Electric Circuit Connections
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E ample! "dentify +odes
Electrically,
any$%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
Electric Circuit Connections
Branch portion of a circuit $it% only t$o e ternalterminals
terminals
Eac% element is a 'ranc%6#%e connection 'et$een and is also a 'ranc%
Eac% element is a 'ranc%6 #%econnection is + # a 'ranc%
'ecause it %as 3 e ternal terminalst1 , t3, and t 9 and t%ree
components connect at t > 6
) 1
v1@2 )
1
) >
v1@2
1t 3t
3t
>t
1t
9t
1t
>t
3t
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E ample 9! "dentify ranc%es
ind 'ranc%es in t%e circuit6
>8/@2
90(
Eac% of t%e components are individually a 'ranc%, 8
components means 8 'ranc%es6
irst identify t%e nodes6
#%ere is a branch containing t%e 90( source and t%e 3 and 1 resistors t%e connection %as t$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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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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>8/-
E ample
>3
9=1 >
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1
23
" 1- 54
Electric Circuit Connections
+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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-
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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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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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 $illgenerally use Convention 1
*olve for i3!
*olve for i3!
1i>i
3i
9i
9
1 > 3 91
4i out of node @5
0 0n
i i i i i=
= + + =3 1 > 9i i i i= +
9
1 > 3 91
4i out of node 250 0n i i i i i= = + =
3 1 > 9i i i i= +
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E ample 7 4cont5! /pplying KCL
Fiven!
ind !
+ote! i2 = -3A meanst%at t%e current actuallyflo$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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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 $illgenerally use Convention 1
*olve for i3!
*olve for i3!
1i
>i
3i
9i
9
1 > 3 91
4i out of node @5
0 0n
i i i i i=
= + + + =
3 1 > 9i i i i=
9
1 > 3 91
4i out of node 25
0 0n
i i i i i=
= =3 1 > 9i i i i=
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) 1@2 ) > ) 3
) 9
v1
E ample :! /pplying KCL in a Circuit
KCL at +ode 1 !
at +ode >!
i1 i2 i3
i4
i s
/pply KCL at eac% node6
"dentify t%e nodes in t%e circuit and la'el6
1 N
> N
3 91
4i out of node @5
0 0S n
i i i i i i=
= + + + = 3 91
4i out of node @5
0 0S n
i i i i i i=
= + =
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E ample : 4cont5! /pplying KCL in a Circuit
,e ne, some actual "alues$
1 > 3
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E ample 10! *olving a Circuit Using KCL
B%at is t%e current t%roug% t%e
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*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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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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Electric Circuit erforms a
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
*imilar Electric Circuit
)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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Cypria
*n)re, (ount?os!ua (ount
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ECE 307 Lecture >KCL and K(L
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 o Energy
Circuit Components )esistor (oltage *ource Current *ource *$itc%
Connections #erminal +ode ranc% Loop -es%
*imilar Electric Circuit
)eductions *ource #ransformation arallel *eries #%evenin +orton
/nalysis #oolsKirc%off s Current La$
+ode (oltage -et%odKirc%off s (oltage La$
-es% Current -et%od*uperposition
vervie$ of DC Electric Circuits
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
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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 =
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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=
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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= +
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E ample > 4cont5! Do2over of E ample 1
*u'stituteand solve!
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 $it% @
v4 + + 1
@2
@
2
v1
@ @@
v3v2
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Kirc%%off s (oltage La$ 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$ise directionare positive
0 N
vn =
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Kirc%%off s (oltage La$ 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$ise direction
are positive
0
N
vn =
1 D
> D
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E ample 3! /pplication of K(L 4E 6 >6=5
D i device could 'e aresistor or somet%ing else
vi are t%e component
voltages
16 )ecogni&e t%at D> and D 3 %ave t%e same t$o nodes parallel
>6 Jo$ many loops in t%e circuit t%ree36 Jo$ many loops include v2 #$o, Loop 1 and Loop >
K(L Loop 1 @ 2v2
@
2
v 4
+ 3 + > + 1
+ 9
2v3@
2v1@
K(L Loop 3
K(L Loop >
Fiven v1 =(,
v4 1(, v s 1>(ind v2
S v
1 D
> D
3 D
9 D
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E ample 3 4cont5! /pplication of K(L 4E 6 >6=5
K(L Loop 1 @ 2v2
@
2
v 4
+ 3 + > + 1
+ 9
2v3@
2v1
@
#raverse Loop 1 in t%e clocA2$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=
= + + + =
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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 >
*u'stituteand solve!
'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=
= + =
=
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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= =
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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
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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
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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)
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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$n element voltages in terms of t%e ar'itrary
assignments made for currents
< a < < c
< )
i1 i2
i3
3>(
9
>
8 >0(1v
+
>v +
+
3v
66ECE 307 Basic Electrical Engineering
E ample 7 4cont5! Use of K(L P KCL #oget%er to *olve a
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E ample 7 4cont5! Use of K(L P KCL #oget%er to olve aCircuit
15 KCL at node '!
>5 K(L around loop 1!
35 rom %m s la$ $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 + + =
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E ample 7 4cont5! Use of K(L P KCL #oget%er to *olve a
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E ample 7 4cont5! Use of K(L P KCL #oget%er to olve aCircuit
!
=5 rom circuit $e see t%at!
75 Com'ine
"- v2"
-v3
" -v1
+ a + ' + c
+ d
3>(
9>
8 >0(
3i
>i1i
> 3drops, CB
>0( 0 N
nv v v= + => > 3 39 i and 8 iv v= =
1 > 340 5 9 8 >0(i i i+ + =
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Jo$ 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
+ = + + =
+ + =
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*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= = =
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*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
71ECE 307 Basic Electrical Engineering
l
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*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
=
72ECE 307 Basic Electrical Engineering
* l i U i i / %
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*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
=
73ECE 307 Basic Electrical Engineering
* l i U i i / % 2 /#L/ * f $
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*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
74ECE 307 Basic Electrical Engineering
* l i U i i / % 2 /#L/ * f $
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*olution Using a -atri /pproac% 2 -/#L/ *oft$are
*GI1 1 -1J 2 0 8J 0 4 8K 1 1 -1 * G 2 0 8 0 4 8
GI0J32J20K 0 G 32 20
in" * Lans G 4&0000 -1&0000 3&0000
Mser >n ut
Mser (*T+*B
out ut 1
>
3
ii
i
75ECE 307 Basic Electrical Engineering
* l i U i i / % 2 * E l
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*olution Using a -atri /pproac% 2 -* E cel
Result
4
-1
3Enter t!e coe icientsan) constants
Nelect an area t!e same siOe as t!e constants 3 ro,s 1 colClic on t!e ormula ar an) enter G mmult min"erse *2$C4 E2$E4T!en simultaneousl. ress Control s!i t an) enter&
1
>
3
i
i
i
76ECE 307 Basic Electrical Engineering
*
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*ummary
%m s La$! 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$s to solve a circuit
77ECE 307 Basic Electrical Engineering
Science As Art at Clemson
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Calc#lated ChaosEric %enimore
#escri tion$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
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ECE 307 Lecture 3o$er and Energy and E.uivalent )esistance
Department of Electrical and ComputerEngineering Clemson University
79ECE 307 Basic Electrical Engineering
i $ f DC El t i Ci it
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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 *$itc%
Connections
#erminal +ode ranc% Loop -es%
*imilar Electric Circuit
)eductions *ource #ransformation arallel *eries #%evenin +orton
/nalysis #oolsKirc%off s Current La$
+ode (oltage -et%od
Kirc%off s (oltage La$ -es% Current -et%od
*uperposition
vervie$ of DC Electric Circuits
DC Lecture 3 o$er andEnergy and E.uivalent)esistance
80ECE 307 Basic Electrical Engineering
o$er in Electric Circuits
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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$er
2
( )B p vi=
v
i
81ECE 307 Basic Electrical Engineering
E ample 1! Using t%e *ign Convention in o$er Calculations
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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 >///
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/merican Bire Fauge 4/BF5
83ECE 307 Basic Electrical Engineering
E ample >! o$er Loss in a Bire
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E ample >! o$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 $ire
'G> G 1* 0&2478 G 0&2478'G'>G 0&2478' 1* G 0&2478
G2 000 t 0&1239 1000 tG 0&2478
1/i =
>
>41/5 :V :B
p i R== =
>v p
R
=
84ECE 307 Basic Electrical Engineering
E ample 3! o$er Calculation Components Can *upply or
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Consume o$er
)4very small,ignore5
)4very small,ignore5
16
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o$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= = = =
86ECE 307 Basic Electrical Engineering
Energy in Electric Circuits
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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 $#V isimplied
s seconds% %ours
( ) ( ) ( )0
0E ( t
( t
p d p t p t vi t = = =
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E ample
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E ample
/ 1>( 'attery is rated at 1(41
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)esistors in eries
)esult >! Using %m s la$ 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+ +=
89ECE 307 Basic Electrical Engineering
)esistors in *eries (oltage Divider
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)esistors in eries (oltage Divider
Eac% resistor voltage isa scaled version of v
v is scaled or divided 'et$een t%e resistors
v
"v1
-"v2-"v3
-
Continuing from previous slide, solve for i
Brite eac% resistor voltage!
#%is result is true for any num'er 4+5 of resistors inseries6 )esistor voltage for t%e n t% resistor is!
(oltage Divider E.uation
1 > 3) ) ) vi = + +
>> >
1 > 3
) )
) @) @) v i v= =
11 1
1 > 3
) )
) ) ) v i v= = + +
1)
>)
3)
1m
i
33 3
1 > 3
) )
) ) ) v i v= = + +
1 > 3
) ) ) )
nnv v= + +
1
)
)
nn N
)
)
v v
=
=
90ECE 307 Basic Electrical Engineering
)esistors in *eries Circuit )eduction
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)esistors in eries Circuit )eduction
(rom the voltage source point of vie!3 t%ere is nodifference 'et$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!
$%ere
#%is result is true for any num'er 4+5 of resistors inseries6 #otal resistance is!
E.uation for E.uivalent)esistance of *eries )esistors
1)
>)
3)
1m
i
) e*
i
1 > 3 1 > 3) ) ) 4) @) @) 5 ) e*v i i i i i= + + = =
1 > 3) ) @) @) e* =
1
) ) N
e* )
) =
=
91ECE 307 Basic Electrical Engineering
E ample =! *eries Lig%t *tring
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E ample ! eries Lig%t tring
*tep 1! ind e.uivalent resistance and solve fort%e current
*tep >! "ndividual 'ul' voltages
C%ecA!
4
"v1-"v2-
"
v1!!-
v1!!
1) 1=
100( >) 1=
100) 1=
i
) e*
i
1 1 > 3 100) 1/41 5 1( 666v i v v v= = =
( )100
1
) ) 1 1 6 1 100e* ) = = + + + = L100(
100(1/
100
i = =
100
1 > 3 1001
100( 666 0() )
v v v v v=
= + + + + + =
92ECE 307 Basic Electrical Engineering
)esistors in arallel
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)esistors in arallel
+ '
+ a
arallel esistors N!are t!e Name 2 31
out of a
0nn
i i i i i=
= + + + =
1 > 3
1 > 3
) ) )
1 1 1@ @
) ) )
v v vi
v
= + +
=
1) v
i
>) 3)
1i >i 3i
1 > 3i i i i= + +
93ECE 307 Basic Electrical Engineering
)esistors in arallel 2 Current Divider
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)esistors in arallel 2 Current Divider
+ '
+ a
Eac% resistor currentis a scaled version ofi
i is scaled ordivided 'et$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
=
94ECE 307 Basic Electrical Engineering
)esistors in arallel 2 Circuit )eduction
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)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* ) ==
95ECE 307 Basic Electrical Engineering
E ample 7! arallel Lig%t *tring
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E ample 7! arallel Lig%t tring
4 "n eac% 'ranc%, $%at is t%e 'ul' resistance sot%at pi 1B
ind e.uivalent resistance of t%e string and solvefor t%e current i
@
2
v R
1)
100+
i
>) 100)
100i
>1 1
1 1 1 11 1
1B) ) v v
p i v v = = = =
> >1
14100(5
) 10A 1B 1Bv= = =
>11
1
100(10 10 / 10m/
) 10A v
i = = = =
100>
1
1 1 1100 10 H
) ) 10A
100() 100 1/
100
) e* )
e* i
=
= = =
= =
>i1i
) e*
i
100(
96ECE 307 Basic Electrical Engineering
E ample 8! Compare *eries and arallel Lig%t *trings
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p p g g
4 4
*tring failure modes! "f one of t%e series 'ul's 'reaAs 4open2circuit5, t%en i 0 for all 'ul's6 "f one of t%e parallel 'ul's 'reaAs 4s%ort2circuit5, t%en v 0 for all 'ul's6
@
2v1
@
2v2
@
2
v1!!*ame current and po$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= = =
97ECE 307 Basic Electrical Engineering
)educing Circuits Using *eries and arallel E.uivalent) i t
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)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$reduction6
)epeat
98ECE 307 Basic Electrical Engineering
E ample :! Use ot% *eries and arallel )eductions to ind )e.
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p ) )
Step1/ R 2 and R3 are in series!
#%e circuit reduces to
1 > 3) ) ) ! ) and ) are in parallele*Step
1e* R
1) e*
99ECE 307 Basic Electrical Engineering
E ample 10! Using )esistor )eductions to *implify Circuit*olution
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*olution
) 1
) 3
1/ ) >
ind i1 and i2
Com'ine R2 and R3 in series!
Use Current Divider!
i1 )e. 1H4) 1@)e. 15 X i
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ind v R2
P v R3
Circuit *olution
) >@
2v R2
@ 2v R3) 1
) 3
1/Use current i > and %m s La$
( ) ( )>
> >) 06= 30 18( Rv i= = =( ) ( )
3 > 3) 06= >0 1>( Rv i= = =
>i
1i
i
101ECE 307 Basic Electrical Engineering
E ample 11! *e.uence of *eries and arallel )eductions to ind)e
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)e.
rom inspection!
B%ere!
1>39 3 9 < =) ) @) ) @) ) @) e* =
$y $ y
$ y= +
ind e* R
102ECE 307 Basic Electrical Engineering
E ample 11 4cont5! *e.uence of *eries and arallel )eductionsto ind )e
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to ind )e.
Fiven!
< =) ) 3 9 < =) ) @) ) @) ) @) > >6= 96=e* = = +
1 > 3 9
< =
) > ) 9 ) < ) ) 3
= = = = = =
103ECE 307 Basic Electrical Engineering
otentiometer
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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
104ECE 307 Basic Electrical Engineering
otentiometer
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(easurementan)
*nalog-#igital
Con"ersion
#igitalCom uterController
105ECE 307 Basic Electrical Engineering
*ummary
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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= =
106ECE 307 Basic Electrical Engineering
*ummary 4cont5
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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* ) ==
107ECE 307 Basic Electrical Engineering
*ummary 4cont5
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)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
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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
109ECE 307 Basic Electrical Engineering
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ECE 307 Lecture 9#%e +ode (oltage -et%od
Department of Electrical and ComputerEngineering Clemson University
110ECE 307 Basic Electrical Engineering
vervie$ of DC Electric Circuits
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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 *$itc%
Connections
#erminal +ode ranc% Loop -es%
*imilar
Electric Circuit
)eductions
*ource #ransformation arallel *eries #%evenin +orton
/nalysis #oolsKirc%off s Current La$
+ode (oltage -et%od
Kirc%off s (oltage La$ -es% Current -et%od*uperposition
DC Lecture 9 #%e +ode(oltage -et%od
111ECE 307 Basic Electrical Engineering
E plosive Fas *ensor
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Fas t%atmay 'e
e plosive
@2
i
Aas lo,st!roug! sensor
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/ 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
113ECE 307 Basic Electrical Engineering
E ample 1! C%oosing a )eference +ode and La'eling +ode(oltages
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( 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$een + 1 and + 9 and all %ave t%e same voltage6
La'el t%e +ode (oltages in t%e Circuit
Z of node voltages Z of nodes 21
1>( >v1v
3v
1 N
3 N
> N
1v1v1v
9 N
114ECE 307 Basic Electrical Engineering
#%e +ode (oltage /pproac%
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Solve theentire circuit
using the ,ode
Voltage Method
(indcomponent
voltages
(indother
electrical5uantities
such ascurrent orpo!er/ll
+ode
(oltages
*pecificComponent
(oltages
Current,
o$er, etc
115ECE 307 Basic Electrical Engineering
+ode (oltage -et%od *ystematic /pplication of KCL
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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 $ill maAe a node dependent6 Count t%e mdependent nodes636 Brite KCL e.uations at eac% of t%e n212m
independent nodes6 Brite m e.uations to relate
t%e dependent node voltages to t%e sourcevoltages6
96 *olve n21 e.uations6
116ECE 307 Basic Electrical Engineering
reliminary 2 Jo$ to Brite t%e Current at a +ode Using +ode(oltages
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( 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=
117ECE 307 Basic Electrical Engineering
reliminary 2 Jo$ to Brite t%e Current at a +ode Using +ode(oltages
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( 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
118ECE 307 Basic Electrical Engineering
reliminary 2 Jo$ to Brite t%e Current at a +ode Using +ode(oltages
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( 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
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*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 = =
120ECE 307 Basic Electrical Engineering
+ode (oltage -et%od
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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$ays need a total of n21
e.ns596 *olve
121ECE 307 Basic Electrical Engineering
E ample >! *traig%tfor$ard +(-, 3 +odes
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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
out of node of interesti
out of node of interesti
out of node of interesti
122ECE 307 Basic Electrical Engineering
E ample 3! our +odes and a Dependent *ource
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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=
123ECE 307 Basic Electrical Engineering
E ample 9! +(- Complete Circuit *olution
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+ 1 + >
/t node 1/t node >
16 La'el all nodes andselect )eference +ode
>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 + =
124ECE 307 Basic Electrical Engineering
E ample 9 4cont5! +(- Complete *olution
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/t node 1
/t node >
/dd
96 *olve
1 >> =v v = 1 >> 3 8v v + =
>> >v =
> 1(v =
1 >6
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/ny component voltage can 'e $ritten using one ort$o node voltages6
Solve theentire circuit
using the ,odeVoltage Method
(indcomponent
voltages
(ind
otherelectrical5uantities
such ascurrent or
po!er
/ll +ode
(oltagesComponent
(oltagesCurrent,
o$er
126ECE 307 Basic Electrical Engineering
Component (oltage from +ode (oltage
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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 $%en
component is attac%ed to reference node6
#%is is avery 'igcircuit
2
100
>00
110
1>>
300 - - - - - -
900
v)100@ 2
v R4!!@
2v R3!!
@
2
@
2
v1!!
@
v110
v4!1
@
@
v4!!
v3!!
127ECE 307 Basic Electrical Engineering
Use Component (oltages to ind /ny t%er Electrical Nuantities
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%m s la$, po$er e.uation, energy, etc6
Solve theentire circuit
using the ,odeVoltage Method
(indcomponent
voltages
(ind
otherelectrical5uantities
such ascurrent or
po!er
/ll +ode
(oltagesComponent
(oltagesCurrent,
o$er
128ECE 307 Basic Electrical Engineering
E ample
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16 La'el all nodes andselect )eference +ode
>6 "dentify dependent nodes 4voltage sources5
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
129ECE 307 Basic Electrical Engineering
E ample =! +(- to ind Component Current, < +odes
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ind current t%oug% t%e 10 resistor6
i10R
16 La'el all nodes andselect )eference +ode
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
131ECE 307 Basic Electrical Engineering
E ample = 4cont5! +(- to ind Component Current, < +odes
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96 *olve 4cont5
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96 *olve 4alternative5 *olve 'y computer or calculator as asystem of 9 e.uations and 9 unAno$ns
#"28:*imultaneous E.uations
/pplication90(av =
out of node '
0> 10 :
b a b b cv v v v vi = + + =
out of node c
0: 9 8
c b c c d v v v v vi = + + =
out of node d
1/ 08
d cv vi = =
9< =9 10 0 00 8 3< : 0
0 0 1 1 8
1 0 0 0 90
a
b
c
d
vv
v
v
=
90301>>0
a
b
c
d
vv
+ v
v
=
133ECE 307 Basic Electrical Engineering
E ample 7! -ore +(- /pplication ractice
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16 La'el all nodes andselect )eference +ode
>6 "dentify dependent nodes 4voltage sources5
36 Brite n 212 m 1 KCL e.ns6 @ 1e.uation to descri'e dependent node
n 3
m 1
ind i and vb
96 *olve(bv = >/9b av vi
= =
i
134ECE 307 Basic Electrical Engineering
E ample 8! Even -ore +(- /pplication ractice
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ind t%e voltage across t%e 90 V resistor usingt%e +(-6 ind t%e current t%roug% t%e 30 V resistor6
a '
c
KCL at node '!
16 La'el all nodes and select)eference +ode
>6 "dentify dependent nodes 4voltage sources5
36 Brite n 212 m 1 KCL e.ns6 @ > e.uations todescri'e dependent nodes
n 9
96 *olve
d
@
2v!
(oltages of Dependent nodes!
10(av =3>(cv =
ut of '
010 90 >0
b a b b cv v v v vi = + + = 109 1968 10 >>
0673/30 30 30
c av vi = = = =0 1968
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C%ecA! Use +ode (oltages to /pply KCL at nodes
a '
c
d
30 0673/i =
9019680 >03> 19680068
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/'sor's o$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
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+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$er etc6 in t%e solved circuit
Solve theentire circuit
using the ,odeVoltage Method
(indcomponent
voltages
(indother
electrical5uantities
such ascurrent or
po!er
138ECE 307 Basic Electrical Engineering
*ummary 4cont5
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*teps of +ode (oltage -et%od16 La'el all nodes and select )eference +ode
Z nodes n
>6 "dentify dependent nodes 4voltage sources5 Z dependent nodes m
36 Brite n212m KCL e.ns6 @ m e.ns6 to descri'edependent nodes 4al$ays need a total of n21e.ns5
96 *olve
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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 $it% a protein for
t%e first time6 #%is novel classof nano'iocomposites isdesigned for 'iomedical andsensor applications6
140ECE 307 Basic Electrical Engineering
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ECE 307 2 Lecture K(L e.ns6
n >m 0
:orm (or resistors in s9ared branc9
:orm (or resistors in ;o'tside< branc9
Feneral orm in a -es% of "nterest!
4imes9 o( interest 2 iad acent 5)
@2
@2 @2
i2i1
1 1 1 > 1 > >drops inCB directionin mes% 1
> > 3 3 > 9 > 1 >drops inCB directionin mes% >
) 4 5) 0
) ) 4 5) 0
n
n
v + i + i i
v + i + i i i
= + + + =
= + + + + =
152ECE 307 Basic Electrical Engineering
E ample 1 4cont5! *et up *olution 'y /pplying t%e *teps of t%e-C-
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96 *olve
K(L E.uations from previous slide!
*olve 'y any means, e ample of matri solution!
i1 i>
@2
@2
@2
4$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
=
153ECE 307 Basic Electrical Engineering
E ample >! *olve t%e circuit 'y applying t%e *teps of t%e -C-6
>V 3V
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>(
1V
V
9V
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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 =
155ECE 307 Basic Electrical Engineering
E ample 9! ne of t%e 'ranc%es %as a current source 4a littlemore difficult5
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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
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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/
157ECE 307 Basic Electrical Engineering
E ample < 4cont5 ! #$o -es%es *%are a Current *ource 4mostdifficult5
9AV
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9AV
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9AV
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/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
ECE 307 Basic Electrical Engineering
Component 4 ranc%5 Current from -es% Current
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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
ECE 307 Basic Electrical Engineering
E ample =! /pply t%e -C- *olution
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%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
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%m s la$, po$er e.uation, energy, etc6
(oltage,o$er
Solve the
entire circuitusing the Mesh
Current Method
(indcomponent
currents
(indother
electrical5uantitiessuch as
voltage orpo!erAll
Mesh
Currents
SpecificComponent
Currents
163
ECE 307 Basic Electrical Engineering
E ample 7! /pply t%e -C- *olution
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Fiven t%at you %ave already solved t%e circuit using -C-!
ind po$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
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Use -C- to find t%e current t%roug% t%e 8 resistor6n G 2
>6 "dentify dependent mes%es 4current sources5
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
ECE 307 Basic Electrical Engineering
E ample :! ull Use of -C- to ind a (oltage of "nterest
Use -C- to find t%e voltage across t%e 8
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resistor6
#"28: #itanium
n 3
>6 "dentify dependent mes%es 4current sources5
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
ECE 307 Basic Electrical Engineering
E ample 10! ull Use of -C- to ind a Current of "nterest4more difficult5Use -C- to find t%e current t%roug%
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t%e 10 resistor6
#"28: #itanium
n 3
>6 "dentify dependent mes%es 4current sources5
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
ECE 307 Basic Electrical Engineering
" %ave to solve a pro'lem, do " use +ode (oltage or -es%Current
L A i i f $ill f $ i
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*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
ECE 307 Basic Electrical Engineering
*ummary
% C t t% d
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-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$er etc6 in t%e solved circuit
Can solve any circuit 'y +ode (oltage -et%odor -es% Current -et%od 'ut one may 'e easier fora given circuit6
Solve theentire circuit
(indcomponent
values
(ind
other electrical5uantities such
as current3voltage3 or
po!er
169
ECE 307 Basic Electrical Engineering
Science As Art at Clemson4%ttp!HHgeo6ces6clemson6eduHgalleryHmain6p%p5
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'istological %rocessing of Cellbased (iber !ith RegenerativeMedicine Application
Billie ;ones
#%is p%otograp% displays five2micron %istological sections of a%ollo$ 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 $ell as t%ecellular distri'ution t%roug%out6
170
ECE 307 Basic Electrical Engineering
ECE 307 2 L
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ECE 307 2 Lecture =E.uivalent Circuits
Department of Electrical and ComputerEngineering
Clemson University
171
ECE 307 Basic Electrical Engineering
C ti ) d
vervie$ of DC Electric Circuits
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Electric Circuit erforms a
function!o rocess
"nformationo #ransfer o$er C%aracteri&ed 'y!o (oltageso Currentso o$er o
Energy
Circuit Components )esistor (oltage *ource Current *ource *$itc%
Connections #erminal +ode ranc% Loop -es%
*imilar
Electric Circuit
)eductions *ource #ransformation arallel *eries #%evenin +orton
/nalysis #oolsKirc%off s Current La$
+ode (oltage -et%odKirc%%off]s (oltage La$
-es% Current -et%od*uperposition
DC Lecture =E.uivalent Circuits
172
ECE 307 Basic Electrical Engineering
#emperature *ensing 2 )esistance #emperature Detectors 4)#Ds5
"n a -etal increasing temperature 4#5 $ill increase t%e resistance 4) 5
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%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
ECE 307 Basic Electrical Engineering
#emperature *ensing 2 )esistance #emperature Detectors 4)#Ds5
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(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
ECE 307 Basic Electrical Engineering
#emperature *ensing 2 #%ermistor "n a +egative #emperature Coefficient 4+#C5 #%ermistor, increasing temperature $illd % i 6
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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
ECE 307 Basic Electrical Engineering
#%evenin and +orton E.uivalents
)eplace part of a circuit $it% a simpler circuit t%at is
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)eplace part of a circuit $it% a simpler circuit t%at is
electrically e.uivalent6
176
ECE 307 Basic Electrical Engineering
Oou Jave /lready Created Electrically E.uivalent Circuits
*eries and arallel reductions create electrically
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*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
ECE 307 Basic Electrical Engineering
Electrically E.uivalent for ur BorA
Electrically E uivalent 2 from t%e load perspective
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Electrically E.uivalent 2 from t%e load perspective,
t%e same (oltage and Current e ist6
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/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
ECE 307 Basic Electrical Engineering
rocedure to find #%evenin E.uivalent Circuit
/6 ind E uivalent )esistance
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/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 $it%out t%e load5
180
ECE 307 Basic Electrical Engineering
E ample 1! ind #%evenin E.uivalent
%in) T!e"enin EPui"alent or circuit to t!e le t o
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/6 ind E.uivalent )esistance)emove Load)emove *ources
( source 2? s%ort circuit
" source 2? open circuit
%in) T!e enin EPui alent or circuit to t!e le t o
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/6 ind E.uivalent)esistance 4continued5
arallel Com'ination
*eries Com'ination
3>1
>13>1 ) ) )
) ) ) ) __) ++=+=
182
ECE 307 Basic Electrical Engineering
E ample 1 4cont5! ind #%evenin E.uivalent
0i =
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6 ind pen Circuit (oltage)emove Load*olve Circuit 2 mes% current
ind voltage at load terminals 4still$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
ECE 307 Basic Electrical Engineering
E ample 1 4cont5! ind #%evenin E.uivalent
>1 )) )
) +=
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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
ECE 307 Basic Electrical Engineering
+orton E.uivalent Circuit
/ny part of t%e circuit $it% t$o terminals 4a one2port
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/ny part of t%e circuit $it% t$o terminals 4a one2port
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
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/6 ind E.uivalent )esistance 4cont5
( )( ) + 1 > 3 9 3 9
1 >3 9
1 >3 9
1 >
) ) __) ) __ ) )
) ) ) __ ) )
) )
) ) ) )
) ) )
) ) ) )
) )
= + + = + + + + + = +
+ ++
+)
188
ECE 307 Basic Electrical Engineering
E ample > 4cont5! ind t%e +orton E.uivalent Circuit
6 ind *%ort Circuit Current *olve 'y +ode (oltage
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)eplace Load $it% a *%ort*olve Circuit 4node voltage, or mes%current5
ind current at t%e load terminals 4still$it% t%e s%ort5 R K(L e.ns/+D m 1 e.uation for t%edependent loop
m 1, Createsuper2mes%
96 *olve
Using -es% Current!
i1 i> i3
mes% 1
> 1 > > 3drops supermes% >
3
1 10 10 0
10 < < 0
061
n
n
v + i i i
v i i i i i
i A
= + + =
= + + + =
=
1 > 3060< , 0 , 061i A i A i A= = = + > 3i 061 sci i i A= + =
197ECE 307 Basic Electrical Engineering
E ample 3 4cont5! )educe Circuit Using *ource #ransformations
10R 5R 5R 5Ra
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10R1' 5'10R0&1*
:ind @*'iva ent (or t9e
circ'it to t9e e(t o( nodes a and b 'sin8 so'rcetrans(ormations
10R
10R
5R 5R
1' 0&
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