Post on 02-Jan-2016
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RESISTIVE CIRCUITS
Here we introduce the basic concepts and laws that are fundamental to circuit analysis
LEARNING GOALS
• OHM’S LAW - DEFINES THE SIMPLEST PASSIVE ELEMENT: THE RESISTOR
• KIRCHHOFF’S LAWS - THE FUNDAMENTAL CIRCUIT CONSERVATION LAWS- KIRCHHOFF CURRENT (KCL) AND KIRCHHOFF VOLTAGE (KVL)
• LEARN TO ANALYZE THE SIMPLEST CIRCUITS• SINGLE LOOP - THE VOLTAGE DIVIDER• SINGLE NODE-PAIR - THE CURRENT DIVIDER• SERIES/PARALLEL RESISTOR COMBINATIONS - A TECHNIQUE TO REDUCE THE COMPLEXITY OF SOME CIRCUITS
• CIRCUITS WITH DEPENDENT SOURCES - (NOTHING VERY SPECIAL)
• WYE - DELTA TRANSFORMATION - A TECHNIQUE TO REDUCE COMMON RESISTOR CONNECTIONS THAT ARE NEITHER SERIES NOR PARALLEL
RESISTORS
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)(tiA resistor is a passive element characterized by an algebraicrelation between the voltage acrossits terminals and the current through it
Resistor afor Model General ))(()( tiFtv
A linear resistor obeys OHM’s Law
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The constant, R, is called theresistance of the component and is measured in units of Ohm )(
From a dimensional point of viewOhms is a derived unit of Volt/Amp
)10(
)10(3
6
Ohm Kilo
Ohm Mega
Ohm of Multiples Standard
k
M
k in resistance in resultingmAVolt
is occurrence commonA
Conductance
If instead of expressing voltage asa function of current one expressescurrent in terms of voltage, OHM’slaw can be written
vR
i1
Since the equation is algebraicthe time dependence can be omitted
Gvi
RG
write andcomponent the of
eConductanc as define We1
The unit of conductance isSiemens
Some practical resistors
Symbol
R
v
i
ation Represent Circuit
Notice passive signconvention
Two special resistor values
Circuit
Short
Circuit
Open
0v
0i
G
R 0
0
G
R
Linear approximation
Actual v-I relationship
Linear range
v
i
Ohm’s Law is an approximation validwhile voltages and currents remainin the Linear Range
“A touch ofreality”
UNITS?
CONDUCTANCE IN SIEMENS, VOLTAGEIN VOLTS. HENCE CURRENT IN AMPERES
][8)( Ati
GIVEN VOLTAGE AND CONDUCTANCEREFERENCE DIRECTIONS SATISFYPASSIVE SIGN CONVENTION
)()( tGvti OHM’S LAW
OHM’S LAW )()( tRitv
][2)()()2(][4 AtitiV UNITS?
V4
THE EXAMPLE COULD BEGIVEN LIKE THIS
)()( tRitv OHM’S LAW
DETERMINE CURRENT AND POWER ABSORBEDBY RESISTOR
mA6
R
VRIVIP
22
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R
VP S
2
)106.3)(1010( 332 WVS
][6VVS
k
V
R
VI
10
][6][6.0 mA
KIRCHHOFF CURRENT LAW
ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLESIN ELECTRICAL ENGINEERING
“CHARGE CANNOT BE CREATED NOR DESTROYED”
NODES, BRANCHES, LOOPS
NODE: point where two, or more, elementsare joined (e.g., big node 1)
LOOP: A closed path that never goestwice over a node (e.g., the blue line)
BRANCH: Component connected between twonodes (e.g., component R4)
The red path is NOT a loop
A NODE CONNECTS SEVERAL COMPONENTS.BUT IT DOES NOT HOLD ANY CHARGE.
TOTAL CURRENT FLOWING INTO THE NODEMUST BE EQUAL TO TOTAL CURRENT OUTOF THE NODE
(A CONSERVATION OF CHARGE PRINCIPLE)
NODE
KIRCHHOFF CURRENT LAW (KCL)SUM OF CURRENTS FLOWING INTO A NODE ISEQUAL TO SUM OF CURRENTS FLOWING OUT OFTHE NODE
A5
A5
node the ofout flowing
negative the to equivalent is
node a into flowingcurrent A
ALGEBRAIC SUM OF CURRENTS FLOWING INTO ANODE IS ZERO
ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OFA NODE IS ZERO
A node is a point of connection of two or more circuit elements.It may be stretched out or compressed for visual purposes…But it is still a node
A GENERALIZED NODE IS ANY PART OF A CIRCUIT WHERE THERE IS NO ACCUMULATIONOF CHARGE
... OR WE CAN MAKE SUPERNODES BY AGGREGATING NODES
0:
0:
7542
461
iiii
iii
3 Leaving
2 Leaving
076521 iiiii:3 & 2 AddingINTERPRETATION: SUM OF CURRENTS LEAVINGNODES 2&3 IS ZEROVISUALIZATION: WE CAN ENCLOSE NODES 2&3INSIDE A SURFACE THAT IS VIEWED AS AGENERALIZED NODE (OR SUPERNODE)
WRITE ALL KCL EQUATIONS
THE FIFTH EQUATION IS THE SUM OF THE FIRST FOUR... IT IS REDUNDANT!!!
1 2 3( ) ( ) ( ) 0i t i t i t
1 4 6( ) ( ) ( ) 0i t i t i t
3 5 8( ) ( ) ( ) 0i t i t i t
FIND MISSING CURRENTS
KCL DEPENDS ONLY ON THE INTERCONNECTION.THE TYPE OF COMPONENT IS IRRELEVANT
KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CIRCUIT
WRITE KCL EQUATIONS FOR THIS CIRCUIT
•THE PRESENCE OF A DEPENDENT SOURCE DOES NOT AFFECT APPLICATION OF KCL KCL DEPENDS ONLY ON THE TOPOLOGY
•THE LAST EQUATION IS AGAIN LINEARLY DEPENDENT OF THE PREVIOUS THREE
Here we illustrate the useof a more general idea of node. The shaded surfaceencloses a section of thecircuit and can be consideredas a BIG node
0NODE BIG LEAVING CURRENTS OF SUM 0602030404 mAmAmAmAI
mAI 704 THE CURRENT I5 BECOMES INTERNAL TO THE NODE AND IT IS NOT NEEDED!!!
mA1
mA4xI2
xIxI FIND
0141 mAmAI
1I
021 XX III
mA3
mA3
bX
Xb
ImAI
mAImAI
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ONVERIFICATI
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KIRCHHOFF VOLTAGE LAW
ONE OF THE FUNDAMENTAL CONSERVATION LAWSIN ELECTRICAL ENGINERING
THIS IS A CONSERVATION OF ENERGY PRINCIPLE“ENERGY CANNOT BE CREATE NOR DESTROYED”
A B V
A B )( V
DROP NEGATIVEA
IS RISE E A VOLTAG
KIRCHHOFF VOLTAGE LAW (KVL)
KVL IS A CONSERVATION OF ENERGY PRINCIPLE
KVL: THE ALGEBRAIC SUM OF VOLTAGEDROPS AROUND ANY LOOP MUST BE ZERO
A B V
A B )( V
DROP NEGATIVEA
IS RISE E A VOLTAG
0321
RRRS VVVV
VVR 181
VVR 122
LOOP abcdefa
THE LOOP DOES NOT HAVE TO BE PHYSICAL
beV
0][3031
VVVV RbeR
PROBLEM SOLVING TIP: KVL IS USEFULTO DETERMINE A VOLTAGE - FIND A LOOP INCLUDING THE UNKNOWN VOLTAGE
be
R3R1
V VOLTAGETHE DETERMINE
KNOWN AREV V:EXAMPLE ,
BACKGROUND: WHEN DISCUSSING KCL WE SAW THAT NOT ALL POSSIBLE KCL EQUATIONSARE INDEPENDENT. WE SHALL SEE THAT THESAME SITUATION ARISES WHEN USING KVL
THE THIRD EQUATION IS THE SUM OF THEOTHER TWO!!
A SNEAK PREVIEW ON THE NUMBER OFLINEARLY INDEPENDENT EQUATIONS
BRANCHES OF NUMBER
NODES OF NUMBER
DEFINE CIRCUIT THE IN
B
N
EQUATIONSKVL
TINDEPENDENLINEARLY
EQUATIONSKCL
TINDEPENDENLINEARLY
)1(
1
NB
N
EXAMPLE: FOR THE CIRCUIT SHOWN WE HAVE N = 6, B = 7. HENCE THERE ARE ONLY TWO INDEPENDENT KVL EQUATIONS
ecae VV , VOLTAGESTHE FIND
GIVEN THE CHOICE USE THE SIMPLEST LOOP
DEPENDENT SOURCES ARE HANDLED WITH THESAME EASE
+-
+-
k10 k5
xV
V25 4
xV
1V
There are no loops with onlyone unknown!!!
The current through the 5k and 10k resistors is the same. Hence the voltage drop across the 5k is one halfof the drop across the 10k!!!
- Vx/2 +
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042
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X
XXX
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4
024
1
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VV
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VVV
X
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SINGLE LOOP CIRCUITSBACKGROUND: USING KVL AND KCL WE CAN WRITE ENOUGH EQUATIONS TO ANALYZE ANYLINEAR CIRCUIT. WE NOW START THE STUDY OF SYSTEMATIC, AND EFFICIENT, WAYS OF USING THE FUNDAMENTAL CIRCUIT LAWS
ab c
def
1
2 3
4
56
6 branches6 nodes1 loop
ALL ELEMEN TS I N SER I ESON LY ON E CUR R EN T
WRITE 5 KCL EQSOR DETERMINE THEONLY CURRENTFLOWING
VOLTAGE DIVISION: THE SIMPLEST CASE
THE PLAN• BEGIN WITH THE SIMPLEST ONE LOOP CIRCUIT• EXTEND RESULTS TO MULTIPLE SOURCE• AND MULTIPLE RESISTORS CIRCUITS
KVL ON THIS LOOP
IMPORTANT VOLTAGEDIVIDER EQUATIONS
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RvR
SUMMARY OF BASIC VOLTAGE DIVIDER
kRkRVVS 30,90,9 21 :EXAMPLE
kR 151
VOLUMECONTROL?
A “PRACTICAL” POWER APPLICATION
HOW CAN ONE REDUCE THE LOSSES?
THE CONCEPT OF EQUIVALENT CIRCUIT
THIS CONCEPT WILL OFTEN BE USED TO SIMPLFYTHE ANALYSIS OF CIRCUITS. WE INTRODUCE ITHERE WITH A VERY SIMPLE VOLTAGE DIVIDER
+-
1R
2R
Sv
i
21 RR
vi S
+-Sv 21 RR
i
AS FAR AS THE CURRENT IS CONCERNED BOTHCIRCUITS ARE EQUIVALENT. THE ONE ON THERIGHT HAS ONLY ONE RESISTOR
1R 2R
21 RR
SERIES COMBINATION OF RESISTORS
IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES
+-
+-
+-
+ -
+-
+ -
FIRST GENERALIZATION: MULTIPLE SOURCES
i(t)
KVL
01542321 vvvvvvv RR
Collect all sources on one side
2154321 RR vvvvvvv
21 RReq vvv eqv
1R
2R
Voltage sources in series can be algebraically added to form an equivalent source.
We select the reference direction to move along the path.Voltage drops are subtracted from rises
1R
2R
1Rv
2Rv
1v
2v
3v
4v
5v
SECOND GENERALIZATION: MULTIPLE RESISTORS
APPLY KVLTO THIS LOOP
VOLTAGE DIVISION FOR MULTIPLE RESISTORS
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APPLY KVLTO THIS LOOP
bdV FOR LOOP
VVIkV bdbd 10(0][2012 KVL)
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RESISTOR30k ON POWER
SINGLE NODE-PAIR CIRCUITS
THESE CIRCUITS ARE CHARACTERIZED BY ALLTHE ELMENTS HAVING THE SAME VOLTAGEACROSS THEM - THEY ARE IN PARALLEL
V
V
EXAMPLE OF SINGLE NODE-PAIR
THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)
BASIC CURRENT DIVIDER
THE CURRENT i(t) ENTERS THE NODE ANDSPLITS - IT IS DIVIDED BETWEEN THECURRENTS i1(t) AND i2(t)
APPLY KCL
USE OHM’S LAW TO REPLACE CURRENTS
DEFINE “PARALLEL RESISTANCE COMBINATION”
THE CURRENT DIVISION
mAI 1)5(41
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WHEN IN DOUBT… REDRAW THE CIRCUIT TOHIGHLIGHT ELECTRICAL CONNECTIONS!!
2*80 IkV24
IS EASIERTO SEE THEDIVIDER
FIRST GENERALIZATION: MULTIPLE SOURCESAPPLY KCL TO THIS NODE
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DEFINE “PARALLEL RESISTANCE COMBINATION”
SOURCE EQUIVALENT
mA10 mA15k3
k6
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SOURCES THEBY SUPPLIED
POWER THE AND VFIND O
kkk
kkRp 2
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mW
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OmA
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APPLY KCL TO THIS NODE
SECOND GENERALIZATION: MULTIPLE RESISTORS
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k
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General current divider
Ohm’s Law at every resistor
SERIES PARALLEL RESISTOR COMBINATIONS
UP TO NOW WE HAVE STUDIED CIRCUITS THATCAN BE ANALYZED WITH ONE APPLICATION OFKVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)
WE HAVE ALSO SEEN THAT IN SOME SITUATIONSIT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT
NOW WE EXAMINE SOME MORE COMPLEX CIRCUITSWHERE WE CAN SIMPLIFY THE ANALYSIS USINGTHE TECHNIQUE OF COMBINING RESISTORS…
… PLUS THE USE OF OHM’S LAW
SERIES COMBINATIONS
PARALLEL COMBINATION
Np GGGG ...21
FIRST WE PRACTICE COMBINING RESISTORS
6k||3k
(10K,2K)SERIES
SERIESk3
kkk 412||6
k12k3
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FIRST REDUCE IT TO A SINGLE LOOP CIRCUITk12kk 12||4
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k
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SECOND: “BACKTRACK” USING KVL, KCL OHM’S
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…OTHER OPTIONS...
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kIO 1)3(
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:DIVIDER CURRENT
LEARNING BY DOING
AN EXAMPLE OF “BACKTRACKING”
A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN ICOMPUTE?”
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OV FIND
DIVIDER VOLTAGEUSE
FIND :STRATEGY 1V
1Vk60
kkk 2060||30
+-
1Vk20
k20
V12
Vkk
k6)12(
2020
20
V6
DIVIDER VOLTAGE
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kVO
V2
SV FIND
THIS IS AN INVERSE PROBLEMWHAT CAN BE COMPUTED?
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mA05.0mA15.0
mAkV 1.0*601
k
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VmAkVS
615.0*20
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SERIESPARALLEL
TIONSTRANSFORMAY
THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL
IF INSTEADOF THIS
WE COULDHAVE THIS
THEN THE CIRCUIT WOULDBECOME LIKE THIS ANDBE AMENABLE TO SERIESPARALLEL TRANSFORMATIONS
http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf
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baab RRR
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321
312 )(
RRR
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RRR
RRRRR cb
321
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RRR
RRRRR ac
SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
321
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321
32
321
21
a
b
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R
R
R
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3
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b
c
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R
RRR
R
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1
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REPLACE IN THE THIRD AND SOLVE FOR R1
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R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
a
accbba
c
accbba
b
accbba
3
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LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION
SI COMPUTE DELTA CONNECTION
a b
c
a b
c
kkkkkkREQ 10)62(||936
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
321
13
321
32
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1R
2R
3R kkk
kk
18612
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mAk
VIS 2.1
12
12
ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ...
AVOV FIND
CIRCUITS WITH DEPENDENT SOURCESGENERAL STRATEGYTREAT DEPENDENT SOURCES AS REGULARSOURCES AND ADD ONE MORE EQUATION FORTHE CONTROLLING VARIABLE
1*2 IkV
A
A CONVENTION ABOUT DEPENDENT SOURCES.UNLESS OTHERWISE SPECIFIED THE CURRENTAND VOLTAGE VARIABLES ARE ASSUMED IN SIUNITS OF Amps AND Volts
FOR THIS EXAMPLE THE MULTIPLIER MUST HAVEUNITS OF OHM
XDIV
CONTROLLINGVARIABLE
DEPENDENTVARIABLE
scalar) (
Siemens) (
scalar) (
SOURCES DEPENDENT OTHER
XD
XD
XD
II
VI
VV
mA IN CURRENT ASSUMES
NDESCRIPTIO IVE ALTERNATAN
mA
VIVXD
2, UNITS ARE EXPLICIT
KVL
A PLAN:SINGLE LOOP CIRCUIT. USE KVL TO DETERMINE CURRENT
0*5*31211 IkVIk
A :KVL
ONE EQUATION, TWO UNKNOWNS. CONTROLLINGVARIABLE PROVIDES EXTRA EQUATION
mAI 21REPLACE AND SOLVE FOR THE CURRENT
1*5 IkV
O V10
USE OHM’S LAW
A PLAN:IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER.TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT
ALGEBRAICALLY, THERE ARE TWO UNKNOWNSAND JUST ONE EQUATION
THE EQUATION FOR THE CONTROLLINGVARIABLE PROVIDES THE ADDITIONAL EQUATION
SUBSTITUTION OF I_0 YIELDS
NOTICE THE CLEVER WAY OF WRITING mA TOHAVE VOLTS IN ALL NUMERATORS AND THESAME UNITS IN DENOMINATOR
6056/* SVkVOLTAGE DIVIDER
VVkk
kV SO )12(
3
2
24
4
OV FINDKCL TO THIS NODE. THEDEPENDENT SOURCE IS JUSTANOTHER SOURCE
OV FINDA PLAN:ONE LOOP PROBLEM. FIND THE CURRENTTHEN USE OHM’S LAW.KVL TO
THIS LOOP
THE DEPENDENT SOURCE IS ONE MORE VOLTAGESOURCE
THE EQUATION FOR THE CONTROLLING VARIABLEPROVIDES THE ADDITIONAL EQUATION
REPLACE AND SOLVE FOR CURRENT I … AND FINALLY
A PLAN:ONE LOOP ON THE LEFT - KVLONE NODE-PAIR ON RIGHT - KCL
KVL
KVL
ALSO A VOLTAGE DIVIDER
KCL
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