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

)(tv

)(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

)()( tRitv

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

])[6])([12( mAVP ][72 mW

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

42

21

ONVERIFICATI

bI

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 +

][20

042

][25

VV

VVVV

X

XXX

][5

4

024

1

1

VV

V

VVV

X

XX

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

)(21

11

tvRR

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

iRv iRi

)30(,, kPVI bd FIND

APPLY KVLTO THIS LOOP

bdV FOR LOOP

VVIkV bdbd 10(0][2012 KVL)

mWARIP 30)10*30()10( 3242

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

11

)5(

51

412 III

)()(

)(1

)(

21

21 tiRR

RRtv

tvR

tip

pR

O21 VI I FIND ,,

WHEN IN DOUBT… REDRAW THE CIRCUIT TOHIGHLIGHT ELECTRICAL CONNECTIONS!!

2*80 IkV24

IS EASIERTO SEE THEDIVIDER

FIRST GENERALIZATION: MULTIPLE SOURCESAPPLY KCL TO THIS NODE

)()(

)(1

)(

21

21 tiRR

RRtv

tvR

ti

O

p

O

DEFINE “PARALLEL RESISTANCE COMBINATION”

SOURCE EQUIVALENT

mA10 mA15k3

k6

OV

SOURCES THEBY SUPPLIED

POWER THE AND VFIND O

kkk

kkRp 2

36

3*6

OVpR

mA5

mW

mAVP

mW

mAVP

VV

OmA

OmA

O

100

)10(

150

)15(

10

6

15

APPLY KCL TO THIS NODE

SECOND GENERALIZATION: MULTIPLE RESISTORS

)()()()(

)()(ti

R

Rti

R

tvti

tiRtv

O

k

p

K

k

k

OP

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

k5

FIRST REDUCE IT TO A SINGLE LOOP CIRCUITk12kk 12||4

k6

kk 6||6

k

VI

12

121 )12(

93

3

aV

SECOND: “BACKTRACK” USING KVL, KCL OHM’S

k

VI a

62 :SOHM'0321 III :KCL

3*3 IkVb :SOHM'

3I

…OTHER OPTIONS...

4

34

*4124

12

IkV

II

b

5

345

*3

0

IkV

III

C

:SOHM'

:KCL

kkk 12||2

VVkk

kVO 1)3(

21

1

:DIVIDER VOLTAGE

kkk 211

AAkk

kIO 1)3(

21

1

:DIVIDER CURRENT

LEARNING BY DOING

AN EXAMPLE OF “BACKTRACKING”

A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN ICOMPUTE?”

4*6 IkVb

k

VI b

33

mA1

432 III

mA5.1

2*2 IkVa

V3

V3

baxz VVV

VVxz 6

k

VI xz

45

mA5.1

521 III

mAI 31

11 *4*6 IkVIkV xzO

VVO 36

mA5.0

OV FIND

DIVIDER VOLTAGEUSE

FIND :STRATEGY 1V

1Vk60

kkk 2060||30

+-

1Vk20

k20

V12

Vkk

k6)12(

2020

20

V6

DIVIDER VOLTAGE

14020

20Vkk

kVO

V2

SV FIND

THIS IS AN INVERSE PROBLEMWHAT CAN BE COMPUTED?

V6

mA05.0mA15.0

mAkV 1.0*601

k

VI

120

61

VmAkVS

615.0*20

V9

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

Y

baab RRR

)(|| 312 RRRRab

321

312 )(

RRR

RRRRR ba

321

213 )(

RRR

RRRRR cb

321

321 )(

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

13

321

32

321

21

a

b

b

a

R

RRR

R

R

R

R 13

3

1 c

b

c

b

R

RRR

R

R

R

R 12

1

2

REPLACE IN THE THIRD AND SOLVE FOR R1

Y

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

Y

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

321

21

1R

2R

3R kkk

kk

18612

612

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

0)(

)( L

Ogm R

tvtvg

KCL

)(

)(

tv

tvG

i

O FIND