KIRCHHOFF CURRENT LAW

ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLES IN ELECTRICAL ENGINEERING

“CHARGE CANNOT BE CREATED NOR DESTROYED”

NODES, BRANCHES, LOOPS

NODE: point where two, or more, elements are joined (e.g., big node 1)

LOOP: A closed path that never goes twice over a node (e.g., the blue line)

BRANCH: Component connected between two nodes (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 NODE MUST BE EQUAL TO TOTAL CURRENT OUT OF THE NODE

(A CONSERVATION OF CHARGE PRINCIPLE)

NODE

KIRCHHOFF CURRENT LAW (KCL)

SUM OF CURRENTS FLOWING INTO A NODE IS EQUAL TO SUM OF CURRENTS FLOWING OUT OF THE NODE

A5

A5

node the ofout flowing

negative the to equivalent is

node a into flowingcurrent A

ALGEBRAIC SUM OF CURRENTS FLOWING INTO A NODE IS ZERO

ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OF A 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 ACCUMULATION OF CHARGE

... OR WE CAN MAKE SUPERNODES BY AGGREGATING NODES

0:

0:

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii:3 & 2 Adding

INTERPRETATION: SUM OF CURRENTS LEAVING NODES 2&3 IS ZERO

VISUALIZATION: WE CAN ENCLOSE NODES 2&3 INSIDE A SURFACE THAT IS VIEWED AS A GENERALIZED NODE (OR SUPERNODE)

PROBLEM SOLVING HINT: KCL CAN BE USED TO FIND A MISSING CURRENT

A5

A3

?XI

a

b

c

d

SUM OF CURRENTS INTO NODE IS ZERO

0)3(5 AIA X

AIX 2

Which way are charges flowing on branch a-b?

...AND PRACTICE NOTATION CONVENTION AT THE SAME TIME...

?

4

3

,2

be

bd

cb

ab

I

AI

AI

AI NODES: a,b,c,d,e BRANCHES: a-b,c-b,d-b,e-b

b

a

c

d

e2A

-3A4A

Ibe = ?

0)2()]3([4 AAAIbe

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 use

of a more general idea of

node. The shaded surface

encloses a section of the

circuit and can be considered

as 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!!!

mAI 501 mAmAmAIT 204010

0410 1 ImAmA

01241 mAmAI03 12 ImAI

1I FindTI Find

1I Find21 I and I Find

mAi

mAii

x

xx

4

04410

01212010 mAmAii xx

+

-

I5I1

I2

I3

I4

I2 = 6mA, I3 = 8mA, I4 = 4mA

I1 = _______

I5 = _______

0345 III

mA5

0123 III

mA14

xi Find

+

-

+ -

1I22I

2I

3I

4I

5I

mAImAImAI 5,3,2 321

5

4

I

I

6I

mAIIII 802 6216

0625 III

0534 III

mA8mA5

mA5

mA2

THE PLAN

MARK ALL THE KNOWN CURRENTS

FIND NODES WHERE ALL BUT ONE CURRENT ARE KNOWN

DETERMINE THE CURRENTS INDICATED

mA1

mA4xI2

xI

xI FIND

0141 mAmAI

1I

021 XX III

mA3

mA3

bX

Xb

ImAI

mAImAI

42

21

ONVERIFICATI

bI

A

B

C

D E

F

G

AIDE 10AIEG 4

EFIA5

xI

xI__ to __ from flowscurrent BD On

EFI

__ to __ from flowscurrent EF On

A3

This question tests KCL and

convention to denote currents

Use sum of currents leaving node = 0

010)3()5( AAAIX

-8A

B D

0104 AAIEF

6A

E F

KCL

KIRCHHOFF VOLTAGE LAW

ONE OF THE FUNDAMENTAL CONSERVATION LAWS IN ELECTRICAL ENGINERING

THIS IS A CONSERVATION OF ENERGY PRINCIPLE “ENERGY CANNOT BE CREATE NOR DESTROYED”

KIRCHHOFF VOLTAGE LAW (KVL)

KVL IS A CONSERVATION OF ENERGY PRINCIPLE

A POSITIVE CHARGE GAINS ENERGY AS IT MOVES TO A POINT WITH HIGHER VOLTAGE AND RELEASES ENERGY IF IT MOVES TO A POINT WITH LOWER VOLTAGE

AV

BB

V)(AB

VVqW

q

ab

V

a b

q

abqVW LOSES

cd

V

c d

q

cdqVW GAINS

AV

BB

V

q

CV

AB

V

B

CV

CA

V

ABqVW

BCqVW

CAqVW

A “THOUGHT EXPERIMENT”

IF THE CHARGE COMES BACK TO THE SAME INITIAL POINT THE NET ENERGY GAIN MUST BE ZERO (Conservative network)

OTHERWISE THE CHARGE COULD END UP WITH INFINITE ENERGY, OR SUPPLY AN INFINITE AMOUNT OF ENERGY

0)( CDBCAB VVVq

KVL: THE ALGEBRAIC SUM OF VOLTAGE DROPS 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 USEFUL TO 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 EQUATIONS ARE INDEPENDENT. WE SHALL SEE THAT THE SAME SITUATION ARISES WHEN USING KVL

THE THIRD EQUATION IS THE SUM OF THE OTHER TWO!!

A SNEAK PREVIEW ON THE NUMBER OF LINEARLY 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 THE SAME EASE

______

______

______

bd

ac

ad

V

V

V

064 acV V10

042 bdV

V6

_______________, ebad VV

06812 adV

01264 ebV

_______bdV

FIRST VFIND MUST1

R

VVVV RRR 1010112111

V11

DEPENDENT SOURCES ARE NOT REALLY DIFFICULT TO ANALYZE

REMINDER: IN A RESISTOR THE VOLTAGE AND CURRENT DIRECTIONS MUST SATISFY THE PASSIVE SIGN CONVENTION

V

V

+

-

+

-

a

b V4 xV

1V

2V

kR 2

VVVV 4,12 21

2k

ab

x

P

resistor 2k the

on disipatedPower

V

V

DETERMINE

SAMPLE PROBLEM

We need to find a closed path where only one voltage is unknown

04124

0412

X

X

X

V

VVV

VFOR

4V

2

2 0

VVV

VVV

Xab

abX

-8V

Remember

past topics

+

-

+

-

k10 k5

xV

V25 4

xV

1V

There are no loops with only

one unknown!!!

The current through the 5k and 10k

resistors is the same. Hence the

voltage drop across the 5k is one half

of 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 CIRCUITS

BACKGROUND: USING KVL AND KCL WE CAN WRITE ENOUGH EQUATIONS TO ANALYZE ANY LINEAR 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 ELEMENTS IN SERIESONLY ONE CURRENT

WRITE 5 KCL EQS OR DETERMINE THE ONLY CURRENT FLOWING

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 VOLTAGE DIVIDER EQUATIONS

)(21

1

1tv

RR

RvR

SUMMARY OF BASIC VOLTAGE DIVIDER

kRkRVVS 30,90,9 21 :EXAMPLE

kR 151

VOLUME CONTROL?

A “PRACTICAL” POWER APPLICATION

HOW CAN ONE REDUCE THE LOSSES?

THE CONCEPT OF EQUIVALENT CIRCUIT

THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT HERE WITH A VERY SIMPLE VOLTAGE DIVIDER

+

-

1R

2R

Sv

i

21RR

vi S

+

-Sv 21

RR

i

AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR

1R

2R

21

RR

SERIES COMBINATION OF RESISTORS

THE DIFFERENCE BETWEEN ELECTRIC CONNECTION AND PHYSICAL LAYOUT

SOMETIMES, FOR PRACTICAL CONSTRUCTION REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY QUITE APART

IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES

COMPONENT SIDE

CONNECTOR SIDE

ILLUSTRATING THE DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS

PHYSICAL NODE

PHYSICAL NODE

SECTION OF 14.4 KB VOICE/DATA MODEM

CORRESPONDING POINTS

+

-

+

-

+

-

+ -

+

-

+ -

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

1R

v

2Rv

1v

2

v

3v

4

v

5v

SECOND GENERALIZATION: MULTIPLE RESISTORS

APPLY KVL TO THIS LOOP

VOLTAGE DIVISION FOR MULTIPLE RESISTORS

iRviRi

)30(,, kPVIbd

FIND

APPLY KVL TO THIS LOOP

bdV FOR LOOP

VVIkV bdbd 10(0][2012 KVL)

mWARIP 30)10*30()10( 3242

RESISTOR30k ON POWER

THE “INVERSE” VOLTAGE DIVIDER

kVS

5003.458220

20220

DIVIDER INVERSE""

+

-

1R

2RS

V

OV

SOV

RR

RV

21

2

VOLTAGE DIVIDER

OSV

R

RRV

2

21

"INVERSE" DIVIDER

SV COMPUTE

APPLY KVL TO THIS LOOP

mAIkIkI 05.004012806

40 12 0 10bd bd

V kI V V V

Find I and Vbd

V3

VVS

9320

201525

INVERSE DIVIDER PROBLEM

If Vad = 3V, find VS

( )i t

: 6 80 * ( ) 12 40 * ( ) 0KVL V k i t V k i t

6( ) 0.05

120

Vi t mA

k

Knowing the current one can compute ALL the remaining voltages and powers

80 * ( )k i t

40 * ( )k i t

Notice use of passive sign convention

+

-

+ -

k30

V9k20

k10

DE

CD

DA

I

V

V

A B C

DE

EXAMPLE

DETERMINE I USING KVL

I

0I*10kI*30k9I*20k12- :KVL

mAk

VI 05.0

60

3

mA05.0

VIk 5.1*30

DA VFORKVL

0*1012 IkVDA :KVL

VVDA 5.11

+

-

+ -

xV3 V4

k4

XV

ab

SV

VVS 12

source dependent theby supplied

or absorbedpower the is )3( VxP

)3( Vx

ab

x

P

V

V

EXAMPLE

APPLY KVL TO THIS LOOP

03412 XX VV :KVL VVX 2

034 Xab VV :KVL VVab 10

0 XSab VVV :KVL

I

)CONVENTION SIGN (PASSIVE IVP XVX

3)3(

mAk

VI 1

4

4 :LAW OHMS'

mWmAVPX

V 2][1*][2)3(

abVKVL HERE

OR KVL HERE

Sometimes you may want to vary a bit

SINGLE NODE-PAIR CIRCUITS

THESE CIRCUITS ARE CHARACTERIZED BY ALL THE ELMENTS HAVING THE SAME VOLTAGE ACROSS THEM - THEY ARE IN PARALLEL

V

V

EXAMPLE OF SINGLE NODE-PAIR

THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)

IN PRACTICE NODES MAY ASSUME STRANGE FORMS

LOW DISTORTION POWER AMPLIFIER

COMPONENT SIDE CONNECTION SIDE

LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW

SAMPLE PHYSICAL NODES

BASIC CURRENT DIVIDER

THE CURRENT i(t) ENTERS THE NODE AND SPLITS - IT IS DIVIDED BETWEEN THE CURRENTS 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 TO HIGHLIGHT ELECTRICAL CONNECTIONS!!

2*80 IkV24

IS EASIER TO SEE THE DIVIDER

LEARNING EXTENSION - CURRENT DIVIDER

CAR STEREO AND CIRCUIT MODEL

mA215mA215

POWER PER SPEAKER

)16(40120

1201

I

01612 II :KCL

mAI 121

mW IN POWER YIELDmA IN CURRENT

,k IN RESISTANCE

:POWER

RI2

WmWP 76.540*144

mAI 4)16(40120

402

DIVIDER CURRENT USING

THERE IS MORE THAN ONE OPTION TO COMPUTE I2

FIRST GENERALIZATION: MULTIPLE SOURCES

APPLY KCL TO THIS NODE

)()(

)(1

)(

21

21 tiRR

RRtv

tvR

ti

O

p

O

DEFINE “PARALLEL RESISTANCE COMBINATION”

SOURCE EQUIVALENT

mA10mA15

k3

k6

OV

SOURCES THEBY SUPPLIED

POWER THE AND VFINDO

kkk

kkR

p2

36

3*6

OVp

R

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

)()()()(

)()(

tiR

Rti

R

tvti

tiRtv

O

k

p

K

k

k

OP

General current divider

Ohm’s Law at every resistor

( )v t

4

v

k 6

v

k 12

v

k

Notice use of passive

sign convention

:6 4 04 6 12

v v vKCL mA mA

k k k

12k

72 3 2 48 0

24 6 0

4

V v v V v

V v

v V

Once v(t) is known

all other variables can

be determined; e.g.,

2

6

162.667

6 6k

v VP mW

k k

mA8

k4 k20 k5

1i

SOURCE THEBY SUPPLIED

POWER THE AND FIND1

i

kkkkkR

p2

1

20

415

5

1

20

1

4

11kR

p2

mA8

k4 k4

1i

AN ALTERNATIVEAPPROACH

mWmAvP

Vikv

mAk

ki

128)8(

16*4

4)8(4

2

1

1

20k||5k

)()()()(

)()(

tiR

Rti

R

tvti

tiRtv

O

k

p

K

k

k

OP

General current divider

FIND THE CURRENT L

I

COMBINE THE SOURCES COMBINE RESISTORS

STRATEGY: CONVERT THE PROBLEM INTO A BASIC CURRENT DIVIDER BY COMBINING SOURCES AND RESISTORS. THE NEXT SECTION EXPLORES IN MORE DETAIL THE IDEA OF COMBINING RESISTORS

NOTICE THE MINUS SIGN

mA1

k3

k3

k6

k6

A

B

C

mA9

A

B

C

k6

k6

k3

k3mA9

k6

k3

k6

k3

A

CB

9mA

I1 I2

I mA mA

I I

1

2 1

3

99 3

[ ]

I1

I2

I1

I2

DIFFERENT LOOKS FOR THE SAME ELECTRIC CIRCUIT

k3k3 k6k6

A

B

C

mA9

k6

k3

k6

k3

A

BC

mA9

I1 I2

I1

I2

REDRAWING A CIRCUIT MAY, SOMETIMES, HELP TO VISUALIZE BETTER THE ELECTRICAL CONNECTIONS

k2 k4 k3

mA20

Determine power

delivered by source

kRp

kkkkRp

13

12

12

436

3

1

4

1

2

11

2)20(* mARpP

WP

AP

13

800.4

][)10*20(*10*13

12 233

+

V

_

SERIES PARALLEL RESISTOR COMBINATIONS

UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)

WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT

NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE 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

SERIES k3

kkk 412||6

k12k3

k5

kkk 612||12

kkk 26||3

)24(||6 kkk

12kIf things get confusing…

EXAMPLES COMBINATION SERIES-PARALLEL

k9

kkk 69||18

kkk 1066

RESISTORS ARE IN SERIES IF THEY CARRY EXACTLY THE SAME CURRENT

RESISTORS ARE IN PARALLEL IF THEY ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES

If the drawing gets confusing… Redraw the reduced circuit and start again

AN “INVERSE SERIES PARALLEL COMBINATION”

AVAILABLE ARERESISTORS ONLY

WHEN600mV BE MUST

1.0

3AIVR

2.03

6.

A

VR REQUIRED 1.01.0R

AVAILABLE ARERESISTORS ONLY

WHEN600mV BE MUST

1.0

9AIVR

0667.09

6.

A

VR REQUIRED

R

SIMPLE CASE

NOT SO SIMPLE CASE

Given the final value Find a proper combination

EFFECT OF RESISTOR TOLERANCE

10% :TOLERANCE RESISTOR

2.7k : VALUERESISTORNOMINAL

RANGES FOR CURRENT AND POWER?

mAI

mAI

115.47.29.0

10

367.37.21.1

10

max

min

:CURRENT MAXIMUM

:CURRENT MINIMUM

THE RANGES FOR CURRENT AND POWER ARE DETERMINED BY THE TOLERANCE BUT THE PERCENTAGE OF CHANGE MAY BE DIFFERENT FROM THE PERCENTAGE OF TOLERANCE. THE RANGES MAY NOT EVEN BE SYMMETRIC

mAI 704.37.2

10

:CURRENTNOMINAL

_

mWP 04.377.2

102

:POWERNOMINAL

:POWER MAXIMUM

:)POWER(VI MINIMUM min

mW

mW

15.41

67.33

CIRCUIT WITH SERIES-PARALLEL RESISTOR COMBINATIONS

COMBINING RESISTORS IN SERIES ELIMINATES ONE NODE FROM THE CIRCUIT. COMBINING RESISTORS IN PARALLEL ELIMINATES ONE LOOP FROM THE CIRCUIT

THE COMBINATION OF COMPONENTS CAN REDUCE THE COMPLEXITY OF A CIRCUIT AND RENDER IT SUITABLE FOR ANALYSIS USING THE BASIC TOOLS DEVELOPED SO FAR.

GENERAL STRATEGY: •REDUCE COMPLEXITY UNTIL THE CIRCUIT BECOMES SIMPLE ENOUGH TO ANALYZE. •USE DATA FROM SIMPLIFIED CIRCUIT TO COMPUTE DESIRED VARIABLES IN ORIGINAL CIRCUIT - HENCE ONE MUST KEEP TRACK OF ANY RELATIONSHIP BETWEEN VARIABLES

FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT k12kk 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

*4

124

12

IkV

II

b

5

345

*3

0

IkV

III

C

:SOHM'

:KCL

kkk 12||2

VVkk

kV

O1)3(

21

1

:DIVIDER VOLTAGE

kkk 211

AAkk

kI

O1)3(

21

1

:DIVIDER CURRENT

LEARNING BY DOING

AN EXAMPLE OF “BACKTRACKING”

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

4*6 IkV

b

k

VI b

33

mA1

432III

mA5.1

2*2 IkV

a

V3

V3

baxzVVV

VVxz

6

k

VI xz

45

mA5.1

521III

mAI 31

11*4*6 IkVIkV

xzO

VVO 36

mA5.0

OV FIND

DIVIDER VOLTAGEUSE

FIND :STRATEGY1

V

1V k60

kkk 2060||30

+

-

1Vk20

k20

V12

Vkk

k6)12(

2020

20

V6

DIVIDER VOLTAGE

1

4020

20V

kk

kV

O

V2

SV FIND

THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED?

V6

mA05.0mA15.0

mAkV 1.0*601

k

VI

120

61

VmAkVS

615.0*20

V9

SERIES PARALLEL

TIONSTRANSFORMAY

THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL

IF INSTEAD OF THIS

WE COULD HAVE THIS

THEN THE CIRCUIT WOULD BECOME LIKE THIS AND BE AMENABLE TO SERIES PARALLEL 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 ADD TO 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

3Rkkk

kk

18612

612

mAk

VIS 2.1

12

12

ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ...

LEARNING EXAMPLE

SHOULD KEEP THESE TWO NODES!

CONVERT THIS Y INTO A DELTA?

IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!

Y

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

3*12 *1236

12

k kk

k

4mA 36k

36k

36k

12k

12kO

V

THE RESULTING CIRCUIT IS A CURRENT DIVIDER

4mA 36k

36 ||12 9k k k

OV

9k

CIRCUIT AFTER PARALLEL RESISTOR REDUCTION

OI

36 84

36 18 3O

kI mA mA

k k

89 9 24

3O O

V k I k mA V NOTICE THAT BY KEEPING THE FRACTION WE PRESERVE FULL NUMERICAL ACCURACY

WYE DELTA

A

VOV FIND

CIRCUITS WITH DEPENDENT SOURCES GENERAL STRATEGY TREAT DEPENDENT SOURCES AS REGULAR SOURCES AND ADD ONE MORE EQUATION FOR THE CONTROLLING VARIABLE

1*2 IkV

A

A CONVENTION ABOUT DEPENDENT SOURCES. UNLESS OTHERWISE SPECIFIED THE CURRENT AND VOLTAGE VARIABLES ARE ASSUMED IN SI UNITS OF Amps AND Volts

FOR THIS EXAMPLE THE MULTIPLIER MUST HAVE UNITS 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

VIV

XD2, UNITS ARE EXPLICIT

KVL

A PLAN: SINGLE LOOP CIRCUIT. USE KVL TO DETERMINE CURRENT

0*5*31211 IkVIk

A :KVL

ONE EQUATION, TWO UNKNOWNS. CONTROLLING VARIABLE PROVIDES EXTRA EQUATION

mAI 21

REPLACE 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 UNKNOWNS AND JUST ONE EQUATION

THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION

SUBSTITUTION OF I_0 YIELDS

NOTICE THE CLEVER WAY OF WRITING mA TO HAVE VOLTS IN ALL NUMERATORS AND THE SAME UNITS IN DENOMINATOR

6056/* SVk

VOLTAGE DIVIDER

VVkk

kV SO )12(

3

2

24

4

OV FINDKCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE

OV FINDA PLAN: ONE LOOP PROBLEM. FIND THE CURRENT THEN USE OHM’S LAW. KVL TO

THIS LOOP

THE DEPENDENT SOURCE IS ONE MORE VOLTAGE SOURCE

THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION

REPLACE AND SOLVE FOR CURRENT I … AND FINALLY

A PLAN: ONE LOOP ON THE LEFT - KVL ONE NODE-PAIR ON RIGHT - KCL

KVL

KVL

ALSO A VOLTAGE DIVIDER

KCL

0)(

)( L

Ogm

R

tvtvg

KCL

)(

)(

tv

tvG

i

O FIND