RESISTIVE CIRCUITS

•MULTI NODE/LOOP CIRCUIT ANALYSIS

DEFINING THE REFERENCE NODE IS VITAL

SMEANINGLES IS4V VSTATEMENT THE 1 UNTIL THE REFERENCE POINT IS DEFINED

BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.

V4

ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT

V2

_____?12 V

12V

THE STRATEGY FOR NODE ANALYSIS 1. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE

2. IDENTIFY KNOWN NODE VOLTAGES

3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION(e.g.,SUM OF CURRENT LEAVING =0)

0:@321 IIIV

a

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

0369

k

VV

k

V

k

VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...

REFERENCE

SV

aV

bV

cV

0:@543 IIIV

b

0:@65 IIV

c

0943

k

VV

k

V

k

VVcbbab

039

k

V

k

VVcbc

SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...

AND PRACTICE WRITING

THESE DIRECTLY

EXAMPLE

@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION

REPEAT THE PROCESS AT NODE 2

03

12

4

122

R

vv

R

vvi

OR VISUALIZE CURRENTS GOING INTO NODE

WRITE THE KCL EQUATIONS

mA6

1I

2I

3I

036

6:@ 222 k

V

k

VmAV 2

12V V

CURRENTS COULD BE COMPUTED DIRECTLYUSING KCL AND CURRENT DIVIDER!!

1

2

3

8

3(6 ) 2

3 66

(6 ) 43 6

I mA

kI mA mA

k kk

I mA mAk k

IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM

NODE EQS. BY INSPECTION

mAVVk

6202

121

mAVkk

V 63

1

6

10 21

116V V

k

VI

k

VI

k

VI

3622

32

21

1

Once node voltages are known

1

1@ : 2 6 0

2

VV mA mA

k

Node analysis

3 nodes plus the reference. In principle one needs 3 equations...

…but two nodes are connected tothe reference through voltage sources. Hence those node voltages are known!!!

…Only one KCL is necessary

012126

12322

k

VV

k

VV

k

V

][5.1][64

0)()(2

22

12322

VVVV

VVVVV

EQUATIONS THE SOLVING

Hint: Each voltage sourceconnected to the referencenode saves one node equation ][6

][12

3

1

VV

VV

THESE ARE THE REMAININGTWO NODE EQUATIONS

Conventional analysis requires all currents at a node

@V_1 06

6 1 SIk

VmA

@V_2 012

4 2 k

VmAIS

2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the source

Math solution: add one equation

][621 VVV

Efficient solutionEfficient solution: enclose the source, and all elements in parallel, inside a surface.

Apply KCL to the surface!!!

04126

6 21 mAk

V

k

VmA

The source current is interiorto the surface and is not required

We STILL need one more equation

][621 VVV Only 2 eqs in two unknowns!!!

SUPERNODE

THE SUPERNODE TECHNIQUE

SI

k 12 / *

][6

046126

21

21

VVV

mAmAk

V

k

V

(2)

(1)

Equations The

ALGEBRAIC DETAILS

...by Multiply Eq(1). in rsdenominato Eliminate 1.

Solution

][6

][242

21

21

VVV

VVV

][10][303 11 VVVV 2 Veliminate to equations Add2.

][4][612 VVVV 2 Vcompute to Eq(2) Use 3.

VV

VV

4

6

4

1

SOURCES CONNECTED TO THE

REFERENCE

SUPERNODE

CONSTRAINT EQUATION VVV 1223

KCL @ SUPERNODE

02

)4(

212

63322

k

V

k

V

k

V

k

Vk2/*

VVV 22332VVV 12

32 add and 3/*

VV 385 3

mAk

VIO

8.32

3 LAW SOHM'

OIV FOR NEEDED NOT IS

2

OI of Value Find

+-

+-

1R 2R

3RV18

V12

2RV 1RV

3RV

Apply node analysis to this circuitThere are 4 non reference nodes

There is one super node

There is one node connected to thereference through a voltage source

We need three equations to compute all node voltages

…BUT THERE IS ONLY ONE CURRENT FLOWING THROUGH ALL COMPONENTS AND IF THAT CURRENT IS DETERMINED ALL VOLTAGES CAN BE COMPUTED WITH OHM’S LAW

I

STRATEGY:1. Apply KVL(sum of voltage drops =0)

0][18][12321

RRRVVVVV

2. Use Ohm’s Law to expressvoltages in terms of the “loop current.”

0][18][12 321 IRVIRIRVRESULT IS ONE EQUATION IN THE LOOP CURRENT!!!

SHORTCUT

Skip this equation

Write this onedirectly

3V2V1V

4V

LOOPS, MESHES AND LOOP CURRENTS

EACH COMPONENTIS CHARACTERIZEDBY ITS VOLTAGEACROSS AND ITSCURRENT THROUGH

A LOOP IS A CLOSED PATH THAT DOES NOTGO TWICE OVER ANY NODE.THIS CIRCUIT HAS THREE LOOPS

1

2 3

4

56

7

A BASIC CIRCUIT

ab c

def

fabcdef

A MESH IS A LOOP THAT DOES NOT ENCLOSEANY OTHER LOOP.fabef, ebcde ARE MESHES

A LOOP CURRENT IS A (FICTICIOUS) CURRENTTHAT IS ASSUMED TO FLOW AROUND A LOOP

fabef ebcde

1I

2I

3I

CURRENTS LOOP ARE321

,, III

A MESH CURRENT IS A LOOP CURRENT ASSOCIATED TO A MESH. I1, I2 ARE MESHCURRENTS

CLAIM: CLAIM: IN A CIRCUIT, THE CURRENT THROUGHANY COMPONENT CAN BE EXPRESSED IN TERMSOF THE LOOP CURRENTS

1

2 3

4

56

7

A BASIC CIRCUIT

ab c

def

FACT:FACT: NOT EVERY LOOP CURRENT IS REQUIREDTO COMPUTE ALL THE CURRENTS THROUGHCOMPONENTS

1I

3I 3

1

31

II

II

III

cb

eb

fa

CURRENTS LOOP

TWO USING

32

21

31

III

III

III

cb

eb

fa

EXAMPLES THE DIRECTION OF THE LOOP

CURRENTS IS SIGNIFICANT

FOR EVERY CIRCUIT THERE IS A MINIMUMNUMBER OF LOOP CURRENTS THAT ARENECESSARY TO COMPUTE EVERY CURRENTIN THE CIRCUIT.SUCH A COLLECTION IS CALLED A MINIMALSET (OF LOOP CURRENTS).

FOR A GIVEN CIRCUIT LETB NUMBER OF BRANCHESN NUMBER OF NODES

THE MINIMUM REQUIRED NUMBER OF LOOP CURRENTS IS

)1( NBL

MESH CURRENTS ARE ALWAYS INDEPENDENT

AN EXAMPLE

2)16(7

6

7

L

N

BTWO LOOP CURRENTS AREREQUIRED.THE CURRENTS SHOWN AREMESH CURRENTS. HENCE THEY ARE INDEPENDENT ANDFORM A MINIMAL SET

KVL ON LEFT MESH

REPLACING AND REARRANGING

DETERMINATION OF LOOP CURRENTS

KVL ON RIGHT MESH

2 4 5 30

Sv v v v USING OHM’S LAW

1 1 1 2 1 2 3 1 2 3

4 2 4 5 2 5

, , ( )

,

v i R v i R v i i R

v i R v i R

+-

+ -

V 1

V 2R 1

R 2 R 3

R 4R 5

WRITE THE MESH EQUATIONS

1I

2I

DRAW THE MESH CURRENTS. ORIENTATIONCAN BE ARBITRARY. BUT BY CONVENTIONTHEY ARE DEFINED CLOCKWISE

NOW WRITE KVL FOR EACH MESH AND APPLYOHM’S LAW TO EVERY RESISTOR.

0)( 51221111 RIRIIRIV

AT EACH LOOP FOLLOW THE PASSIVE SIGN CONVENTION USING LOOP CURRENT REFERENCE DIRECTION

0)( 21242322 RIIRIRIV

DEVELOPING A SHORTCUT

WHENEVER AN ELEMENTHAS MORE THAN ONELOOP CURRENT FLOWINGTHROUGH IT WE COMPUTENET CURRENT IN THE DIRECTION OF TRAVEL

SHORTCUT: SHORTCUT: POLARITIES ARE NOT NEEDED.APPLY OHM’S LAW TO EACH ELEMENT AS KVLIS BEING WRITTEN

1I @KVL

2I @KVL

396

12612

21

21

kIkI

kIkIREARRANGE

add and 2/*

mAIkI 5.0612 22

mAIkIkI4

561212 121

EXPRESS VARIABLE OF INTEREST AS FUNCTIONOF LOOP CURRENTS

21 IIIO

1I @KVL

1IIO NOWTHIS SELECTION IS MORE EFFICIENT

996

12612

21

21

kIkI

kIkIREARRANGE

substract and 2/*

3/*

mAIkI4

31824 11

EXAMPLE: FIND Io

AN ALTERNATIVE SELECTION OF LOOP CURRENTS

2I @KVL

1. DRAW THE MESH CURRENTS

1I 2I

2. WRITE MESH EQUATIONS

MESH 1 ][32)242( 21 VkIIkkk

MESH 2 )36()62(2 21 VVIkkkI DIVIDE BY 1k. GET NUMBERS FOR COEFFICIENTSON THE LEFT AND mA ON THE RHS

][982

][328

21

21

mAII

mAII

add and 4/*][3330 2 mAI ][

5

336 2 VkIVO

3. SOLVE EQUATIONS

THERE IS NO RELATIONSHIP BETWEEN V1 ANDTHE SOURCE CURRENT! HOWEVER ...

MESH 1 CURRENT IS CONSTRAINED

MESH 1 EQUATION mAI 21

MESH 2

VkIkI 282 21 “BY INSPECTION”

][2

96

4

3

8

2)2(222 VkIVmA

k

VmAkI O

CURRENT SOURCES THAT ARE NOT SHAREDBY OTHER MESHES (OR LOOPS) SERVE TO DEFINE A MESH (LOOP) CURRENT AND REDUCE THE NUMBER OF REQUIRED EQUATIONS

TO OBTAIN V1 APPLY KVL TO ANY CLOSEDPATH THAT INCLUDES V1

KVL

CURRENT SOURCES SHARED BY LOOPS - THE SUPERMESH APPROACH

1. SELECT MESH CURRENTS

2. WRITE CONSTRAINT EQUATION DUE TOMESH CURRENTS SHARING CURRENT SOURCES

mAII 432

3. WRITE EQUATIONS FOR THE OTHER MESHES

mAI 21

4. DEFINE A SUPERMESHSUPERMESH BY (MENTALLY)REMOVING THE SHARED CURRENT SOURCE

SUPERMESH

5. WRITE KVL FOR THE SUPERMESH

0)(1)(2216 131223 IIkIIkkIkI

NOW WE HAVE THREE EQUATIONS IN THREE

UNKNOWNS. THE MODEL IS COMPLETE

-+

1R2R

3R

4R

1SI 2SI

3SISV

RESISTORS ACROSS VOLTAGESFIND

Three independent current sources.Four meshes.One current source shared by twomeshes.

For loop analysis we notice...

Careful choice of loop currents should make only one loop equationnecessary. Three loop currents canbe chosen using meshes and notsharing any source.

1I 2I

3I

)( 43111 IIIRV

)( 1222 IIRV

SOLVE FOR THE CURRENT I4.USE OHM’S LAW TO C0MPUTE REQUIREDVOLTAGES

Now we need a loop current that doesnot go over any current source and passes through all unused components.

HINT: IF ALL CURRENT SOURCES ARE REMOVEDTHERE IS ONLY ONE LOOP LEFT

4I

11 sII 22 SII

33 SII

MESH EQUATIONS FOR LOOPS WITHCURRENT SOURCES

KVL OF REMAINING LOOP

)( 4233 IIRV

4V

1V

2V

3V

0)()()( 3441341243 IIRIIIRIIRVS