The second systematic technique to determine all currents and voltages in a circuit

IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUITAND THEN IT USES OHM’S LAW TO COMPUTE NECESSARY VOLTAGES

THERE ARE SITUATION WHERE NODE ANALYSIS IS NOT AN EFFICIENT TECHNIQUEAND WHERE THE NUMBER OF EQUATIONS REQUIRED BY THIS NEW METHOD ISSIGNIFICANTLY SMALLER

LOOP ANALYSIS

+-

+-

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

23

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

THESE ARE LOOP EQUATIONS FOR THECIRCUIT

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

IN MATRIX FORM

1 2 3 3 11

3 3 4 5 22

S

S

R R R R vi

R R R R vi

WRITE KVL ON EACH MESH

WRITE THE MESH EQUATIONS

IDENTIFY ALL VOLTAGE DROPS 4Rv

1Rv

2Rv

5Rv

3Rv

BOOKKEEPINGBRANCHES = 8NODES = 7

LOOP CURRENTS NEEDED = 2

02211 RSRS vvvv :MESH TOP

033452 RSRRR vvvvv :BOTTOM

USE OHM’S LAW

11Ri

52Ri

42Ri

221 )( Rii

32Ri

AND WE ARE TOLD TOUSE MESH CURRENTS!THIS DEFINES THE LOOPCURRENTS TO BE USED

+-

+ -

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

2I @KVL

1IIO NOWTHIS SELECTION IS MORE EFFICIENT

996

12612

21

21

kIkI

kIkIREARRANGE

substract and 2/*

3/*

mAIkI4

31824 11

LEARNING EXAMPLE: FIND Io USING LOOP ANALYSII

AN ALTERNATIVE SELECTION OF LOOP CURRENTS

A PRACTICE EXAMPLEIF THE CIRCUIT CONTAINS ONLY INDEPENDENTSOURCE THE MESH EQUATIONS CAN BE WRITTEN“BY INSPECTION”

THE RIGHT HAND SIDE IS THE ALGEBRAIC SUMOF VOLTAGE SOURCES AROUND THE LOOP(VOLTAGE RISES - VOLTAGE DROPS)

MUST HAVE ALL MESH CURRENTS WITH THESAME ORIENTATION

IN LOOP KIN LOOP K

THE COEFFICENT OF Ik IS THE SUM OF RESISTANCES AROUND THE LOOP.

THE COEFFICIENT OF Ij IS THE SUM OFRESISTANCES COMMON TO BOTH k AND j ANDWITH A NEGATIVE SIGN.

LOOP 1 12612 21 kIkI

LOOP 2 396 21 kIkI

LOOP 1 kkI 641 oft coefficien02 I oft coefficienkI 63 oft coefficien

LOOP 2 01 I oft coefficienkkI 392 oft coefficien

kI 33 oft coefficien

][6V RHS

][6VRHS

Loop 31 2 3

(6 ) (3 ) (3 6 12 ) 0k I k I k k k I

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

LEARNINGEXTENSION

3. SOLVE EQUATIONS

12k

6k

4k 4k

2k

12V

9V

WRITE THE MESH EQUATIONS

BOOKKEEPING: B = 7, N = 4

1I

2I

3I

4I 1. DRAW MESH CURRENTS

2. WRITE MESH EQUATIONS. USE KVL

0)(61212 311 IIkVkI :1 MESH0)(4)(412 3242 IIkIIkV :2 MESH

0)(4)(69 2313 IIkIIkV :3 MESH02)(49 424 kIIIkV :4 MESH

EQUATIONS BY INSPECTION

VkIkI 12618 31 VkIkIkI 12448 432 VkIkIkI 91046 321

VkIkI 964 42

CHOOSE YOUR FAVORITE TECHNIQUETO SOLVE THE SYSTEM OF EQUATIONS

CIRCUITS WITH INDEPENDENT CURRENT SOURCES

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

ANALYSISMESH USING COMPUTE OV

TWO MESH CURRENTS ARE DEFINED BY CURRENTSOURCES

mAImAI 24 21 MESH 3

“BY INSPECTION” VkIkIkI 31242 321

mAk

mAkmAkVI

4

1

12

)2(4)4(233

KVL FOR Vo

USE KVL TOCOMPUTE Vo

LEARNING EXAMPLE

WE ACTUALLY NEED THE CURRENT ON THE RIGHT MESH. HENCE, USE MESH ANALYSIS

1I2I

MESH 1: mAI 41

MESH 2: 06)(4][5 212 kIIIkV

mAmAmAI 1144510 2

][5

336 2 VkIVO

1I

2I

MESH 1: mAI 41

MESH 2: 0124 21 kIkI

mAI3

4

12

162

][86 2 VkIVO

LEARNING EXTENSIONS

+-

+

VO

_

6k

4k

2k2k

IS

VS

IS = 2mA, V S = 6V

Determine VO

P ro b lem 3.46 (6th Ed )

1. Select loop currents.

In this case we use meshes. We note that the current source could define one mesh.

1I

2I

3I

2. Write loop equations.

Loop 1SII 1

Loop 2 0)(2)(4 1232 IIkIIkVSLoop 3 0)(26)(4 13323 IIkkIIIk

Since we need to compute Vo it is efficient to solve for I3 only.

HINT: Divide the loop equations by 1k. Coefficientsbecome numbers and voltage source becomes mA.

mAII28

323421028 33

VkIVO 7

486 3

3 non-reference nodes. 3 meshes

One current source, one super node

SELECTING THE SOLUTION METHOD

BOTH APPROACHES SEEM COMPARABLE. CHOOSELOOP ANALYSIS

We use the fact that I1 = Is

])[46(21

46 132 mAIk

VII S Loop 2

mAIII S 42124 32 Loop 3

2/*

eqs add and 3/*

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

CURRENT SOURCES SHARED BY MESHES - THE GENERAL LOOP APPROACH

THE STRATEGY IS TO DEFINE LOOP CURRENTS THAT DO NOT SHARE CURRENT SOURCES -

EVEN IF IT MEANS ABANDONING MESHES

FOR CONVENIENCE START USING MESH CURRENTSUNTIL REACHING A SHARED SOURCE. AT THATPOINT DEFINE A NEW LOOP.

IN ORDER TO GUARANTEE THAT IF GIVES ANINDEPENDENT EQUATION ONE MUST MAKE SURETHAT THE LOOP INCLUDES COMPONENTS THATARE NOT PART OF PREVIOUSLY DEFINED LOOPS

A POSSIBLE STRATEGY IS TO CREATE A LOOPBY OPENING THE CURRENT SOURCE

mAI

mAI

4

2

2

1

THE LOOP EQUATIONS FOR THE LOOPS WITHCURRENT SOURCES ARE

THE LOOP EQUATION FOR THE THIRD LOOP IS

0)(1)(2)(21][6 13123233 IIkIIIkIIkkIV

THE MESH CURRENTS OBTAINED WITH THISMETHOD ARE DIFFERENT FROM THE ONESOBTAINED WITH A SUPERMESH. EVEN FORTHOSE DEFINED USING MESHES.

-+

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

0 ) ( ) ( ) (3 4 4 1 3 4 1 2 4 3 I I R I I I R I I R VSKVL OF REMAINING LOOP

)( 4233 IIRV

)( 4344 IIRV

4V

1V

2V

3V

-+

1R2R

3R

4R

1SI 2SI

3SISV

A COMMENT ON METHOD SELECTIONThe same problem can be solved by node analysis but it requires 3 equations

1V

2V3V

4V

0

0

0

1

14

1

41

1

43

2

13

3

233

212

31

2

R

VV

R

VI

R

VV

R

VV

R

VVI

IIR

VV

VV

S

S

SS

S

CIRCUITS WITH DEPENDENT SOURCES Treat the dependent source as thoughit were independent.Add one equation for the controllingvariable

k

VI

mAI

X

2

4

2

1

SOURCESBY DETERMINED

CURRENTS MESH

0)(1)(21 4313 IIkIIkkI x :3 MESH

012)(1)(1 2434 VIIkIIk :4 MESH

)(2 1324 IIkVIII xx VARIABLESGCONTROLLIN

COMBINE EQUATIONS. DIVIDE BY 1k

122

823

0

4

432

432

321

1

III

III

III

I

122

823

0

4

432

432

321

1

III

III

III

I

SOLVE USING MATLAB

PUT IN MATRIX FORM

12

8

0

4

2110

2310

0111

0001

4

3

2

1

I

I

I

I

» R=[1,0,0,0; %FIRST ROW1,1, -1, 0; %SECOND ROW0,1,3,-2; %THIRD ROW0,-1,-1,2] %FOURTH ROW

R = 1 0 0 0 1 1 -1 0 0 1 3 -2 0 -1 -1 2

DEFINE THE MATRIX

DEFINE THE RIGHT HAND SIDE VECTOR» V=[4;0;8;12]

V =

4 0 8 -12

Since we divided by1k the RHS is mA andall the coefficientsare numbers

SOLVE AND GET THE ANSWER » I=R\V

I =

4 -6 -2 -10

The answers are in mA

>> is the MATLAB prompt. Whatfollows is the command entered

LEARNING EXTENSION: Dependent Sources

We treat the dependent source as one more voltage source

and solve...

The selection of loop currentssimplifies expression for Vx and computation of Vo.

USING LOOP CURRENTSUSING MESH CURRENTS

MESH 1 0)(422 211 IIkkIVx LOOP 1 04)(22 121 kIIIkVx

MESH 2 0)(463 122 IIkkI LOOP 2 063)(22 221 kIIIkVxNOW WE EXPRESS THE CONTROLLING VARIABLE IN TERMS OF THE LOOP CURRENTS

14kIVx )(4 21 IIkVx

3104

042

21

21

kIkI

kIkI REPLACE AND REARRANGE

386

066

21

21

kIkI

kIkI

mAImAI 5.1,3 21 mAImAI 5.1,5.1 21 SOLUTIONS

][96 2 VkIVO

NOTICE THE DIFFERENCE BETWEEN MESH CURRENT I1 AND LOOP CURRENT I1 EVENTHOUGH THEY ARE ASSOCIATED TO THESAME PATH

Find Vo

DEPENDENT CURRENT SOURCE. CURRENT SOURCES NOT SHARED BY MESHES

We treat the dependent source as a conventional source

Then KVL on the remaining loop(s)

And express the controlling variable, Vx, in terms of loop currents

Equations for meshes with current sources

WE ARE ASKED FOR Vo. WE ONLY NEEDTO SOLVE FOR I3

REPLACE AND REARRANGE

mAIIIIkV

kIV

x

x 42)(4

221

21

1

mAIkIkI8

11238 323

][4

336 3 VkIVO

1I

DRAW MESH CURRENTS

2I

WRITE MESH EQUATIONS.

0)(4212 122 IIkkI :2 MESH

CONTROLLING VARIABLE IN TERMS OF LOOP CURRENTS

2II x

0)(422 211 IIkkIkI x :1 MESH

REPLACE AND REARRANGE

1264

066

21

21

kIkI

kIkI

SOLVE FOR I2

mAIkI 6122 22

][122 2 VkIVO

In the following we shall solve using loopanalysis two circuits that had previously beensolved using node analysis

This is one circuit.we recap first the node analysisapproach and then we solve usingloop analysis

LEARNING EXAMPLE FIND THE VOLTAGE Vo

AT SUPER NODE

1 22

XV V V

2 3 1 32 14

2 01 1 1 1

V V V VV V VmA

k k k k

3 2 3 1

3@ : 2 0

1 1

V V V VV mA

k k

4 4@ : 4V V V

CONTROLLING VARIABLE2X

V V

1k

1k

SOLVE EQUATIONS NOW

1

1 3

1 3

3

2 2 6

2 2

X

X

X

V V

V V V V

V V V V

VARIABLE OF INTEREST 1 3OV V V

RECAP OF NODE ANALYSIS

USING LOOP ANALYSISDETERMINE Vo

1I

2I

3I

START SELECTION USING MESHES

4I

SELECT A GENERAL LOOP TO AVOIDSHARING A CURRENT SOURCE

1

3

1: 2

3 : 2

Loop I mA

Loop I mA

Write loop equations

2 2 32 : 2 1 1 ( ) 0

XLoop V kI k I I

4 3 1 44 : 4 1 ( ) 2 1 0

XLoop V k I I I V kI

Controlling variable: 1 3 41 ( )

XV k I I I

2 4

2 4

4

2 2 61 , 2

4 8

kI kII mA I mA

kI

21

OV kIVariable of Interest

LEARNING EXAMPLE Find the current Io

2 2@ : 12V V V

3 3@ : 2

XV V V

4 1

1 3 4 3 4 51 2 4

@super node:

6 (constraint eq.)

2 01 1 1 1 1X

V V V

V V V V V VV V VI

k k k k k

5 4 5

5@ : 2 0

1 1X

V V VV I

k k

1 2

4

CONTROLLING VARIABLES

1

X

X

V V V

VI

k

7 eqs in 7 variables

VARIABLE OF INTEREST 5

1O

VI

k

RECAP OF NODE ANALYSIS

Find the current Io using mesh analysis

Write loop/mesh equations

1 1 2 1 4Loop1: 1 1 ( ) 1 ( ) 0kI k I I k I I

2 1 2 5Loop 2: 1 ( ) 6 1 ( ) 0k I I V k I I

3Loop 3: 2

XI I

4 1Loop 4: 12 1 ( ) 2 0

XV k I I V

5 2 5 6Loop 5: 1 ( ) 1 ( ) 2 0

Xk I I k I I V

6 3 6 6 5Loop 6: 1 ( ) 1 1 ( ) 0k I I kI k I I

1

5 6

Controlling variables

1X

X

V kI

I I I

8 eqs in 8 unknowns

Variable of interest:6O

I I

Select mesh currents