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 ANALYSIS1. 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 =++− IIIVa

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

0369

=−

++−

kVV

kV

kVV baasa AND GET ALGEBRAIC EQUATIONS IN

THE NODE VOLTAGES ...

REFERENCE

SV aV bV cV

0:@ 543 =++− IIIVb

0:@ 65 =+− IIVc

0943

=−

++−

kVV

kV

kVV cbbab

039

=+−

kV

kVV cbc

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 =

−+

−+−

Rvv

Rvvi

OR VISUALIZE CURRENTS GOING INTO NODE

WRITE THE KCL EQUATIONS

mA6

1I2I

3I

⇒=++− 036

6:@ 222 k

Vk

VmAV 2 12V V=

CURRENTS COULD BE COMPUTED DIRECTLYUSING KCL AND CURRENT DIVIDER!!

1

2

3

83 (6 ) 2

3 66 (6 ) 4

3 6

I mAkI mA mA

k kkI mA mA

k k

= −

= =+

= =+

IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM

NODE EQS. BY INSPECTION

( ) ( )mAVVk

62021

21 +−=+

( ) mAVkk

V 631

610 21 =⎟

⎠⎞

⎜⎝⎛ ++

1 16V V⇒ = −

kVI

kVI

kVI

3622

32

21

1 ===

Once node voltages are known

11@ : 2 6 0

2VV mA mAk+ + =

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 =−

+−

+kVV

kVV

kV

][5.1][640)()(2

22

12322

VVVVVVVVV

=⇒==−+−+

EQUATIONSTHESOLVING

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

][12

3

1

VVVV

−==

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

Apply KCL to the surface!!!

04126

6 21 =+++− mAk

Vk

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

][6

04612621

21

VVV

mAmAk

Vk

V

=−

=+−+

(2)

(1)

Equations The

ALGEBRAIC DETAILS

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

][6][242

21

21

VVVVVV

=−=+

][10][303 11 VVVV =⇒=2Veliminateto equations Add2.

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

VVVV4

6

4

1

−== SOURCES CONNECTED TO THE

REFERENCE

SUPERNODE

CONSTRAINT EQUATION VVV 1223 =−

KCL @ SUPERNODE

02

)4(212

6 3322 =−−

+++−

kV

kV

kV

kV

k2/*

VVV 223 32 =+VVV 1232 =+− addand 3/*VV 385 3 =

mAk

VIO 8.32

3 == LAW SOHM'

OIV FORNEEDED NOT IS 2

OI of Value Find

+-

+-

1R 2R

3RV18

V12

−+ 2RV−+ 1RV

−+ 3RV

Apply node analysis to this circuit

There 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][12 321 =−+++− RRR VVVVV

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

CURRENTS LOOP ARE321 ,, III

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

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

−

1I

−

3I3

1

31

−

−

−−

=

=

−−=

II

II

III

cb

eb

fa

CURRENTS LOOPTWOUSING

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(767

=−−===

LNB

TWO 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 3 0Sv 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 Rv i R v i R

= = = −

= =

+-

+ -

V1

V2R1

R2 R3

R4R5

WRITE THE MESH EQUATIONS

1I2I

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

1I @KVL

2I @KVL

39612612

21

21

−=+−=−

kIkIkIkIREARRANGE

addand 2/*mAIkI 5.0612 22 =⇒=

mAIkIkI4561212 121 =⇒+=

EXPRESS VARIABLE OF INTEREST AS FUNCTIONOF LOOP CURRENTS

21 IIIO −=

1I@KVL

1IIO = NOWTHIS SELECTION IS MORE EFFICIENT

99612612

21

21

=+=+

kIkIkIkIREARRANGE

substractand 2/*3/*

mAIkI431824 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

mAIImAII

=+−−=−

addand 4/*][3330 2 mAI = ][

5336 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”

][296

43

82)2(2

22 VkIVmAk

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 SUPERMESH 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