Transcript NODE ANALYSIS

NODAL AND LOOP ANALYSIS TECHNIQUES
LEARNING GOALS
NODAL ANALYSIS
LOOP ANALYSIS
Develop systematic techniques to determine all the voltages
and currents in a circuit
NODE ANALYSIS
• One of the systematic ways to
determine every voltage and
current in a circuit
The variables used to describe the circuit will be “Node Voltages”
-- The voltages of each node with respect to a pre-selected
reference node
IT IS INSTRUCTIVE TO START THE PRESENTATION WITH
A RECAP OF A PROBLEM SOLVED BEFORE USING SERIES/
PARALLEL RESISTOR COMBINATIONS
COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT
4k || 12k 12k
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
6k
I3
Va
KCL : I1  I 2  I 3  0
6k
OHM' S : Vb  3k * I 3 …OTHER OPTIONS...
12
I

I3
4
6k || 6k
4  12
FIRST REDUCE TO A SINGLE LOOP CIRCUIT
Vb  4k * I 4
OHM' S : I 2 
KCL : I 5  I 4  I 3  0
OHM' S : VC  3k * I 5
I1 
12V
12k
Va 
3
(12)
39
THE NODE ANALYSIS PERSPECTIVE
VS
 V1  V
a
KVL
 V3  V
b
KVL
THERE ARE FIVE NODES.
IF ONE NODE IS SELECTED AS
REFERENCE THEN THERE ARE
FOUR VOLTAGES WITH RESPECT
TO THE REFERENCE NODE
 V5  V
c
KVL
 Vc  V5  Vb  0
 VS  V1  Va  0  Va  V3  Vb  0
V1  VS  Va
V3  Va  Vb
V5  Vb  Vc
REFERENCE
WHAT IS THE PATTERN???
THEOREM: IF ALL NODE VOLTAGES WITH
RESPECT TO A COMMON REFERENCE NODE
ARE KNOWN THEN ONE CAN DETERMINE
ANY OTHER ELECTRICAL VARIABLE FOR
THE CIRCUIT
ONCE THE VOLTAGES ARE
KNOWN THE CURRENTS CAN
BE COMPUTED USING OHM’S
LAW
v R  vm  v N
 vR 
DRILL QUESTION
Vca  ______
A GENERAL VIEW
THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT
 vR

v R'
USING THE LEFT-RIGHT REFERENCE DIRECTION
THE VOLTAGE DROP ACROSS THE RESISTOR MUST
HAVE THE POLARITY SHOWN
v  vN
OHM' S LAW i  m
R
i  i
'
PASSIVE SIGN CONVENTION RULES!


i'
IF THE CURRENT REFERENCE DIRECTION IS
REVERSED ...
THE PASSIVE SIGN CONVENTION WILL ASSIGN
THE REVERSE REFERENCE POLARITY TO THE
VOLTAGE ACROSS THE RESISTOR
OHM' S LAW i ' 
v N  vm
R
DEFINING THE REFERENCE NODE IS VITAL
 V12 

4V


2V

THE STATEMENT V1  4V IS MEANINGLESS
UNTIL THE REFERENCE POINT IS DEFINED
BY CONVENTION THE GROUND SYMBOL
SPECIFIES THE REFERENCE POINT.
ALL NODE VOLTAGES ARE MEASURED WITH
RESPECT TO THAT REFERENCE POINT
V12  _____?
THE STRATEGY FOR NODE ANALYSIS
VS
Va
Vb
Vc
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)
@Va :  I1  I 2  I 3  0
Va  Vs Va Va  Vb


0
9k
6k
3k
@Vb :  I3  I 4  I5  0
REFERENCE
4. REPLACE CURRENTS IN TERMS OF
NODE VOLTAGES
AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
Vb  Va Vb Vb  Vc


0
3k
4k
9k
SHORTCUT: SKIP WRITING
THESE EQUATIONS...
@Vc :  I5  I 6  0
Vc  Vb Vc

0
9k
3k
AND PRACTICE WRITING
THESE DIRECTLY
WHEN WRITING A NODE EQUATION...
AT EACH NODE ONE CAN CHOSE ARBITRARY
DIRECTIONS FOR THE CURRENTS
a
Va
Vb
R1
b
R3
c
d
CURRENTS LEAVING  0
Va  Vb Vb  Vd Vb  Vc


0
R1
R2
R3
CURRENTS INTO NODE  0
I1  I 2  I 3  0 
I 3'
2
Vd
R3
c
Vc
I 2'
d
I2
AND SELECT ANY FORM OF KCL.
WHEN THE CURRENTS ARE REPLACED IN TERMS
OF THE NODE VOLTAGES THE NODE EQUATIONS
THAT RESULT ARE THE SAME OR EQUIVALENT
 I1  I 2  I 3  0  
b
I1' R
I3
R2
Vd

Va
R1
Vc
I1

a
Vb
Va  Vb Vb  Vd Vb  Vc


0
R1
R2
R3

CURRENTS LEAVING  0
I1'  I 2'  I 3'  0 

Vb  Va Vb  Vd Vc  Vb


0
R1
R2
R3
CURRENTS INTO NODE  0
 I1'  I 2'  I 3'  0  
Vb  Va Vb  Vd Vc  Vb


0
R1
R2
R3
WHEN WRITING THE NODE EQUATIONS
WRITE THE EQUATION DIRECTLY IN TERMS
OF THE NODE VOLTAGES.
BY DEFAULT USE KCL IN THE FORM
SUM-OF-CURRENTS-LEAVING = 0
THE REFERENCE DIRECTION FOR THE
CURRENTS DOES NOT AFFECT THE NODE
EQUATION
CIRCUITS WITH ONLY INDEPENDENT SOURCES
HINT: THE FORMAL MANIPULATION OF
EQUATIONS MAY BE SIMPLER IF ONE
USES CONDUCTANCES INSTEAD OF
RESISTANCES.
@ NODE 1
USING RESISTANCE S  i A 
v1 v1  v2

0
R1
R2
WITH CONDUCTANCES  i A  G1v1  G2 (v1  v2 )  0
REORDERING TERMS
@ NODE 2
REORDERING TERMS
THE MODEL FOR THE CIRCUIT IS A SYSTEM
OF ALGEBRAIC EQUATIONS
THE MANIPULATION OF SYSTEMS OF ALGEBRAIC
EQUATIONS CAN BE EFFICIENTLY DONE
USING MATRIX ANALYSIS
EXAMPLE
WRITE THE KCL EQUATIONS
@ NODE 1 WE VISUALIZE THE CURRENTS
LEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
 i2 
v2  v1 v2  v1

0
R4
R3
OR VISUALIZE CURRENTS GOING INTO NODE
ANOTHER EXAMPLE OF WRITING NODE EQUATIONS
V
BB
MARK THE NODES
(TO INSURE THAT
NONE IS MISSING)
15mA
A
VA
8k
2k
8k
2k
C
WRITE KCL AT EACH NODE IN TERMS OF
NODE VOLTAGES
SELECT AS
REFERENCE
VA VA

 15mA  0
2k 8k
V V
@ B B  B  15mA  0
8k 2k
@A
A MODEL IS SOLVED BY MANIPULATION OF
EQUATIONS AND USING MATRIX ANALYSIS
NUMERICAL MODEL
LEARNING EXAMPLE
USE GAUSSIAN ELIMINATION
i A  12mA , i B  4mA
THE NODE EQUATIONS
R1  12k, R2  R3  6k
ALTERNATIVE MANIPULATION
THE MODEL
* / 12k
REPLACE VALUES AND SWITCH NOTATION
TO UPPER CASE
RIGHT HAND
SIDE IS
VOLTS.
COEFFS ARE
NUMBERS
* / 6k
3V1  2V2  12
 V1  2V2  24 * / 3 (and add equations)
4V2  60[V ]
ADD EQS 2V  12[V ]
1
SOLUTION USING MATRIX ALGEBRA
PLACE IN MATRIX FORM
AND DO THE MATRIX ALGEBRA ...
USE MATRIX ANALYSIS TO SHOW SOLUTION
PERFORM THE MATRIX MANIPULATIONS
Adj ( A)
A 
| A|
1
FOR THE ADJOINT REPLACE
EACH ELEMENT BY ITS
COFACTOR
SAMPLE
2  1  10
V1  18k 

3k
3
 4  103 


6k

AN EXAMPLE OF NODE ANALYSIS
Rearranging terms ...
@v1
@ v2
COULD WRITE EQUATIONS BY INSPECTION
@ v3
  CONDUCTANCES CONNECTED TO NODE
  CONDUCTANCES BETWEEN 1 & 2
  CONDUCTANCES BETWEEN 1 & 3
  CONDUCTANCES BETWEEN 2 & 3
WRITING EQUATIONS “BY INSPECTION”
FOR CIRCUITS WITH ONLY INDEPENDENT
SOURCES THE MATRIX IS ALWAYS SYMMETRIC
THE DIAGONAL ELEMENTS ARE POSITIVE
THE OFF-DIAGONAL ELEMENTS ARE NEGATIVE
Conductances connected to node 1
Conductances between 1 and 2
Conductances between 1 and 3
Conductances between 2 and 3
VALID ONLY FOR CIRCUITS
WITHOUT DEPENDENT
SOURCES
LEARNING EXTENSION
V1 V1  V2 USING

6k
12k
V V V
@V2 : 2mA  2  2 1  0
6k
12k
@V1 :  4mA 
BY “INSPECTION”
1 
1
 1

V

V2  4mA

 1
12k
 6k 12k 
1  1
1 

 
V2  2mA
12k  6k 12k 
KCL
LEARNING EXTENSION
6mA
I3
I1
I2
Node analysis
V
@ V1 : 1  2mA  6mA  0  V1  16V
2k
V V
@V :  6mA  2  2  0  V2  12V
2
6k
IN MOST CASES THERE
ARE SEVERAL DIFFERENT
WAYS OF SOLVING A
PROBLEM
NODE EQS. BY INSPECTION
1
V1  0V2  2  6mA
2k
0V1   1  1 V2  6mA
 6k 3k 
3k
I 1  8mA
3k
I2 
(6mA)  2mA
3k  6k
6k
I3 
(6mA)  4mA
3k  6k
CURRENTS COULD BE COMPUTED DIRECTLY
USING KCL AND CURRENT DIVIDER!!
Once node voltages are known
I1 
V1
2k
I2 
V2
6k
I3 
V2
3k
CIRCUITS WITH DEPENDENT SOURCES
NUMERICAL EXAMPLE
LEARNING EXAMPLE
CIRCUITS WITH DEPENDENT SOURCES CANNOT
BE MODELED BY INSPECTION. THE SYMMETRY
IS LOST.
A PROCEDURE FOR MODELING
•WRITE THE NODE EQUATIONS USING DEPENDENT
SOURCES AS REGULAR SOURCES.
•FOR EACH DEPENDENT SOURCE WE ADD
ONE EQUATION EXPRESSING THE CONTROLLING
VARIABLE IN TERMS OF THE NODE VOLTAGES
1 
1 
 1
 2

v



 1 
v 2  0
12
k
6
k
3
k
6
k




1
1 
 1
 v1  
 v2  2mA
6k
 12k 3k 
* / 4k
* / 6k
V1  2V2  0
v v v
 io  1  1 2  0
R1
R2
v
v v
 iA  2  2 1  0
R3
R2
MODEL FOR
CONTROLLING VARIABLE
io 
v2
R3
 V1  3V2  12[V ]
REPLACE AND REARRANGE

1
1 
1 
  v1    v2  0
 R1 R2 
 R3 R2 
 1
1
1 
 v1    v2  i A
R2
 R2 R3 
ADDING THE EQUATIONS
V1  
24
V
5
5V2  12[V ]
LEARNING EXAMPLE: CIRCUIT WITH VOLTAGE-CONTROLLED CURRENT
REPLACE AND REARRANGE
CONTINUE WITH GAUSSIAN ELIMINATION...
WRITE NODE EQUATIONS. TREAT DEPENDENT
SOURCE AS REGULAR SOURCE
OR USE MATRIX ALGEBRA
EXPRESS CONTROLLING VARIABLE IN TERMS OF
NODE VOLTAGES
FOUR EQUATIONS IN OUR UNKNOWNS. SOLVE
USING FAVORITE TECHNIQUE
USING MATLAB TO SOLVE THE NODE EQUATIONS
R1  1k, R2  R3  2k,
R4  4k, i A  2mA, iB  4mA,
  2[ A / V ]
DEFINE THE COMPONENTS OF THE CIRCUIT
DEFINE THE MATRIX G
Entries in a row are
separated by commas
(or plain spaces).
Rows are separated by
semi colon
» R1=1000;R2=2000;R3=2000;
R4=4000; %resistances in Ohm
» iA=0.002;iB=0.004; %sources in Amps
» alpha=2; %gain of dependent source
» G=[(1/R1+1/R2), -1/R1, 0; %first row of the matrix
-1/R1, (1/R1+alpha+1/R2), -(alpha+1/R2); %second row
0, -1/R2, (1/R2+1/R4)], %third row. End in comma to have the echo
G=
0.0015 -0.0010
0
-0.0010 2.0015 -2.0005
0 -0.0005 0.0008
DEFINE RIGHT HAND SIDE VECTOR
» I=[iA;-iA;iB]; %end in ";" to skip echo
» V=G\I % end with carriage return and get the echo
SOLVE LINEAR EQUATION
V=
11.9940
15.9910
15.9940
LEARNING EXTENSION: FIND NODE VOLTAGES
REARRANGE AND MULTIPLY BY 10k
2V1  V2  40[V ] * / 2 and add eqs.
V1  2V2  0
5V1  80V  V1  16V
NODE EQUATIONS
V1
V V
 4mA  1 2  0
10k
10k
V V
V
@V2 : 2 1  2 IO  2  0
10k
10k
@V1 :
CONTROLLING VARIABLE (IN TERMS ON NODE
VOLTAGES)
IO 
V1
10k
REPLACE
V1
V V
 4mA  1 2  0
10k
10k
V2  V1
V
V
2 1  2 0
10k
10k 10k
V2  
V1
 V2  8V
2
FIND THE VOLTAGE VO
NODE EQUATIONS
LEARNING EXTENSION
NOTICE REPLACEMENT OF DEPENDENT SOURCE
IN TERMS OF NODE VOLTAGE
Vx Vx

 0 * / 6k
3k 6k
V
V
V
 x  O  O  0 * / 12k
6k 12k 12k
 2mA 
3Vx  12[V ]  Vx  4[V ]
2VO  2Vx  0  VO  4[V ]
CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES
3 nodes plus the reference. In
principle one needs 3 equations...
…but two nodes are connected to
the reference through voltage
sources. Hence those node
voltages are known!!!
…Only one KCL is necessary
Hint: Each voltage source
connected to the reference
node saves one node equation
V2 V2  V3 V2  V1


0
6k
12k
12k
V1  12[V ] THESE ARE THE REMAINING
V3  6[V ] TWO NODE EQUATIONS
SOLVING THE EQUATIONS
2V2  (V2  V3 )  (V2  V1 )  0
One more example ….
4V2  6[V ]  V2  1.5[V ]
Problem 3.67 (6th Ed) Find V_0
V4
I S1
V1
R1
IS2
R2
V2
 VO 
V3
R3
R4
IS3
+

VS 1
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k
Is1 =2mA, Is2 = 4mA, Is3 = 4mA,
Vs = 12 V
KNOWN NODE VOLTAGE
@V3 : V3  VVS  12[V ]
V V V
@V1 :  I S1  1 2  1  0
R1
R4
 2[mA ] 
IDENTIFY AND LABEL ALL NODES
V1  V2 V1

0
1k
2k
WRITE THE NODE EQUATIONS
@V2 :  I S 3 
NOW WE LOOK WHAT IS BEING
ASKED TO DECIDE THE SOLUTION
STRATEGY.
 4[mA ] 
V0  V1  V2
ONLY V1,V2 ARE NEEDED FOR VO
V2  V1 V2  V3 V2  V4


0
R1
R3
R2
V2  V1 V2  12 V2  V4


0
1k
1k
2k
@V4 : I S1  I S 2 
V4  V2
0
R2
V4  V2
2[mA ]  4[mA ] 
0
2k
TO SOLVE BY HAND ELIMINATE DENOMINATORS
*/2k
V1  V2 V1
 2[mA] 

0
1k
2k
V  V V  12 V2  V4
 4[mA]  2 1  2

0
1k
1k
2k
V4  V2
2[mA]  4[mA] 
0
2k
3V1  2V2  4[V ]
*/2k
*/2k
Add 2+3
ALTERNATIVE: USE LINEAR ALGEBRA
 3  2 0  V1   4 
 2 5  1 V   32

 2   
 0  1 1  V3   4 
(1)
 2V1  5V2  V4  32[V ]
 V2  V4  4[V ]
(2)
(3)
 2V1  4V2  36[V ]
3V1  2V2  4[V ] * / 2 and add
4V1  40[V ]  V1  10[V ]
4V2  56[V ]  V2  14[V ]
FINALLY!!
V0  V1  V2  4[V ]
So. What happens when sources are connected between two non
reference nodes?
THE SUPERNODE TECHNIQUE
We will use this example to introduce the concept of a SUPERNODE
SUPERNODE
IS
Conventional node analysis
requires all currents at a node
@V_1
@V_2
V
 6mA  1  I S  0
6k
V2
 I S  4mA 
0
12k
Efficient solution: enclose the
source, and all elements in
parallel, inside a surface.
Apply KCL to the surface!!!
 6mA 
V1 V2

 4mA  0
6k 12k
The source current is interior
to the surface and is not required
We STILL need one more equation
2 eqs, 3 unknowns...Panic!!
The current through the source is not
related to the voltage of the source
1
2
V  V  6[V ]
Math solution: add one equation
V1  V2  6[V ]
Only 2 eqs in two unknowns!!!
ALGEBRAIC DETAILS
The Equations
* / 12k
V1 V2
(1)

 6mA  4mA  0
6k 12k
(2) V1  V2  6[V ]
Solution
1. Eliminate denominato rs in Eq(1). Multiply by ...
2V1  V2  24[V ]
V1  V2  6[V ]
2. Add equations to eliminate V2
3V1  30[V ]  V1  10[V ]
3. Use Eq(2) to compute V2
V2  V1  6[V ]  4[V ]
Is2
FIND THE NODE VOLTAGES
AND THE POWER SUPPLIED
BY THE VOLTAGE SOURCE
R3 I
V1
V2

R1
VS
V
I s1
R2
R1  R2  10k, R3  4k
VS  20[V ], I s1  10[mA], I s 2  6[mA]
 V1  V2  20[V ]
V2 V1  20
V1
V
* / 10k  V1  V2  100[V ]
 2  10mA  0
10k 10k
adding : V2  60[V ]
V1  100  V2  40[V ]
TO COMPUTE THE POWER SUPPLIED BY VOLTAGE SOURCE
WE MUST KNOW THE CURRENT THROUGH IT
IV 
V1
V  V2
 6mA  1
 8mA
10k
10k
P  20[V ]  8[mA ]  160mW
BASED ON PASSIVE SIGN CONVENTION THE
POWER IS RECEIVED BY THE SOURCE!!
LEARNING EXAMPLE
WRITE THE NODE EQUATIONS
@v1
@ SUPERNODE
CONSTRAINT : v2  v3  v A
KCL (leaving supernode) :
THREE EQUATIONS IN THREE UNKNOWNS
SUPERNODE
LEARNING EXAMPLE
V3  12
FIND I O
V2  6V ,V4  12V
KNOWN NODE VOLTAGES
SUPERNODE CONSTRAINT  V1  V3  12
KCL @ SUPERNODE
LEARNING EXTENSION
SUPERNODE
V1  6V
V4  4V
SOURCES CONNECTED TO THE
REFERENCE
CONSTRAINT EQUATION
V3  V2  12V
KCL @ SUPERNODE
V2  6 V2 V3 V3  (4)
 

 0 * / 2k
2k
1k 2k
2k
V2 IS NOT NEEDED FOR IO 3V2  2V3  2V
 V2  V3  12V * / 3 and add
5V3  38V
V
OHM' S LAW I O  3  3.8mA
2k
Supernodes can be more complex
WRITE THE NODE EQUATIONS
supernode
V2
R4
R2
V1
+ +
-
KCL@V_3
V4
R7
V5
R3
V3  V2 V3  V4 V3


0
R4
R5
R7
KCL @SUPERNODE
(Careful not to omit any current)
R5
+
-
R1
V3
R6
V2  V1 V5  V1 V5 V4 V4  V3 V2  V3





0
R1
R2
R3 R6
R5
R4
CONSTRAINTS DUE TO VOLTAGE SOURCES
V1  VS1
Identify all nodes, select a
reference and label nodes
V2  V5  VS 2
Nodes connected to reference through
a voltage source
V5  V4  VS 3
Voltage sources in between nodes
and possible supernodes
EQUATION BOOKKEEPING:
KCL@ V_3, KCL@ supernode,
2 constraints equations
and one known node
5 EQUATIONS IN FIVE UNKNOWNS.
CIRCUITS WITH DEPENDENT SOURCES
PRESENT NO SIGNIFICANT ADDITIONAL
COMPLEXITY. THE DEPENDENT SOURCES
ARE TREATED AS REGULAR SOURCES
WE MUST ADD ONE EQUATION FOR EACH
CONTROLLING VARIABLE
LEARNING EXAMPLE
FIND IO
VOLTAGE SOURCE CONNECTED TO REFERENCE
V1  3V
V V V
KCL@ V2 : 2 1  2  2 I x  0
3k
6k
CONTROLLING VARIABLE IN
TERMS OF NODE VOLTAGES
Ix 
V2  V1 V2
V

 2 2  0 * / 6k
3k
6k
6k
V2  2V1  0  V2  6V
IO 
V1  V2
 1mA
3k
REPLACE
V2
6k
SUPER NODE WITH DEPENDENT SOURCE
VOLTAGE SOURCE CONNECTED TO REFERENCE
V3  6V
SUPERNODE CONSTRAINT
V1  V2  2Vx
CONTROLLING VARIABLE IN TERMS OF NODES
KCL AT SUPERNODE
Vx  V2  V1  3V2
* / 12k
2(V1  6)  V1  2V2  V2  6  0
3V1  3V2  18  4V1  18
CURRENT CONTROLLED VOLTAGE SOURCE
CONSTRAINT DUE TO SOURCE
V2  V1  2kI x
CONTROLLING VARIABLE IN TERMS OF NODES
V1
2k
V
V
SUPERNODE
 4mA  1  2mA  2  0
2k
2k
V1  V2  4[V ] * / 2 and add
 2V1  V2  0
3V2  8[V ]
 V1  2kI x  V2  2V1
KCL AT
IO 
V2 4
 mA
2k 3
Ix 
An example with dependent sources
2k
V1

VS

VX 3k
V2

2k

Ix
6k
1000 aI x
IDENTIFY AND LABEL NODES
2 nodes are connected to the
reference through voltage sources
V1  VS
V2  1000aI X
KCL @ Vx
V X  VS V X V X  v 2


0
2k
2k
3k
EXPRESS CONTROLLING VARIABLE IN
TERMS OF NODE VOLTAGES
VX
IX 
2k
‘a’ has units of [Volt/Amp]
REPLACE Ix IN V2
1k * aVX
V2 
2k
aV X
V2 
2
REPLACE V2 IN KCL
3(VX  VS )  3VX  2(VX  aVX / 2)  0
(8  a )V X  3VS
What happens when a=8?
LEARNING EXAMPLE
FIND THE VOLTAGE Vo
@V4 : V4  4V
AT SUPER NODE
V1  V2  2VX
V2 V2  V3 V1  V3 V1  4V

2
mA




0
1k 
1k
1k
1k
1k
1k  @V :  2mA  V  V  V  V  0
3
3
2
1k
CONTROLLING VARIABLE
3
1
1k
VX  V2
SOLVE EQUATIONS NOW
V1  3VX
2V1  2VX  V3  6V
V1  VX  2V3  2V
VARIABLE OF INTEREST
VO  V1  V3
LEARNING EXAMPLE
Find the current Io
@V2 : V2  12V
@V3 : V3  2VX
@ super node:
V4  V1  6V (constraint eq.)
V  V3 V4  V5 V4
V1  V2 V1  V3

 2I X  4


0
1k
1k
1k
1k
1k
FIND NODES – AND SUPER NODES
V5  V4 V5

0
1k
1k
CONTROLLING VARIABLES
VX  V1  V2
V
IX  4
7 eqs in 7 variables
1k
@V5 :  2 I X 
VARIABLE OF INTEREST
IO 
V5
1k