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
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
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 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
4V2  6[V ]  V2  1.5[V ]
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 ]
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