This document discusses node analysis, which is a method to determine all voltages and currents in a circuit. It involves selecting a reference node and writing Kirchhoff's Current Law (KCL) equations at each other node in terms of the node voltages. Once the node voltages are solved, the currents can be determined using Ohm's Law. The document provides examples of writing KCL equations at nodes and using techniques like super nodes to solve circuits with dependent sources.

1. 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

2. 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

3. SECOND: “BACKTRACK” USING KVL, KCL OHM’S
k
V
I a
6
2 
:
S
OHM' 0
3
2
1 

 I
I
I
:
KCL
3
*
3 I
k
Vb 
:
S
OHM' …OTHER OPTIONS...
4
3
4
*
4
12
4
12
I
k
V
I
I
b 


5
3
4
5
*
3
0
I
k
V
I
I
I
C 



:
S
OHM'
:
KCL
k
12
k
k 12
||
4
k
6
k
k 6
||
6
k
V
I
12
12
1 
)
12
(
9
3
3


a
V
3
I
FIRST REDUCE TO A SINGLE LOOP CIRCUIT

4. THE NODE ANALYSIS PERSPECTIVE
THERE ARE FIVE NODES.
IF ONE NODE IS SELECTED AS
REFERENCE THEN THERE ARE
FOUR VOLTAGES WITH RESPECT
TO THE REFERENCE NODE
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
a
S
a
S
V
V
V
V
V
V






1
1
0
b
a
b
a
V
V
V
V
V
V






3
3 0
______

ca
V
QUESTION
DRILL
REFERENCE
S
V a
V b
V c
V

 1
V
KVL

 3
V
KVL

 5
V
WHAT IS THE PATTERN???
KVL
0
5 


 b
c V
V
V
ONCE THE VOLTAGES ARE
KNOWN THE CURRENTS CAN
BE COMPUTED USING OHM’S
LAW
A GENERAL VIEW

 R
v
N
m
R v
v
v 

5 b c
V V V
 

5. THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT
'
i

 R
v
USING THE LEFT-RIGHT REFERENCE DIRECTION
THE VOLTAGE DROP ACROSS THE RESISTOR MUST
HAVE THE POLARITY SHOWN
R
v
v
i N
m 

LAW
S
OHM'
IF THE CURRENT REFERENCE DIRECTION IS
REVERSED ...

 '
R
v
THE PASSIVE SIGN CONVENTION WILL ASSIGN
THE REVERSE REFERENCE POLARITY TO THE
VOLTAGE ACROSS THE RESISTOR
R
v
v
i m
N 

'
LAW
S
OHM'
'
i
i 

PASSIVE SIGN CONVENTION RULES!

6. DEFINING THE REFERENCE NODE IS VITAL
S
MEANINGLES
IS
4V
V
STATEMENT
THE 1 
UNTIL THE REFERENCE POINT IS DEFINED
BY CONVENTION THE GROUND SYMBOL
SPECIFIES THE REFERENCE POINT.


V
4
ALL NODE VOLTAGES ARE MEASURED WITH
RESPECT TO THAT REFERENCE POINT


V
2
_____?
12 
V

 12
V

7. 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
:
@ 3
2
1



 I
I
I
Va
4. REPLACE CURRENTS IN TERMS OF
NODE VOLTAGES
0
3
6
9





k
V
V
k
V
k
V
V b
a
a
s
a
AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
REFERENCE
S
V a
V b
V c
V
0
:
@ 5
4
3



 I
I
I
Vb
0
:
@ 6
5


 I
I
Vc
0
9
4
3





k
V
V
k
V
k
V
V c
b
b
a
b
0
3
9



k
V
k
V
V c
b
c
SHORTCUT: SKIP WRITING
THESE EQUATIONS...
AND PRACTICE WRITING
THESE DIRECTLY

8. a b c
d
a
V
b
V
c
V
d
V
1
R 3
R
2
R
WHEN WRITING A NODE EQUATION...
AT EACH NODE ONE CAN CHOSE ARBITRARY
DIRECTIONS FOR THE CURRENTS
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
0
0
3
2
1
3
2
1 













R
V
V
R
V
V
R
V
V
I
I
I c
b
d
b
b
a
0
LEAVING
CURRENTS
0
0
3
2
1
3
2
1 











R
V
V
R
V
V
R
V
V
I
I
I c
b
d
b
b
a
0
NODE
INTO
CURRENTS
2
I
3
I
1
I
a b c
d
a
V
b
V
c
V
d
V
1
R 3
R
2
R '
2
I
'
3
I
'
1
I
0
0
3
2
1
'
3
'
2
'
1 











R
V
V
R
V
V
R
V
V
I
I
I b
c
d
b
a
b
0
LEAVING
CURRENTS
0
0
3
2
1
'
3
'
2
'
1 













R
V
V
R
V
V
R
V
V
I
I
I b
c
d
b
a
b
0
NODE
INTO
CURRENTS
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

9. CIRCUITS WITH ONLY INDEPENDENT SOURCES
HINT: THE FORMAL MANIPULATION OF
EQUATIONS MAY BE SIMPLER IF ONE
USES CONDUCTANCES INSTEAD OF
RESISTANCES.
@ NODE 1
0
2
2
1
1
1





R
v
v
R
v
iA
S
RESISTANCE
USING
0
)
( 2
1
2
1
1 



 v
v
G
v
G
iA
ES
CONDUCTANC
WITH
REORDERING TERMS
@ NODE 2
REORDERING TERMS
THE MODEL FOR THE CIRCUIT IS A SYSTEM
OF ALGEBRAIC EQUATIONS

10. EXAMPLE
@ NODE 1 WE VISUALIZE THE CURRENTS
LEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
0
3
1
2
4
1
2
2 





R
v
v
R
v
v
i
OR VISUALIZE CURRENTS GOING INTO NODE
WRITE THE KCL EQUATIONS

11. mA
15
A
B
C
k
8 k
8
k
2 k
2
SELECT AS
REFERENCE
A
V
B
V
MARK THE NODES
(TO INSURE THAT
NONE IS MISSING)
ANOTHER EXAMPLE OF WRITING NODE EQUATIONS
WRITE KCL AT EACH NODE IN TERMS OF
NODE VOLTAGES 0
15
8
2
@ 

 mA
k
V
k
V
A A
A
0
15
2
8
@ 

 mA
k
V
k
V
B B
B

12. k
V
V
k
V
mA
V
12
6
4
:
@ 2
1
1
1



 USING KCL
0
12
6
2
:
@ 1
2
2
2 



k
V
V
k
V
mA
V
BY “INSPECTION”
mA
V
k
V
k
k
4
12
1
12
1
6
1
2
1 








mA
V
k
k
k
2
12
1
6
1
12
1
2 










LEARNING EXTENSION

13. mA
6
1
I
2
I
3
I




 0
3
6
6
:
@ 2
2
2
k
V
k
V
mA
V 2
12
V V

CURRENTS COULD BE COMPUTED DIRECTLY
USING KCL AND CURRENT DIVIDER!!
1
2
3
8
3
(6 ) 2
3 6
6
(6 ) 4
3 6
I mA
k
I mA mA
k k
k
I mA mA
k k
 
 

 

LEARNING EXTENSION
IN MOST CASES THERE
ARE SEVERAL DIFFERENT
WAYS OF SOLVING A
PROBLEM
NODE EQS. BY INSPECTION
   mA
V
V
k
6
2
0
2
1
2
1 



  mA
V
k
k
V 6
3
1
6
1
0 2
1 








1
16
V V
  
k
V
I
k
V
I
k
V
I
3
6
2
2
3
2
2
1
1 


Once node voltages are known
1
1
@ : 2 6 0
2
V
V mA mA
k
  
Node analysis

14. 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
0
12
12
6
1
2
3
2
2





k
V
V
k
V
V
k
V
]
[
5
.
1
]
[
6
4
0
)
(
)
(
2
2
2
1
2
3
2
2
V
V
V
V
V
V
V
V
V








EQUATIONS
THE
SOLVING
Hint: Each voltage source
connected to the reference
node saves one node equation
CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES
]
[
6
]
[
12
3
1
V
V
V
V


 THESE ARE THE REMAINING
TWO NODE EQUATIONS

15. We will use this example to introduce the concept of a SUPERNODE
Conventional node analysis
requires all currents at a node
@V_1 0
6
6 1



 S
I
k
V
mA
@V_2 0
12
4 2




k
V
mA
IS
2 eqs, 3 unknowns...Panic!!
The current through the source is not
related to the voltage of the source
Math solution: add one equation
]
[
6
2
1 V
V
V 

Efficient solution: enclose the
source, and all elements in
parallel, inside a surface.
Apply KCL to the surface!!!
0
4
12
6
6 2
1




 mA
k
V
k
V
mA
The source current is interior
to the surface and is not required
We STILL need one more equation
]
[
6
2
1 V
V
V 

Only 2 eqs in two unknowns!!!
SUPERNODE
THE SUPERNODE TECHNIQUE
S
I

16. k
12
/
*
]
[
6
0
4
6
12
6
2
1
2
1
V
V
V
mA
mA
k
V
k
V






(2)
(1)
Equations
The
ALGEBRAIC DETAILS
...
by
Multiply
Eq(1).
in
rs
denominato
Eliminate
1.
Solution
]
[
6
]
[
24
2
2
1
2
1
V
V
V
V
V
V




]
[
10
]
[
30
3 1
1 V
V
V
V 


2
V
eliminate
to
equations
Add
2.
]
[
4
]
[
6
1
2 V
V
V
V 


2
V
compute
to
Eq(2)
Use
3.

18. V
V
V
V
4
6
4
1


 SOURCES CONNECTED TO THE
REFERENCE
SUPERNODE
CONSTRAINT EQUATION V
V
V 12
2
3


KCL @ SUPERNODE
0
2
)
4
(
2
1
2
6 3
3
2
2







k
V
k
V
k
V
k
V
k
2
/
*
V
V
V 2
2
3 3
2


V
V
V 12
3
2


 add
and
3
/
*
V
V 38
5 3 
mA
k
V
IO
8
.
3
2
3


LAW
S
OHM'
O
I
V FOR
NEEDED
NOT
IS
2