Transcript NODE ANALYSIS - Carleton University

RESISTIVE CIRCUITS
•MULTI NODE/LOOP CIRCUIT ANALYSIS
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 :  I 3  I 4  I 5  0
Vb  Va Vb Vb  Vc


0
3k
4k
9k
@Vc :  I 5  I 6  0
Vc  Vb Vc

0
9k
3k
REFERENCE
4. REPLACE CURRENTS IN TERMS OF
NODE VOLTAGES
AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
SHORTCUT: SKIP WRITING
THESE EQUATIONS...
AND PRACTICE WRITING
THESE DIRECTLY
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
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
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
SUPERNODE
IS
Conventional analysis
requires all currents at a node
@V_1
@V_2
V
 6mA  1  I S  0
6k
V2
 I S  4mA 
0
12 k
Efficient solution: enclose the
source, and all elements in
parallel, inside a surface.
Apply KCL to the surface!!!
 6mA 
V1 V2

 4mA  0
6k 12 k
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 denominators 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 ]
Find Value of I O
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 I O 3V2  2V3  2V
 V2  V3  12V * / 3 and add
5V3  38V
V
OHM'S LAW I O  3  3.8mA
2k
Apply node analysis to this circuit
There are 4 non reference nodes
 VR2
V1  VR1 V2
+
-
R1
I
R2
 V3
+
-
R3
12V
 VR3 
There is one super node
18V
V4
There is one node connected to the
reference 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
STRATEGY:
1. Apply KVL
(sum of voltage drops =0)
 12[V ]  VR1  VR 2  18[V ]  VR 3  0
Skip this equation
2. Use Ohm’s Law to express
voltages in terms of the “loop current.”
12[V ]  R1I  R2 I  18[V ]  R3 I  0
RESULT IS ONE EQUATION IN THE LOOP CURRENT!!!
SHORTCUT
Write this one
directly
LOOPS, MESHES AND LOOP CURRENTS
a
1
2
I1
b
7
3
c
I2
e
d
f
6
5
A BASIC3 ICIRCUIT
EACH COMPONENT
IS CHARACTERIZED
4 BY ITS VOLTAGE
ACROSS AND ITS
CURRENT THROUGH
A LOOP IS A CLOSED PATH THAT DOES NOT
GO TWICE OVER ANY NODE.
THIS CIRCUIT HAS THREE LOOPS
CLAIM: IN A CIRCUIT, THE CURRENT THROUGH
ANY COMPONENT CAN BE EXPRESSED IN TERMS
OF THE LOOP CURRENTS
EXAMPLES
I a f   I1  I 3
I b e  I1  I 2
I b c  I 2  I3
FACT: NOT EVERY LOOP CURRENT IS REQUIRED
TO COMPUTE ALL THE CURRENTS THROUGH
COMPONENTS
a
fabef
ebcde
fabcdef
A MESH IS A LOOP THAT DOES NOT ENCLOSE
ANY OTHER LOOP.
fabef, ebcde ARE MESHES
A LOOP CURRENT IS A (FICTICIOUS) CURRENT
THAT IS ASSUMED TO FLOW AROUND A LOOP
I1 , I 2 , I 3 ARE LOOP CURRENTS
A MESH CURRENT IS A LOOP CURRENT
ASSOCIATED TO A MESH. I1, I2 ARE MESH
CURRENTS
THE DIRECTION OF THE LOOP
CURRENTS IS SIGNIFICANT
1
2

I1
b
3
USING TWO
LOOP CURRENTS
c

7
4

Ia f   I 1  I3

e
d
f
6
5
A BASIC CIRCUIT 
I3
Ib e  I 1

Ib c  I 3
FOR EVERY CIRCUIT THERE IS A MINIMUM
NUMBER OF LOOP CURRENTS THAT ARE
NECESSARY TO COMPUTE EVERY CURRENT
IN THE CIRCUIT.
SUCH A COLLECTION IS CALLED A MINIMAL
SET (OF LOOP CURRENTS).
DETERMINATION OF LOOP CURRENTS
FOR A GIVEN CIRCUIT LET
B
NUMBER OF BRANCHES
N
NUMBER OF NODES
THE MINIMUM REQUIRED NUMBER OF
LOOP CURRENTS IS
L  B  ( N  1)
MESH CURRENTS ARE ALWAYS INDEPENDENT
KVL ON LEFT MESH
KVL ON RIGHT MESH
v S 2  v4  v5  v 3  0
USING OHM’S LAW
v1  i1 R1 , v 2  i1 R2 , v 3  ( i1  i2 ) R3
AN EXAMPLE
v4  i2 R4 , v5  i2 R5
REPLACING AND REARRANGING
B7
N 6
L  7  (6  1)  2
TWO LOOP CURRENTS ARE
REQUIRED.
THE CURRENTS SHOWN ARE
MESH CURRENTS. HENCE
THEY ARE INDEPENDENT AND
FORM A MINIMAL SET
DEVELOPING A SHORTCUT
WRITE THE MESH EQUATIONS
V2
R1
+ -
V1
+
-
I1
R5
R2
I2
R3
WHENEVER AN ELEMENT
HAS MORE THAN ONE
LOOP CURRENT FLOWING
THROUGH IT WE COMPUTE
NET CURRENT IN THE
DIRECTION OF TRAVEL
R4
DRAW THE MESH CURRENTS. ORIENTATION
CAN BE ARBITRARY. BUT BY CONVENTION
THEY ARE DEFINED CLOCKWISE
NOW WRITE KVL FOR EACH MESH AND APPLY
OHM’S LAW TO EVERY RESISTOR.
AT EACH LOOP FOLLOW THE PASSIVE SIGN
CONVENTION USING LOOP CURRENT REFERENCE
DIRECTION
 V1  I1R1  ( I1  I 2 ) R2  I1R5  0
V2  I 2 R3  I 2 R4  ( I 2  I1 ) R2  0
EXAMPLE: FIND Io
AN ALTERNATIVE SELECTION OF LOOP CURRENTS
SHORTCUT: POLARITIES ARE NOT NEEDED.
APPLY OHM’S LAW TO EACH ELEMENT AS KVL
IS BEING WRITTEN
KVL @ I1
KVL @ I2
REARRANGE
KVL @ I1
KVL @ I2
12kI1  6kI 2  12
 6kI1  9kI 2  3 * / 2 and add
12kI 2  6  I 2  0.5mA
5
12kI1  12  6kI 2  I1  mA
4
EXPRESS VARIABLE OF INTEREST AS FUNCTION
OF LOOP CURRENTS
I O  I1  I 2
NOW I O  I1
THIS SELECTION IS MORE EFFICIENT
REARRANGE
12kI1  6kI 2  12 * / 3
6kI1  9kI 2  9 * / 2 and substract
3
24kI1  18  I1  mA
4
1. DRAW THE MESH CURRENTS
I1
I2
2. WRITE MESH EQUATIONS
MESH 1
(2k  4k  2k ) I1  2kI 2  3[V ]
MESH 2
 2kI1  (2k  6k ) I 2  (6V  3V )
DIVIDE BY 1k. GET NUMBERS FOR COEFFICIENTS
ON THE LEFT AND mA ON THE RHS
3. SOLVE EQUATIONS
8 I1  2 I 2  3[mA ]
 2 I1  8 I 2  9[mA ] * / 4 and add
33
30 I 2  33[mA ]
VO  6kI 2  [V ]
5
KVL
THERE IS NO RELATIONSHIP BETWEEN V1 AND
THE SOURCE CURRENT! HOWEVER ...
MESH 1 CURRENT IS CONSTRAINED
MESH 1 EQUATION
I1  2mA
CURRENT SOURCES THAT ARE NOT SHARED
BY OTHER MESHES (OR LOOPS) SERVE TO
DEFINE A MESH (LOOP) CURRENT AND
REDUCE THE NUMBER OF REQUIRED EQUATIONS
MESH 2
 2kI1  8kI 2  2V
2k  (2mA)  2V 3
9
I2 
 mA  VO  6kI 2  [V ]
8k
4
2
“BY INSPECTION”
TO OBTAIN V1 APPLY KVL TO ANY CLOSED
PATH THAT INCLUDES V1
CURRENT SOURCES SHARED BY LOOPS - THE SUPERMESH APPROACH
2. WRITE CONSTRAINT EQUATION DUE TO
MESH CURRENTS SHARING CURRENT SOURCES
I 2  I 3  4mA
3. WRITE EQUATIONS FOR THE OTHER MESHES
I1  2mA
4. DEFINE A SUPERMESH BY (MENTALLY)
REMOVING THE SHARED CURRENT SOURCE
5. WRITE KVL FOR THE SUPERMESH
 6  1kI3  2kI 2  2k ( I 2  I1 )  1k ( I 3  I1 )  0
1. SELECT MESH CURRENTS
SUPERMESH
NOW WE HAVE THREE EQUATIONS IN THREE
UNKNOWNS. THE MODEL IS COMPLETE
FIND VOLTAGESACROSSRESISTORS
I1
I S1

V4


V
R2

R1
 V1 
R4
I3
I2
2
IS2
 V3 
HINT: IF ALL CURRENT SOURCES ARE REMOVED
THERE IS ONLY ONE LOOP LEFT
MESH EQUATIONS FOR LOOPS WITH
CURRENT SOURCES
I1  I s1
R3
I S3 I
4
Now we need a loop current that does
not go over any current source and
passes through all unused components.
+
VS
I2  IS 2
I3  I S 3
KVL OF REMAINING LOOP
VS  R3 ( I 4  I 2 )  R1 ( I 4  I 3  I1 )  R4 ( I 4  I3 )  0
For loop analysis we notice...
Three independent current sources.
Four meshes.
One current source shared by two
meshes.
Careful choice of loop currents
should make only one loop equation
necessary. Three loop currents can
be chosen using meshes and not
sharing any source.
SOLVE FOR THE CURRENT I4.
USE OHM’S LAW TO C0MPUTE REQUIRED
VOLTAGES
V1  R1 ( I1  I 3  I 4 )
V2  R2 ( I 2  I1 )
V3  R3 ( I 2  I 4 )