07-Nodal Analysis
Text: 3.1 – 3.4
ECEGR 210
Electric Circuits I
Overview
• Introduction
• Nodal Analysis
• Nodal Analysis with Voltage Sources
Dr. Louie
2
Introduction
• Basic Circuit Laws
 Ohm’s Law
 Kirchhoff’s Voltage Law (KVL)
 Kirchhoff’s Current Law (KCL)
• Now we seek to analyze any linear circuit
systematically using
 Nodal analysis
 Mesh analysis
Dr. Louie
3
Introduction
• Express circuit as a system of linear equations
• Solve linear equations





Substitution
Gaussian elimination
Cramer’s Rule
Numerical methods
Others
 y11 y12
y
 21 y22
y31 y32
y13   x1  b1 
y23  x2   b2 
y33   x3  b3 
• Remember: for a unique solution there must be
an equal number of independent equations as
unknowns
Dr. Louie
4
Introduction
• Let X be a vector of variables to be solved for
• Nodal Analysis:
 X: node voltages
 Y: conductances
 b: currents
 y11 y12
y
 21 y22
y31 y32
• Mesh Analysis:
 X: loop currents
 Y: resistances
 b: voltages
Dr. Louie
y13   x1  b1 
y23  x2   b2 
y33   x3  b3 
5
Nodal Analysis
Steps to solve a circuit with N nodes
1. Assign a reference node
2. Assign a variable to the voltages at all nodes wrt
the reference node (N-1 variables)
3. Apply KCL to each of the N-1 non-reference
nodes to generate N-1 equations
4. Use Ohm’s Law to express currents as functions
of node voltages
5. Solve resulting simultaneous equations
Dr. Louie
6
Nodal Analysis
• Let I1, I2, R1, R2, R3 be known (given)
• Solve for the voltages at each node
I2
R2
I1
R1
Dr. Louie
R3
7
Step 1: Assign Reference Node
• Recall: all voltage is a relative measure
• Reference node is also known as ground
 In power systems it is actually the ground (earth)
 In electronics it could be the metallic chassis
I2
 Or arbitrary
R2
I1
R1
R3
(usually the “bottom” of circuit)
Dr. Louie
8
Step 2: Assign Variables to Nodes
• There are two remaining nodes (variables)
• Assign voltages and polarities wrt reference node
• No need to write voltage across R2
 Does not add an independent equation (no new
information)
I2
 Can solve later as V1 – V2
I1
Dr. Louie
+
V1
-
x
R2
R1
x
+
R3 V2
-
9
Step 3: Apply KCL
• KCL gives one equation per node (two equations)
• I1 = I2 + i1 + i2
• I2 = i3 – i2
I2
i2
I1
i1
Dr. Louie
i3
10
Step 4: Apply Ohm’s Law
• Need to write i1, i2, i3 in terms of the node
voltages
• Remember sign convention
• i1 = (V1 - 0)/R1
I2
• i2 = (V1 – V2)/R2
• i3 = (V2 - 0)/R3
V1
x
I1
i2
i1
Dr. Louie
x V2
i3
11
Step 5: Solve Equations
• Recap
 two variables (V1, V2)
 two KCL equations
• KCL equations use intermediate variables i1, i2, i3
 I1 = I2 + i1 + i2
 I2 = i3 – i2
• Use substitution to express KCL in terms of
voltages
 i1 = (V1 - 0)/R1
 i2 = (V1 – V2)/R2
 i3 = (V2 - 0)/R3
Dr. Louie
12
Step 5: Solve Equations
• Via substitution:
 I1 = I2 + V1/R1 + (V1 – V2)/R2
 I2 = V2/R3 – (V1 – V2)/R2
• Can we solve this system of equations?
• Yes!
 Two independent equations, two unknowns
 The rest is just math
Dr. Louie
13
Step 5: Solve Equations
• In matrix form
 I1 = I2 + V1/R1 + (V1 – V2)/R2
 I2 = V2/R3 – (V1 – V2)/R2
1
1

R
R2
 1

1


R2

1
R2

  V  I  I 
  1   1 2
1
1   V2   I2 

R 3 R 2 

• Many methods of solving linear equations
Dr. Louie
14
Example
• Let





I1 = 10A
I2 = 5A
R1 = 6Ω
R2 = 4Ω
R3 = 2Ω
I2
• Find all voltages
I1
i2
i1
Dr. Louie
i3
15
Example
• Equations:
 I1 = I2 + V1/R1 + (V1 – V2)/R2
 I2 = V2/R3 – (V1 – V2)/R2
• Using circuit values:
 10 = 5 + V1/6 + (V1 – V2)/4
 5 = V2/2 – (V1 – V2)/4
Dr. Louie
16
Example
• Starting with the second equation
 5x4 = [V2/2 – (V1 – V2)/4]x4
Yields
20 = 2V2 – (V1 – V2) = 3V2 – V1
Now the first equation:
10x12 = [5 + V1/6 + (V1 – V2)/4]x12
120=60 + 2V1+ 3V1 – 3V2
60 = 5V1 – 3V2
Dr. Louie
17
Example
 20 = -V1 + 3V2
 60 = 5V1 – 3V2
 Solve using elimination
• 20x5 = [-V1 + 3V2]x5 = 100 = -5V1 + 15V2
60 = 5V1 – 3V2
160 = 0 +12V2
 Yields:
• V2 = 13.333V
• V1 = 20V
Dr. Louie
18
Example
• Solving in matrix form:
1
1

R
R2
 1

1


R2

1 1
6  4

 1
 4
1
R2

  V  I  I 
  1   1 2
1
1   V2   I2 

R 3 R 2 

1 
4   V1   5
   
1 1   V2  5

2 4 

Can use Matlab, Gaussian elimination
Cramer’s Rule, matrix inversion
Dr. Louie
19
Solution by Matrix Inversion
1 1
6  4

 1
 4
1 

4   V1   5
   
1 1   V2  5

2 4 
G (matrix)
 V1  5
G    
 V2  5
 V1 
 20 
1 5

G

V 
5 13.3
  

 2
1
where G
1 
3


1
1.67


For small matrices use
“inv()” command
in Matlab
Dr. Louie
20
Nodal Analysis with Voltage Sources
• Nodal analysis requires knowledge of current
through elements
• Recall that the current through a voltage source
is not readily known
• Example:
 Current through resistor is I = (Va – Vb)/R
 But what is the current through a voltage source?
How is the “R” in the equation to be interpreted?
V
+
-
+
-
Dr. Louie
21
Nodal Analysis with Voltage Sources
• If the voltage source is connected to the
reference node, then it can be easily included in
nodal analysis
I2
i2
I1
i1
Dr. Louie
+
-
V3
22
Nodal Analysis with Voltage Sources
• Only one node with unknown voltage
• I1 = I2 + V1/R1 + (V1 – V3)/R2
• Compare this to earlier example (2 eqns, 2 unknowns)
I2
I1
+
V1
-
x
R2
R1
Dr. Louie
+
-
V3
23
Nodal Analysis with Voltage Sources
• If voltage source not connected to reference:
 Move the reference so it is! (only works if there is
one voltage source)
Or
 Use Supernode concept
Dr. Louie
24
Nodal Analysis with Voltage Sources
I
I
-
I
+
node
-
I
+
• Supernode: a closed surface containing a voltage
source and its two nodes AND any elements in
parallel with it
Supernodes
• Current into supernode = current out of
supernode
• Also applies to dependent voltage sources
Dr. Louie
25
Nodal Analysis with Voltage Sources
• Properties of a supernode
 Voltage source inside the supernode provides a
constraint equation needed to solve for node
voltages
 A supernode has no voltage of its own
 There is only one KCL equation for the supernode,
but two unknown node voltages
 Additional equation is found from application of KVL
Dr. Louie
26
Nodal Analysis with Voltage Sources
• Find the node voltages
10W
2A
2W
-
+
2V
4W
7A
assume reference is here
Dr. Louie
27
Nodal Analysis with Voltage Sources
• Make a supernode out of the voltage source
• Write KCL for supernode
 2 = i1 + i2 + 7 (only one equation since it is a
supernode)
10W
2A
2W
-
+
2V
i1 4W
i2
7A
assume reference is here
Dr. Louie
28
Nodal Analysis with Voltage Sources
• Write i1, i2 in terms of node voltage
• i1 = (v1 – 0)/2
• i2 = (v2 – 0)/4
10W
2A
2W
-
+
v1
2V
i1 4W
v2
i2
7A
assume reference is here
Dr. Louie
29
Nodal Analysis with Voltage Sources
• Substitute into supernode KCL
• 2 = i1 + i2 + 7 (supernode KCL)
 2 = 0.5v1 + 0.25v2 + 7 (after substitution)
 8 = 2v1 + v2 +28 (multiplying by 4)
10W
2A
2W
-
+
v1
2V
i14W
Dr. Louie
v2
i2
7A
30
Nodal Analysis with Voltage Sources
• Apply KVL around loop
 -v1 + 2 + v2 = 0 (second equation)
10W
2A
2W
-
+
v1
2V
i14W
Dr. Louie
v2
i2
7A
31
Nodal Analysis with Voltage Sources
• How many independent equations?
 8 = 2v1 + v2 +28
 -v1 + 2 + v2 = 0
• How many unknowns?
 v1, v2
10W
2A
2W
-
v1
2V
+
Solve!
i14W
Dr. Louie
v2
i2
7A
32
Nodal Analysis with Voltage Sources
• Solving…
 8 = 2v1 + v1 - 2 +28
 v1 = -5.333V
 v2 = -7.333V
10W
2A
2W
-
+
v1
2V
i14W
Dr. Louie
v2
i2
7A
33
Example
• Find the node voltages in the circuit shown
-
+
10V
2W
2W
5A
4W
Dr. Louie
8W
34
Example
• 3 unknown voltages: V1, V2, V3
• Need three independent equations (that do not
introduce new variables)
2W
V1
-
+
10V
V2
2W
5A
4W
Dr. Louie
V3
8W
39
Example
• Identify the supernode
V1
-
+
10V
2W V
2
2W
5A
4W
Dr. Louie
V3
8W
40
Example
• Write KCL for the supernode
 i5 = i1+i4+i3 (supernode)
• Write KCL for the other node
 5+i4 = i5 (node 2)
V1
4W
-
+
10V
i4 2W V 2W i5
2
i1
5A i3
Dr. Louie
V3
8W
41
Example
i1
i3
i4
i5
=
=
=
=
(V1
(V3
(V1
(V2
–
–
–
–
0)/4
0)/8
V2)/2
V3)/2
+




V1
4W
Dr. Louie
-
10V
• Express currents in terms of voltages
i4 2W V 2W i5
2
i1
V3
5A i3
8W
42
Example
• Have two equations
• Need one more
independent equation
• Do KVL around top loop
+
 KCL for node 2, supernode
V1
4W
-
10V
i4 2W V 2W i5
2
i1
V3
5A i3
8W
 10 = 2i4 + 2i5
Dr. Louie
43
Example
• 5 + i4 = i5 (node 2)
• i6 = i1 + i5 + i3 (supernode)




i1
i3
i4
i5
=
=
=
=
(V1
(V3
(V1
(V2
–
–
–
–
0)/4 = 0.25V1
0)/8 = 0.125V3
V2)/2 = 0.5V1 – 0.5V2
V3)/2 = 0.5V2 – 0.5V3
5 + 0.5V1 – 0.5V2 = 0.5V2 – 0.5V3 (solving node 2 eqn)
10 + V1 – V2 = V2 – V3
0.5V2 - 0.5V3 = 0.25V1 + 0.5V1 – 0.5V2 + 0.125V3 (supernode eqn)
4V2 - 4V3 = 2V4 + 4V1 – 4V2 + V3
10 = 2i4 + 2i5 (KVL of top loop)
10 = V1 – V2 + V2 – V3
Dr. Louie
44
Example
10 + V1 – V2 = V2 + V3 (node 2)
10 = -V1 + 2V2 + V3 (after rearranging)
4V2 - 4V3 = 2V1 + 4V1 – 4V2 - V3 (supernode)
0 = 6V1 -8V2 + 3V3 (after rearranging)
10 = V1 – V2 + V2 – V3 (KVL of top loop)
10 = V1 + 0V2– V3 (after
Solving
 1 2
V1 = 12.22V
 6 8
V2 = 10V

 1 0
V3= 2.22V
rearranging)
1   V1  10
3   V2    0 
1  V3  10 
Dr. Louie
45
Example
• Find Vx
2A
10W
+
Vx
-
Dr. Louie
20W
0.2Vx
46
Example
• One node, write equation from KCL
 2 =I1 + I2 + 0.2Vx
• From Ohm’s Law
 I1 = Vx/10
 I2 = Vx/20
• Solving:
 2 = 0.1Vx + 0.05Vx + 0.2Vx
 Vx = 5.71V
2A
10W
I1
+
Vx
-
Dr. Louie
20W
I2
0.2Vx
47