EGR 2201 Unit 3
Nodal Analysis



Read Alexander & Sadiku, Appendix A
and Sections 3.1 to 3.3.
Homework #3 and Lab #3 due next
week.
Quiz next week.
A Systematic Approach



With the laws from Chapter 2 (Ohm’s
law, KCL, and KVL), we can analyze a
wide variety of circuits containing
resistors and sources.
However, those laws do not constitute
a systematic method for analyzing
circuits. Different people may apply the
laws in different orders.
We want a systematic, step-by-step
method. Nodal analysis is one such
method.

Mesh analysis, which we’ll study next
week, is another one.
Simultaneous Linear Equations


Nodal analysis and mesh analysis both
involve writing down a set of equations,
based on an inspection of the circuit
you’re analyzing, and then solving
those equations simultaneously.
In your math classes you have studied
how to solve simultaneous equations,
so let’s start with a review of the math.
Math Review: Solving
Simultaneous Linear Equations

Recall that a linear equation in one
variable has a unique solution.
 Example: The linear equation
2x = 8

has a unique solution, namely x = 4.
In contrast, a linear equation in two
variables does not have a unique
solution.
 Example: The linear equation
3x  4y = 7
does not have a unique solution.
Simultaneous Linear Equations

Although a single linear equation in two
variables doesn’t have a unique
solution, a pair of independent linear
equations in two variables does have a
unique solution.
 Example: The pair of equations
3x  4y = 7
and
2x + 8y = 26
has a unique solution, namely x = 5 and
y = 2.
Generalizing

More generally, for any positive integer
n, a set of n independent linear
equations in n variables does have a
unique solution.
 Example of a set of three equations in
three variables:
3x  y  2z = 1
x + 6y  3z = 0
2x  3y + 6z = 6
 How do we find the solution?
Many Methods for Solving
Simultaneous Equations

Mathematicians have found many ways
of attacking this problem:
1.
2.
3.
4.
5.
Substitution
Gaussian elimination
Cramer’s rule
Matrix inversion
…
How Many Equations Do You
Have?
In theory, there’s no limit to the size of
the problems that you can attack with
the methods listed on the previous
slide.



Example: Eight equations in eight variables.
But from a practical standpoint, all of
these methods are cumbersome if you
have more than three equations in three
variables. In such cases, you can use a
powerful calculator (such as a TI-89) or
software (such as MATLAB).
Matrices
We’ll focus on the methods that use
Cramer’s rule and matrix inversion.
Both of these methods rely on matrices
and determinants, so let’s do a quick
review of those topics.



The textbook’s Appendix A also gives a nice
review.
Subscripted Variables


When working with matrices, we
generally prefer to use variable names
like x1, x2, x3,… rather than x, y, z,….
Rewriting our earlier set of two
equations in two variables:
3x1  4x2 = 7
2x1 + 8x2 = 26
A 2-by-2 Matrix Equation


Suppose we have two equations in
two variables:
a11x1 + a12x2 = b1
a21x1 + a22x2 = b2
We can write this in matrix form as
 a11 a12   x1   b1 
 
a



 21 a22   x2  b2 
or
AX=B
Example of a 2-by-2 Matrix Equation


Using our two equations in two
variables from earlier:
3x1  4x2 = 7
2x1 + 8x2 = 26
We can write this in matrix form as
3  4  x1   7 
2 8   x   26

 2   
or
AX=B
Solution Using Cramer’s Rule

Cramer’s rule says that the solution of
the matrix equation A X = B is given
by:
n
1
2
x1  , x2 
, ... , xn 



where  is the determinant of A, and i
is the determinant of the matrix formed
by replacing the ith column of A by B.
Determinant of a 2-by-2 Matrix

Suppose we have a 2-by-2 matrix A:
 a11 a12 
A

a21 a22 

This matrix’s determinant  is given
by:
 = a11a22  a12a21
Example of the Determinant of a 2by-2 Matrix

In the earlier example, our 2-by-2
matrix A was:
 3  4
A

2 8 

This matrix’s determinant  is equal to
32, since :
 = 3×8  (4×2) = 24  (8) = 32
Example of Solution Using Cramer’s
Rule

Continuing with our 2-by-2 example:
3  4  x1   7 
2 8   x   26

 2   





We’ve just seen that  = 32 for the matrix
above.
 7  4
Also, 1 is the determinant of 

26
8


So 1 = 160.
3 7 
And 2 is the determinant of 

2
26


So 2 = 64.
Example of Solution Using Cramer’s
Rule (Continued)


We’ve found that, for our example,
 = 32, 1 = 160, and 2 = 64.
Therefore, Cramer’s rule tells us that
1 168
x1 

5

32
 2 64
x2 

2
 32

This agrees with our earlier solution.
A 3-by-3 Matrix Equation


Suppose we have three equations in three
variables:
a11x1 + a12x2 + a13x3 = b1
a21x1 + a22x2 + a23x3 = b2
a31x1 + a32x2 + a33x3 = b3
We can write this in matrix form as
 a11
a
 21
a31
a12
a22
a32
a13   x1   b1 
a23   x2   b2 
a33   x3  b3 
or
AX=B
Example of a 3-by-3 Matrix Equation


Consider the following three equations in
three variables:
3x1  x2  2x3 = 1
x1 + 6x2  3x3 = 0
2x1  3x2 + 6x3 = 6
We can write this in matrix form as
 3  1  2  x1  1
  1 6  3   x   0 

  2  
 2  3 6   x3  6
or
AX=B
Determinant of a 3-by-3 Matrix

Suppose we have a 3-by-3 matrix A:
 a11
A  a21
a31


a12
a22
a32
a13 
a23 
a33 
This matrix’s determinant is given by:
a11a22a33 + a21a32a13 + a31a12a23 
a13a22a31  a23a32a11  a33a12a21
See page A-2 for a way to remember
this.
How Much of This Do You Need to
Know?


On exams, I’ll expect you to be able to
use Cramer’s rule or matrix inversion to
solve a problem involving two equations
in two variables—in other words,
involving a 2-by-2 matrix. You’ll need
to show every step in the solution.
But for more complicated exam
problems, you can use a calculator or
MATLAB to solve the equations.
Solution Using Matrix Inversion



Suppose again that we have the
matrix equation
AX=B
We can solve for X by rewriting this
equation as
X = A-1 B
where A-1 is a matrix called the
inverse of A.
The question now is: How do we
find this inverse matrix A-1?
Inverse of a 2-by-2 Matrix

Suppose we have a matrix A given by
 a11 a12 
A

a21 a22 

This matrix’s inverse is given by
1
A 
A
1
 a22  a12 
 a

 21 a11 
where |A| is the determinant of A.
Example of the Inverse of a 2-by-2
Matrix

In the earlier example, our 2-by-2
matrix A was:
 3  4
A

2 8 


And we found that this matrix’s
determinant |A| is equal to 32.
Therefore this matrix’s inverse is
0.125 
1  8 4  0.25
A 




32  2 3  0.0625 0.09375
-1
Example of Solution Using Matrix
Inversion

Now that we’ve found A-1, we write
X = A-1 B, or
0.125  7  5 
 0.25
X
 



 0.0625 0.09375 26 2


(That’s because
0.25 × 7 + 0.125 × 26 = 5
0.0625 × 7 + 0.09375 × 26 = 2)
So x1 = 5 and x2 = 2, in agreement with
our earlier solutions.
Inverse of a 3-by-3 Matrix?
If you have three equations in three
variables, you’ll need to find the inverse
of at 3-by-3 matrix.



This gets a little ugly. See page A-6 in the
textbook.
If you have four (or more) equations in
four (or more) variables, don’t try it by
hand. Use a calculator or MATLAB.
A Quick Introduction to MATLAB
The main window
is the Command
Window, in which
you type commands
at the >> prompt.
Examples:









>>
>>
>>
>>
>>
>>
>>
3*5
3+5^2
(3+5)^2
a=10
b=2
c=a/b
sqrt(36)
A Few MATLAB Tips



You can’t move up to a previous line to
re-execute a command, but you can use the
up-arrow and down-arrow keys to retrieve old
commands.
The clc command clears the command window.
MATLAB has many built-in math functions,
Function
Description
including:
sqrt(x)
Square root
exp(x)
Exponential (ex)
abs(x)
Absolute value
log(x)
Natural logarithm
log10(x)
Base-10 logarithm
cos(x)
Cosine (x in radians)
cosd(x)
Cosine (x in degrees)
Solving Our 2-by-2 Example with
MATLAB


First, define the coefficient matrix A and
the vector of constants B:
>> A = [3 -4; 2 8]
>> B = [7; 26]
Second, multiply A’s inverse times B:
>> X = inv(A)*B
Solving Our 3-by-3 Example with
MATLAB


First, define the coefficient matrix A and
the vector of constants B:
>> A = [3 -1 -2; -1 6 -3; -2 -3 6]
>> B = [1; 0; 6]
Second, multiply A’s inverse times B:
>> X = inv(A)*B
Simplifying Before You Can Use
MATLAB


Often you’ll have a set of equations that you
must simplify using basic algebra to put it into
a form that you can enter in MATLAB.
𝑥1 𝑥1 − 𝑥2
Example:
𝑥1 − 10 +
5
+
2
=0
𝑥2 − 𝑥1 𝑥2
+
−2=0
2
10

This simplifies to:
17𝑥1 − 5𝑥2 = 100
−5𝑥1 + 6𝑥2 = 20
Nodal Analysis


We’ve finished our review of the math.
Now let’s see how to use the math to
analyze circuits.
Nodal analysis, which is based on
Kirchhoff’s current law (KCL), is a
process for finding the voltages at the
nodes in a circuit.

Once we’ve found these node voltages, we
can use KVL to find any other voltages in
the circuit and use Ohm’s law to find any
currents in the circuit.
Reference Node



To perform nodal analysis on a circuit,
we must first select one of the circuit’s
nodes to be the reference node.
This means that we’ll only talk about
voltages that are relative to this node.
Thinking in terms of a multimeter, this
means that we’ll keep the meter’s
black lead at the reference node when
making any voltage measurements.

See the next two slides for a picture….
Voltage Measurements without a
Reference Node


In Lab #1 you built
this three-node
circuit and then
measured several
voltages: vab, vac,
vbc, vba, vcb, and vca.
Recall that to measure any of these
voltages, you touched the meter’s red
lead to the node named first and the
black lead to the node named second.

Example: To measure vac, you touched
the meter’s red lead to node a and the
black lead to node c.
Voltage Measurements with a
Reference Node

When doing nodal analysis,
we’ll single out one node as
the reference node and
then measure all voltages
relative to that node.


Example: Suppose that in this circuit, we
select node c as the reference node. Then
we’ll keep the meter’s black lead on node c
for all voltage measurements.
We can therefore drop the letter c from the
names of the voltages.
 Instead of writing vac, we’ll just write va.
 And instead of vbc, we’ll write vb.
Schematic Symbol for the
Reference Node


In schematic diagrams, the symbol
identifies the reference node.
So if we’ve selected node c in the
previous circuit as the reference node,
we can redraw the circuit as shown
here. (And we no longer label the
reference node with a letter.)
“Datum Node” and “Ground”



Two other names for the reference
node are datum node and ground.
“Ground” is a confusing term that is
used in several different ways,
including “chassis ground” and “earth
ground.”
Also, there are several different ground
symbols, which different authors use in
different ways.
We’ll always use this one.
How to Select the Reference
Node?


In many of the textbook’s problems, the
authors have already selected a reference
node and identified it with the
symbol.
If not, you should select the node
connected to the most branches as your
reference node. This is often the lowest
node in the diagram, as in the two cases
below.
Steps in Performing Nodal Analysis on
a Circuit with No Voltage Sources

Given a circuit with n nodes, with no
voltage sources, follow these steps:
1.
2.
3.
Select a node as the reference node.
Assign voltages v1, v2, …, vn-1 to the
remaining n-1 nodes. These voltages are
relative to the reference node.
Apply KCL to each of the n-1 nonreference nodes. Use Ohm’s law to express
the branch currents in terms of node
voltages. Then simplify the equations.
Solve the resulting n-1 simultaneous
equations to obtain the unknown node
voltages.
Example: Step 1 (Identify the
Nodes)


Consider this circuit
from the book’s
Example 3.1.
Step 1 has
already been
performed for
us, since the
reference node is identified and the
other two nodes are labeled 1 and 2.
We’ll call the voltages at these nodes
v1 and v2.
Example: Step 2 (Apply KCL & Ohm’s
Law)
Part 1 of 4


Voltages and
currents (with
assumed
directions) are
labeled here.
If one or more
of our assumed
current directions
are wrong, that’s no problem. The
math will still work out just fine.
Example: Step 2 (Apply KCL & Ohm’s
Law)
Part 2 of 4



Apply KCL at node 1:
i1 = i2 + i 3
Apply KCL at node 2:
i2 + i4 = i 1 + i5
Note that we
already know the values of some currents
(i1 and i4), but not of the others.
Example: Step 2 (Apply KCL & Ohm’s
Law)
Part 3 of 4


Trickiest part!
Rewrite the KCL
equations using
Ohm’s law and the
known currents.
At node 1:
i1=i2+i3

becomes
5=
𝑣1 −𝑣2
4
+
𝑣1 −0
2
At node 2:
i2+i4=i1+i5 becomes
𝑣1 −𝑣2
4
+ 10 = 5 +
𝑣2 −0
6
Example: Step 2 (Apply KCL & Ohm’s
Law)
Part 4 of 4


Second-trickiest part! Next we use
algebra to simplify our equations.
At node 1:
5=

+
𝑣1 −0
2
becomes
3𝑣1 − 𝑣2 = 20
At node 2:
𝑣1 −𝑣2
4

𝑣1 −𝑣2
4
+ 10 = 5 +
𝑣2 −0
becomes
6
−3𝑣1 + 5𝑣2 = 60
We now have our two equations in two
variables.
Example: Step 3 (Solve)


Next we use any of our methods—
substitution, Cramer’s rule, matrix
inversion, MATLAB—to solve our two
equations in two variables.
Using MATLAB, the solution to
3𝑣1 − 𝑣2 = 20
−3𝑣1 + 5𝑣2 = 60
is
v1 = 13.33 V
and
v2 = 20 V
Example: Extending the Analysis



Nodal analysis has
given us the values
of the node voltages
v1 and v2.
We can find all
other voltages
and currents in the
circuit once we
know these node voltages.
Example: Knowing that v1 = 13.33 V
and v2 = 20 V, how would we find i2?
Review: Steps in Performing Nodal
Analysis on a Circuit with No Voltage
Sources

Given a circuit with n nodes, with no
voltage sources, follow these steps:
1.
2.
3.
Select a node as the reference node.
Assign voltages v1, v2, …, vn-1 to the
remaining n-1 nodes. These voltages are
relative to the reference node.
Apply KCL to each of the n-1 nonreference nodes. Use Ohm’s law to express
the branch currents in terms of node
voltages. Then simplify the equations.
Solve the resulting n-1 simultaneous
equations to obtain the unknown node
voltages.
Online Alternative to MATLAB



WolframAlpha (wolframalpha.com) is a
powerful free online math tool.
An earlier example using MATLAB:
>> A = [3 -4; 2 8]
>> B = [7; 26]
>> X = inv(A)*B
Same example using WolframAlpha:
inv({{3, -4},{2, 8}})*{{7},{26}}
What About Circuits with Voltage
Sources?



As described above, our procedure
applies only to circuits without
voltage sources.
But it’s not hard to extend the
procedure to circuits with voltage
sources.
The way you handle a voltage source
depends on whether the source is
connected to the reference node….
Is the Voltage Source Connected
to the Reference Node?
Case 1. Voltage sources connected
to the reference node, like this one,
are handled one way.
Case 2. Voltage
sources not
connected to
the reference
node, like this
one, are
handled
another way.
Case 1: A Voltage Source That Is
Connected to the Reference Node

A voltage source
connected to the
reference node
is easy to handle,
because it
immediately
reveals the voltage
at one of the
non-reference nodes.
 Example:
In the circuit shown, we can
immediately see that v1 = 10 V.
Case 2: A Voltage Source That Is Not
Connected to the Reference Node
A voltage source
not connected to
the reference
node is trickier.
 To handle it,
we treat the
voltage source
and its two
nodes (along with
any elements in parallel with the voltage
source), as a supernode.

How to Handle a Supernode
We apply KCL
and KVL to the
supernode to
get two
equations.
 Example: In
the circuit shown,
KCL gives
i1 + i4 = i2 + i3


And KVL gives
v2 = 5 + v3
We Still Get Enough Equations
If this circuit did
not have a supernode, we would
get one equation
by applying KCL
to node 2 and
another by applying
KCL to node 3.
 With the supernode, we get one equation
by applying KCL to the supernode and
another by applying KVL to the supernode.
