HIGHER TECHNOLOGICAL INSTITUTE (HTI)
Circuit theory
(MTE 111)
Prerequisite, PHY 002
Dr. Ahmed Said
Dr. Tarek Abbas
Lecture 3
Nodal Analysis: The Concept.
• Every circuit has n nodes with one of the nodes being designated as a
reference node.
• We designate the remaining n – 1 nodes as voltage nodes and give each
node a unique name, vi.
• At each node we write Kirchhoff’s current law in terms of the node
voltages.
• We form n-1 linear equations at the n-1 nodes in terms of the node
voltages.
•We solve the n-1 equations for the n-1 node voltages.
From the node voltages we can calculate any branch current or any
voltage across any element.
Concept Illustration:
v1
v2
v3
•
•
R4
R2
R1
I
reference node
V1 − V2
R2
+
V1
R1
+
V1
R3
+
V1 − V3
R4
=I
R3
Clearing the previous equation gives,
 1
 1 
 1 
1
1
1 
 +
+
+ V1 −  V2 −  V3 = I
 R2 
 R4 
 R1 R2 R3 R4 
We would need two additional equations, from the remaining
circuit, in order to solve for V1, V2, and V3
Ex. 1
Given the following circuit. Set-up the equations to solve for V1 and
V2. Also solve for the voltage V6.
R2
v1
R3
•
v2
R5
•
+
R1
I1
R4
Solution
v6
_
R6
V1
R1 + R2
+
V1 − V2
R3
R2
R3
+
V2
R4
+
R3
•
= I1
v2
R5
•
+
R1
V2 − V1
v1
V2
R5 + R6
I1
R4
= 0

 1 
1
1 

R +R + R 
V1 − 
R 
V2 = I 1
 1
 3
2
3 
 1 
 1

1
1
−
R 
V1 + 
R + R + R +R 
V2 = 0
 3
 3
4
5
6 
v6
_
R6
Ex. 2 Find V1 and V2.
2A
Solution
•
v2
5
At v1:
V1
10
At v2:
v1 •
+
V1 − V2
5
V2 − V1
5
+
V2
20
10 
=2
20 
= −6
V1 + 2V1 – 2V2 = 20
3V1 – 2V2 = 20
Solution: V1 = -20 V,
4V2 – 4V1 + V2 = -120
-4V1 + 5V2 = -120
V2 = -40 V
4A
Ex. 3 Given the following circuit. Set-up the equations to solve for
V1 and V2.
I
Solution
R1
At V1:
V1 − E V1 V1 − V2
+
+
=I
R1
R2
R3
At V2:
v1
v2
R3
E
+
_
V2 − V
1 = −I
+
R
R
4
3
V2
 1
 1 
1
1 
E




+
+
V1 − 
V2 = I +


R1
 R1 R2 R3 
 R3 
 1
 1 
1 
V2 = − I
V1 + 
− 
+
 R2 
 R3 R4 
R2
R4
Ex. 4 Find V1 and V2.
Solution
4
v2
•
At v1:
V1 V1 + 10 − V2
+
= −5
10
4
10 V
v1
_
•
+
6
10 
5A
At v2:
V2 V2 − 10 − V1
+
=0
6
4
4V1 + 10V1 + 100 – 10V2 = -200
14V1 – 10V2 = -300
V1 = -30 V, V2 = -12 V
4V2 + 6V2 – 60 – 6V1 = 0
-6V1 + 10V2 = 60
Ex. 5
Given the following circuit. Solve for the indicated nodal voltages.
super node
2
v1
5
6A
x
x
v2
_
v3
+
10 V
x
x
4
10 
Solution
2
v2
v1
_
5
6A
At V1
At super
node
+
v3
Constraint Equation
10 V
4
10 
V1 − V2 V1 − V3
+
=6
5
2
V2 − V1 V2 V3 V3 − V1
+
+
+
=0
5
4
10
2
V3 – V2 = 10
7V1 – 2V2 – 5V3 = 60
-14V1 + 9V2 + 12V3 = 0
V3 – V2 = 10
Solving gives:
V1 = 30 V, V2 = 14.29 V, V3 = 24.29 V
Ex. 6
Consider the circuit below. We desire to solve for the node voltages V1 and V2.
2
_
v1
10 
10 V
+
_
Vx +
•
•
2A
v2
4
5
5Vx
In this case we have a dependent source, 5Vx, that must be reckoned with. Actually,
there is a constraint equation of
V2 − Vx =V1
2
_
v1
10 
10 V
+
_
Vx +
•
•
2A
v2
4
5
At node V1 V1 − 10 + V1 + V1 − V2 = 2
10
5
2
The constraint equation:
5Vx
At node V2
Vx = V2 − V1
V2 − V1 V2 − 5Vx
+
= −2
2
4
8V1 − 5V2 = 30
3V1 − 2V2 = − 8
which yields,
V1 = 100V ,
V2 = 154V