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Node Equations
Introduction
The circuits in this problem set consist of resistors and independent sources. We will analyze
these circuits by writing and solving a set of node equations. This analysis is accomplished by
doing two things:
1. Express the element currents and voltages in terms of the node voltages.
2. Apply KCL at nodes of the circuit.
In particular, we will express the resistor currents in terms of the node voltages. Consider these
two resistors.
(b)
(a)
In (a) a resistor is connected between two nodes, node 1 and node 2. The corresponding node
voltages are v1 and v 2 . Apply KVL in (a) to get
v R + v 2 − v1 = 0 ⇒ v R = v1 − v 2
Then apply Ohm’s law to get
iR =
vR
25
=
v1 − v 2
25
In (b) a resistor is connected between node 2 and the reference node. The node voltage at node 2
is v 2 . Apply Ohm’s law in (b) to get
i2 =
1
v2
9
2
To summarize:
1. Consider a resistor having resistance R connected between node i and node j. The current
vi − v j
directed from node i to node j is
.
R
2. Consider a resistor having resistance R connected between node i and the reference node.
vi
The current directed from node i to the reference node is
.
R
We will also express the voltages of voltage sources currents in terms of node voltages. Consider
these two independent voltage sources.
(b)
(a)
In (a) a voltage source is connected between two nodes, node 1 and node 2. The corresponding
node voltages are v1 and v 2 . Apply KVL in (a) to get
25 + v 2 − v1 = 0 ⇒ v1 − v 2 = 25 V
There is no easy way to express the voltage source current in terms of the node voltages.
In (b) a voltage source is connected between node 2 and the reference node. The node voltage at
node 2 is v 2 . We see that
v2 = 9 V
Again, there is no easy way to express the voltage source current in terms of node voltages.
3
To summarize:
1. Consider a voltage source having voltage V s connected between node i and node j.
Suppose the + node of the voltage source is connected to node i. Then v i − v j = V s .
2. Consider a voltage source having voltage V s connected between node i and the reference
node. Suppose the + node of the voltage source is connected to node i. Then v i = V s .
3. Consider a voltage source having voltage V s connected between node i and the reference
node. Suppose the + node of the voltage source is connected to the reference node. Then
v i = −V s .
4. There is no easy way to express the voltage source current in terms of the node voltages.
Worked Examples
Example:
Analyze this circuit by writing and solving a set of node equations.
Solution:
Emphasize and label the nodes:
4
Noticing the 24 V source connected between node 3 and the reference node, we determine that
node voltage at node 3 is
v 3 = 24 V
Apply KCL at node 1 to get
v1
8
In this equation
v1
+
v1 − v 2
25
+2=0
is the current directed downward in the 8 Ω resistor and
v1 − v 2
is the
8
25
current directed from left to right in the 25 Ω resistor. We will simplify this equation by doing
two things:
1. Multiplying each side by 8 × 25 = 200 to eliminate fractions.
2. Move the terms that don’t involve node voltages to the right side of the equation.
The result is
33 v1 − 8 v 2 = −400
Next, apply KCL at node 2 to get
v2
9
In this equation
v2
9
+
v 2 − 24
14
=
v1 − v 2
25
is the current directed downward in the 9 Ω resistor,
directed from left to right in the 14 Ω resistor and
v1 − v 2
v 2 − 24
14
is the current
is the current directed from left to
25
right in the 25 Ω resistor. We will simplify this equation by doing two things:
1. Multiplying each side by 8 × 25 ×14 = 2800 to eliminate fractions.
2. Move the terms that involve node voltages to the left side of the equation and move the
terms that don’t involve node voltages to the right side of the equation.
The result is
− ( 9 ×14 ) v1 + ( 9 ×14 + 25 ×14 + 25 × 9 ) v 2 = 24 × 25 × 9 ⇒ − 126 v1 + 701v 2 = 5400
The simultaneous equations can be written in matrix form
33 v1 − 8 v 2 = −400
−8 ⎤ ⎡ v1 ⎤ ⎡ −400⎤
⎡ 33
⇒ ⎢
⎥⎢ ⎥ = ⎢
⎥
−126 v1 + 701v 2 = 5400
⎣ −126 701⎦ ⎣v 2 ⎦ ⎣ 5400 ⎦
Using MATLAB to solve the matrix equation
Then
⎡ v1 ⎤ ⎡ −10.7209 ⎤
⎢v ⎥ = ⎢
⎥
⎣ 2 ⎦ ⎣ 5.7763 ⎦
That is, the node voltages are v 1 = −10.7209 V and v 2 = 5.7763 V .
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6
Example:
This circuit is represented by the simultaneous equations:
a11 v1 + a12 v 2 = −40
a 21 v1 + a 22 v 2 = −228
Determine the values of the coefficients a 11 , a 12 , a 21 and a 22 .
Solution:
Emphasize and label the nodes:
Noticing the 10 V source connected between node 3 and the reference node, we determine that
node voltage at node 3 is
v 3 = −10 V
Apply KCL at node 1 to get
v1
v1 − v 2
+ 0.4 +
=0
10
10
In this equation
v1
is the current directed downward in the vertical 10 Ω resistor and
v1 − v 2
10
10
the current directed from left to right in the horizontal 10 Ω resistor. We will simplify this
equation by doing two things:
7
is
1. Multiplying each side by 10 to eliminate fractions.
2. Move the terms that don’t involve node voltages to the right side of the equation.
The result is
2 v 1 − v 2 = −4
Next, apply KCL at node 2 to get
v2
19
In this equation
v2
19
+
v 2 − ( −10 )
22
=
v1 − v 2
10
+ 0.4
is the current directed downward in the 19 Ω resistor,
current directed from left to right in the 22 Ω resistor and
v1 − v 2
v 2 − ( −10 )
22
is the
is the current directed from left
10
to right in the horizontal 10 Ω resistor. We will simplify this equation by doing two things:
1. Multiplying each side by 19 × 22 ×10 = 4180 to eliminate fractions.
2. Move the terms that involve node voltages to the left side of the equation and move the
terms that don’t involve node voltages to the right side of the equation.
The result is
− (19 × 22 ) v1 + (19 × 10 + 22 × 10 + 19 × 22 ) v 2 = −10 × 10 × 19 + 0.4 × 19 × 22 × 10
⇒ − 418 v1 + 828 v 2 = −228
Comparing our equations to the given equations, we see that we need to multiply both sides of
our first equation by 10. Then
20 v1 − 10 v 2 = −40
−418 v1 + 828 v 2 = −228
Comparing coefficients gives
a 11 = 20 , a 12 = −10 , a 21 = −418 and a 22 = 828 .