no de equations

7/31/2019 No de Equations

1/7

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 thesetwo resistors.

(a) (b)

In (a) a resistor is connected between two nodes, node 1 and node 2. The corresponding nodevoltages are and . Apply KVL in (a) to get

1v

2v

R 2 1 R 10v v v v v v+ = = 2

Then apply Ohms law to get

R 1

R25 25

v v vi

2= =

In (b) a resistor is connected between node 2 and the reference node. The node voltage at node 2

is . Apply Ohms law in (b) to get2

v

2

29

vi =

1


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2

To summarize:

1. Consider a resistor having resistance R connected between node i and node j. The currentdirected from node i to node j is

i jv v

R

.

2. Consider a resistor having resistance R connected between node i and the reference node.The current directed from node i to the reference node is

iv

R.

We will also express the voltages of voltage sources currents in terms of node voltages. Consider

these two independent voltage sources.

(a) (b)

In (a) a voltage source is connected between two nodes, node 1 and node 2. The corresponding

node voltages are and . Apply KVL in (a) to get1v 2v

2 1 1 225 0 25 Vv v v 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 . We see that2

v

2 9 Vv =

Again, there is no easy way to express the voltage source current in terms of node voltages.


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3

s

To summarize:

1. Consider a voltage source having voltage connected between node i and node j.Suppose the + node of the voltage source is connected to node i. Then .

sV

i jv v V =

2. Consider a voltage source having voltage connected between node i and the referencenode. Suppose the + node of the voltage source is connected to node i. Then .

sV

i sv V=

3. Consider a voltage source having voltage connected between node i and the referencenode. Suppose the + node of the voltage source is connected to the reference node. Then

.

sV

i sv V=

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:


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Noticing the 24 V source connected between node 3 and the reference node, we determine that

node voltage at node 3 is

324 Vv =

Apply KCL at node 1 to get

1 1 22 0

8 25

v v v+ + =

In this equation1

8

vis the current directed downward in the 8 resistor and

1 2

25

v vis the

current directed from left to right in the 25 resistor. We will simplify this equation by doingtwo things:

1.

Multiplying each side by 8 to eliminate fractions.25 200 =

2. Move the terms that dont involve node voltages to the right side of the equation.The result is

1 233 8 400v v =

Next, apply KCL at node 2 to get

2 2 124

9 14 25

v v v v 2 + =

In this equation2

9

vis the current directed downward in the 9 resistor,

224

14

v is the current

directed from left to right in the 14 resistor and1 2

25

v vis the current directed from left to

right in the 25 resistor. We will simplify this equation by doing two things:


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5

2v

1. Multiplying each side by 8 to eliminate fractions.25 14 2800 =

2. Move the terms that involve node voltages to the left side of the equation and move theterms that dont involve node voltages to the right side of the equation.

The result is

( ) ( )1 2 19 14 9 14 25 14 25 9 24 25 9 126 701 5400v v v + + + = + =

The simultaneous equations can be written in matrix form

1 2 1

1 2 2

33 8 400 33 8 400

126 701 5400 126 701 5400

v v v

v v v

= =

+ =

Using MATLAB to solve the matrix equation

Then

1

2

10.7209

5.7763

v

v

=

That is, the node voltages are and1

10.7209 Vv = 2 5.7763 Vv = .


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Example:

This circuit is represented by the simultaneous equations:

11 1 12 2

21 1 22 2

40

228

a v a v

a v a v

+ =

+ =

Determine the values of the coefficients , , and .11

a12

a21

a22

a

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

3 10 Vv =

Apply KCL at node 1 to get

1 1 20.4 0

10 10

v v v+ + =


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In this equation1

10

vis the current directed downward in the vertical 10 resistor and

1 2

10

v vis

the current directed from left to right in the horizontal 10 resistor. We will simplify thisequation by doing two things:

1. Multiplying each side by 10 to eliminate fractions.2. Move the terms that dont involve node voltages to the right side of the equation.

The result is

1 22 4v v =

Next, apply KCL at node 2 to get

( )2 2 1 2100.4

19 22 10

v v v v + = +

In this equation2

19

vis the current directed downward in the 19 resistor,

( )2 10

22

v is the

current directed from left to right in the 22 resistor and1 2

10

v vis the current directed from left

to right in the horizontal 10 resistor. We will simplify this equation by doing two things:

1. Multiplying each side by 19 to eliminate fractions.22 10 4180 =

2. Move the terms that involve node voltages to the left side of the equation and move theterms that dont involve node voltages to the right side of the equation.

The result is

( ) ( )1 2

1 2

19 22 19 10 22 10 19 22 10 10 19 0.4 19 22 10

418 828 228

v v

v v

+ + + = +

+ =

Comparing our equations to the given equations, we see that we need to multiply both sides ofour first equation by 10. Then

1 2

1 2

20 10 40

418 828 228

v v

v v

=

+ =

Comparing coefficients gives

1120a = , 12 10a = , 21 418a = and 22 828a = .

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