07-Nodal Analysis - · PDF fileNodal Analysis with Voltage Sources • Properties of a...

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

• Remember: for a unique solution there must be an equal number of independent equations as unknowns

Dr. Louie 4

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

y y y x b

y y y x b

y y y x b

Introduction

• Let X be a vector of variables to be solved for

• Nodal Analysis:

X: node voltages

Y: conductances

b: currents

• Mesh Analysis:

X: loop currents

Y: resistances

b: voltages

Dr. Louie 5

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

y y y x b

y y y x b

y y y x b

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

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Nodal Analysis

• Let I1, I2, R1, R2, R3 be known (given)

• Solve for the voltages at each node

Dr. Louie 7

I1

I2

R1 R3

R2

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

Or arbitrary

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I1

I2

R1 R3

R2

(usually the “bottom” of circuit)

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)

Can solve later as V1 – V2

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I1

I2

R1 R3

R2

+

- V1

+

- V2

x x

Step 3: Apply KCL

• KCL gives one equation per node (two equations)

• I1 = I2 + i1 + i2

• I2 = i3 – i2

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I1

I2

i1 i3

i2

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 = (V1 – V2)/R2

• i3 = (V2 - 0)/R3

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I1

I2

i1 i3

i2

x x V2 V1

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

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• 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

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• In matrix form

I1 = I2 + V1/R1 + (V1 – V2)/R2

I2 = V2/R3 – (V1 – V2)/R2

• Many methods of solving linear equations

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1 2 2 1 1 2

2 2

2 3 2

1 1 1

R R R V I I

V I1 1 1

R R R

Example

• Let

I1 = 10A

I2 = 5A

R1 = 6Ω

R2 = 4Ω

R3 = 2Ω

• Find all voltages

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I1

I2

i1 i3

i2

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

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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

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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

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Example

• Solving in matrix form:

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1 2 2 1 1 2

2 2

2 3 2

1

2

1 1 1

R R R V I I

V I1 1 1

R R R

1 1 1V 56 4 4

V 51 1 1

4 2 4

Can use Matlab, Gaussian elimination Cramer’s Rule, matrix inversion

Solution by Matrix Inversion

Dr. Louie 20

1

2

1 1 1V 56 4 4

V 51 1 1

4 2 4

1

2

1 1

2

V 5

V 5

V 5 20

V 5 13.3

G

G

G (matrix)

For small matrices use “inv()” command in Matlab

1 3 1

1 1.67

Gwhere

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?

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+

- V

+

-


• If the voltage source is connected to the reference node, then it can be easily included in nodal analysis

Dr. Louie 22

I1

I2

i1

i2

+

- V3


• Only one node with unknown voltage

• I1 = I2 + V1/R1 + (V1 – V3)/R2

• Compare this to earlier example (2 eqns, 2 unknowns)

Dr. Louie 23

I1

I2

+

- V3

+

- V1

R2

R1

x


• 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

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• Supernode: a closed surface containing a voltage source and its two nodes AND any elements in parallel with it

• Current into supernode = current out of supernode

• Also applies to dependent voltage sources

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+ -

Supernodes

I I

node

+ -

I I


• 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

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• Find the node voltages

Dr. Louie 27

+ -

2V

10W

2A 7A 4W 2W

assume reference is here


• Make a supernode out of the voltage source

• Write KCL for supernode

2 = i1 + i2 + 7 (only one equation since it is a supernode)

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+ -

2V

10W

2A 7A 4W 2W


i1 i2


• Write i1, i2 in terms of node voltage

• i1 = (v1 – 0)/2

• i2 = (v2 – 0)/4

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+ -

2V

10W

2A 7A 4W 2W


i1 i2

v1 v2


• 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)

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+ -

2V

10W

2A 7A 4W 2W i1 i2

v1 v2


• Apply KVL around loop

-v1 + 2 + v2 = 0 (second equation)

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+ -

2V

10W

2A 7A 4W 2W i1 i2

v1 v2


• How many independent equations?

8 = 2v1 + v2 +28

-v1 + 2 + v2 = 0

• How many unknowns?

v1, v2

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Solve!

+ -

2V

10W

2A 7A 4W 2W i1 i2

v1 v2


• Solving…

8 = 2v1 + v1 - 2 +28

v1 = -5.333V

v2 = -7.333V

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+ -

2V

10W

2A 7A 4W 2W i1 i2

v1 v2

Example

• Find the node voltages in the circuit shown

Dr. Louie 34

+ -

10V

2W

4W

2W

8W 5A

Example

• 3 unknown voltages: V1, V2, V3

• Need three independent equations (that do not introduce new variables)

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+ -

10V

2W

4W

2W

8W

V1 V2 V3

5A

Example

• Identify the supernode

Dr. Louie 40

+ -

10V

2W

4W

2W

8W

V1 V2 V3

5A

Example

• Write KCL for the supernode

i5 = i1+i4+i3 (supernode)

• Write KCL for the other node

5+i4 = i5 (node 2)

Dr. Louie 41

+ -

10V

2W

4W

2W

8W

V1 V2 V3

5A i1 i3

i4 i5

Example

• Express currents in terms of voltages

i1 = (V1 – 0)/4

i3 = (V3 – 0)/8

i4 = (V1 – V2)/2

i5 = (V2 – V3)/2

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+ -

10V

2W

4W

2W

8W

V1 V2 V3

5A i1 i3

i4 i5

Example

• Have two equations

KCL for node 2, supernode

• Need one more independent equation

• Do KVL around top loop

10 = 2i4 + 2i5

Dr. Louie 43

+ -

10V

2W

4W

2W

8W

V1 V2 V3

5A i1 i3

i4 i5

Example

• 5 + i4 = i5 (node 2)

• i6 = i1 + i5 + i3 (supernode)

i1 = (V1 – 0)/4 = 0.25V1

i3 = (V3 – 0)/8 = 0.125V3

i4 = (V1 – V2)/2 = 0.5V1 – 0.5V2

i5 = (V2 – 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 rearranging)

Solving

V1 = 12.22V

V2 = 10V

V3= 2.22V

Dr. Louie 45

1

2

3

1 2 1 10

6 8 3 0

1 0 1 10

V

V

V

Example

• Find Vx

Dr. Louie 46

2A

+

-

Vx 20W 10W 0.2Vx

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

Dr. Louie 47

2A

+

-

Vx

20W 10W 0.2Vx I1 I2

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