module i lecture 2

NODAL AND MESH ANALYSIS

Galgotias University

Basic CircuitsNodal Analysis: The Concept.

• Every circuit has n nodes with one of the nodes beingdesignated as a reference node.

• We designate the remaining n – 1 nodes as voltage nodesand give each node a unique name, vi.

• At each node we write Kirchhoff’s current law in termsof the node voltages.

1Galgotias University

Basic CircuitsNodal Analysis: The Concept.

• We form n-1 linear equations at the n-1 nodesin 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 branchcurrent or any voltage across any element.

2 Galgotias University

Basic CircuitsNodal Analysis: Concept Illustration:

reference node

v1v2 v3

R2

R1R3

R4

I

Figure 1: Partial circuit used to illustrate nodal analysis.

IR

VV

R

V

R

V

R

VV

4

31

3

1

1

1

2

21 Eq 1


Basic Circuits

Nodal Analysis: Concept Illustration:

Clearing the previous equation gives,

IVR

VR

VRRRR

3

4

2

2

1

4321

111111 Eq 2

We would need two additional equations, from theremaining circuit, in order to solve for V1, V2, and V3


Basic Circuits

Nodal Analysis: Example 1

Given the following circuit. Set-up the equations to solve for V1 and V2. Also solve for the voltage V6.

R2 R3

R1 R4

R5

R6I1

v1 v2

+

_

v6

Figure 2: Circuit for Example 1.


Basic Circuits

Nodal Analysis: Example 1, the nodal equations.

R2 R3

R1 R4

R5

R6I1

v1 v2

+

_

v6

0

65

2

4

2

3

12

13

21

21

1

RR

V

R

V

R

VV

IR

VV

RR

VEq 3

Eq 4


Basic CircuitsNodal Analysis: Example 1: Set up for solution.

0

65

2

4

2

3

12

13

21

21

1

RR

V

R

V

R

VV

IR

VV

RR

V

01111

111

2

6543

1

3

12

3

1

321

VRRRR

VR

IVR

VRRR

Eq 3

Eq 4

Eq 5

Eq 6


Nodal Analysis: Example 2, using circuit values.

v1v2

10

5

20 4 A

2 A

Figure 3: Circuit forExample 2.

Find V1 and V2.

At v1:

2510

211

VVV

At v2:

6205

212 VVV

Eq 7

Eq 8

Basic Circuits


Nodal Analysis: Example 2: Clearing Equations;

From Eq 6.7:

V1 + 2V1 – 2V2 = 20

or

3V1 – 2V2 = 20

From Eq 6.8:

4V2 – 4V1 + V2 = -120

or

-4V1 + 5V2 = -120

Eq 9

Eq 10

Solution: V1 = -20 V, V2 = -40 V

Basic Circuits


Basic CircuitsNodal Analysis: Example 3: With voltage source.

R1

R3

I

v2v1

+_ R2 R4E

Figure 4: Circuit forExample 3.

At V1:

IR

VV

R

V

R

EV

3

21

2

1

1

1

At V2:

IR

VV

R

V

3

12

4

2

Eq 11

Eq 12


Basic CircuitsNodal Analysis: Example 3: Continued.

Collecting terms in Equations (11) and (12) gives

12

3

11

3

1

2

1

1

1

R

EIV

RV

RRR

IVRR

VR

2

4

1

3

11

2

1

Eq 13

Eq 14


Basic CircuitsNodal Analysis: Example 4: Numerical example with voltagesource.

v2v1

6

4

10 5 A

+_

10 V


What do we do first?


Basic CircuitsNodal Analysis: Example 4: Continued

v2v1

6

4

10 5 A

+_

10 V

At v1:

54

2101

10

1

VVV

At v2:

04

1102

6

2

VVV

Eq 15

Eq 16


Basic CircuitsNodal Analysis: Example 4: Continued

Clearing Eq 6.15

4V1 + 10V1 + 100 – 10V2 = -200

or

14V1 – 10V2 = -300

Clearing Eq 6.16

4V2 + 6V2 – 60 – 6V1 = 0

or

-6V1 + 10V2 = 60

Eq 17

Eq 18

V1 = -30 V, V2 = -12 V, I1 = -2 A14


Basic CircuitsNodal Analysis: Example 5: Voltage super node.

Given the following circuit. Solve for the indicated nodal voltages.

+_

6 A

5

4

2

10

v1v2 v3

10 V

x

x

xx


When a voltage source appears between two nodes, an easy way to

handle this is to form a super node. The super node encircles the

voltage source and the tips of the branches connected to the nodes.

super node



+_

6 A

5

4

2

10

v1v2 v3

10 V

At V1 62

31

5

21

VVVV

At super

node0

2

13

10

3

4

2

5

12

VVVVVV

Constraint Equation

V2 – V3 = -10 Eq 19

Eq 20

Eq 21



Clearing Eq 19, 20, and 21:

7V1 – 2V2 – 5V3 = 60

-14V1 + 9V2 + 12V3 = 0

V2 – V3 = -10

Eq 22

Eq 23

Eq 24

Solving gives:

V1 = 30 V, V2 = 14.29 V, V3 = 24.29 V


Basic CircuitsNodal Analysis: Example 6: With Dependent Sources.

Consider the circuit below. We desire to solve for the node voltages

V1 and V2.

10

2

4

5

2 A

+_10 V

5Vx

v1 v2

Vx+_

Figure 6: Circuit for

Example 6.

In this case we have a dependent source, 5Vx, that must be reckoned

with. Actually, there is a constraint equation of

012 VVV x Eq 25



10

2

4

5

2 A

+_10 V

5Vx

v1 v2

Vx+_

At node V1 22510

10 2111

VVVV

At node V2 24

5

2

212

xVVVV

The constraint equation: 21 VVVx 19



Clearing the previous equations and substituting

the constraint VX = V1 - V2 gives,

887

3058

21

21

VV

VV Eq 26

Eq 27

which yields,

VVVV 03.5,9.6 21


module i lecture 2

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