Lesson 6

Basic Laws of Electric Circuits

Nodal Analysis

Basic CircuitsNodal 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.

1

Basic CircuitsNodal Analysis: The Concept.

• 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.

2

Basic CircuitsNodal Analysis: Concept Illustration:

r e fe r e n c e n o d e

v 1v 2 v 3

R 2

R 1R 3

R 4

I

Figure 6.1: Partial circuit used to illustrate nodal analysis.

IR

VV

R

V

R

V

R

VV

4

31

3

1

1

1

2

21 Eq 6.1

3

Basic Circuits

Nodal Analysis: Concept Illustration:

Clearing the previous equation gives,

IVR

VR

VRRRR

3

42

21

4321

111111 Eq 6.2

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

4

Basic Circuits

Nodal Analysis: Example 6.1

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

R 2 R 3

R 1 R 4

R 5

R 6I1

v 1 v 2

+

_

v 6

Figure 6.2: Circuit for Example 6.1.

5

Basic Circuits

Nodal Analysis: Example 6.1, the nodal equations.

R 2 R 3

R 1 R 4

R 5

R 6I1

v 1 v 2

+

_

v 6

065

2

4

2

3

12

13

21

21

1

RR

V

R

V

R

VV

IR

VV

RR

VEq 6.3

Eq 6.46

Basic CircuitsNodal Analysis: Example 6.1: Set up for solution.

065

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 6.3

Eq 6.4

Eq 6.5

Eq 6.6

7

Nodal Analysis: Example 6.2, using circuit values.

v1v2

1 0

5

2 0 4 A

2 A

Figure 6.3: Circuit for Example 6.2.

Find V1 and V2.

At v1:

2510

211

VVV

At v2:

6205

212 VVV

Eq 6.7

Eq 6.8

Basic Circuits

8

Nodal Analysis: Example 6.2: 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 6.9

Eq 6.10

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

Basic Circuits

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Basic CircuitsNodal Analysis: Example 6.3: With voltage source.

R 1

R 3

I

v2v1

+_ R 2 R 4E

Figure 6.4: Circuit for Example 6.3.

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 6.11

Eq 6.12

10

Basic CircuitsNodal Analysis: Example 6.3: Continued.

Collecting terms in Equations (6.11) and (6.12) gives

12

3

11

3

1

2

1

1

1

R

EIV

RV

RRR

IVRR

VR

2

4

1

3

11

2

1

Eq 6.13

Eq 6.14

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Basic CircuitsNodal Analysis: Example 6.4: Numerical example with voltagesource.

v2v1

6

4

1 0 5 A

+ _

1 0 V

Figure 6.5: Circuit for Example 6.4.

What do we do first?

12

Basic CircuitsNodal Analysis: Example 6.4: Continued

v2v1

6

4

1 0 5 A

+ _

1 0 V

At v1:

54

2101101

VVV

At v2:

04

110262

VVV

Eq 6.15

Eq 6.16

13

Basic CircuitsNodal Analysis: Example 6.4: 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 6.17

Eq 6.18

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

Basic CircuitsNodal Analysis: Example 6.5: Voltage super node.

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

+_

6 A

5

4

2

1 0

v 1v 2 v 3

1 0 V

x

x

xx

Figure 6.6: Circuit for Example 6.5.

When a voltage source appears between two nodes, an easy way to handle this is to form a super node. The super node encircles thevoltage source and the tips of the branches connected to the nodes.

super node

15

Basic CircuitsNodal Analysis: Example 6.5: Continued.

+_

6 A

5

4

2

1 0

v 1v 2 v 3

1 0 V

At V1 62

315

21

VVVV

At supernode

02

1310

342

512

VVVVVV

Constraint Equation

V2 – V3 = -10 Eq 6.19

Eq 6.20

Eq 6.21

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Basic CircuitsNodal Analysis: Example 6.5: Continued.

Clearing Eq 6.19, 6.20, and 6.21:

7V1 – 2V2 – 5V3 = 60

-14V1 + 9V2 + 12V3 = 0

V2 – V3 = -10

Eq 6.22

Eq 6.23

Eq 6.24

Solving gives:

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

17

Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.

Consider the circuit below. We desire to solve for the node voltagesV1 and V2.

1 0

2

4

5

2 A

+_1 0 V

5 V x

v1 v2

V x +_

Figure 6.7: Circuit for Example 6.6.

In this case we have a dependent source, 5Vx, that must be reckonedwith. Actually, there is a constraint equation of

012 VVV x Eq 6.25

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Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.

1 0

2

4

5

2 A

+_1 0 V

5 V x

v1 v2

V x +_

At node V1 22510

10 2111

VVVV

At node V22

4

5

2212

xVVVV

The constraint equation: 21 VVVx 19

Basic CircuitsNodal Analysis: Example 6.6: With Dependent Sources.

Clearing the previous equations and substituting the constraint VX = V1 - V2 gives,

887

3058

21

21

VV

VV Eq 6.26

Eq 6.27

which yields,

VVVV 03.5,9.6 21

20

circuits

End of Lesson 6Nodal Analysis