ELEC Chap3

7/30/2019 ELEC Chap3

1/42

UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

Chapter (3)

Nodal and loop analysis

Consider the following circuit:

+

-

+

-

R2

R4

R5

R7

R6

R3

R1

V1

V2

I

i1

i5 i4

i3

i6

i2

a b

cd

e

fg


2/42


Definitions:

Node :

A point where two or more circuit elements join

Ex. a,b,c,d,e,f,g

Essential Node:A node where three or more circuit element join

Ex. b,c,e,g

Path:

A trace of adjoining elements with no elements included more

than once

1. V1-R1-R5-R62. R5-R6-R4-V2 ,etc


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Branch:

A path that connects any two nodes.

Ex. R1 , V1 , R1-R5 , etc

Essential Branch:

A path that connects two essential nodes without passing throughan essential node.

Ex. V1-R1 , R5 , R2-R3 , V2-R4 ,

Loop :

A path whose last node is the same as its starting node

Ex. (1) V1-R1-R5-R3-R2

(2) V1

-R1

-R5

-R6

-R4

-V2


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Mesh :

A loop that doesnt enclose any other loops.

Ex. V1-R1-R5-R3-R2Ex. V2-R2-R3-R6-R4 ,

In chapter (2) we studied circuits containing a single loop or a

single node-pair

Such circuits can be solved easily by one algebraic equation.

Here , we will study circuit containing multiple node andmultiple loops

Hence we will introduce (2) analysis techniques :

1. Nodal analysis2. Loop analysis


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(1) Nodal Analysis :

Nodal analysis : is a technique in which KCL is used to determine

the nodes voltages at all essential nodes with respect to the

reference node.

Here , node voltage is defined as the voltage of a given node with

respect to a reference node

+

-+-

i5i3i1 i4i2Vm = 10

R1= 2 R3= 2 R5= 1

R4 = 2R2= 1

1 2

3

Vn = 5

Example:


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Essential nodes : 1,2,3

Consider (3) to be the reference node (ground)

At (1) apply KCL:

( )110V4V

02

V2V5

02

V

2

VV

2

V5

02

VV

1

V

2

V10

0R

VV

R

V

R

VV

0iii

21

21

211

1

2111

3

21

2

1

1

1m

321

KK=

=+

=+

=

=

=


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( ) ( )

( )210V4V

05V22

V

01

5V

2

V

2

VV

21

21

2221

KK=

=

=+

0R

VV

R

V

R

VV

0iii

:KCLApply,(2)At

5

n2

4

2

3

21

543

=

=

2V

2VBAVBVA

10

10

V

V

41

14

2

1

1

2

1

=

===

=


8/42


Note :

Number of equations = N-1

Where :N is the number of essential nodes

Example :Circuit with only independent current source

iR1

i1

R3= 5

R1= 60

R4 = 2R2= 15

1 2

3iR2

5 A = I S2iR4

V 1V 2

+

-

+

-

15 A = I S1


9/42


Find V1,V2 and i

# of essential node = N=3Select (3) as ground (reference node)

# of KCL equations = N-1 = 2

At node (1) apply KCL

05

V

5

V

15

V

60

V

15

0R

VVRV

RV15

0iiiI

2111

3

21

2

1

1

1

1R2R11s

=+

=

=


10/42


(1)15V

5

1V

60

17

15V51V

601412

15V5

1V

60

1

15

1

5

1

21

21

21

KK=

=++

=

++

(2)50V7V2

050V5V2V2

052

V

5

VV

05RV

RVV

0Iii

21

221

221

4

2

3

21

s2R41

KK=

==

=

=At node (2)


11/42


A105

1060R

VVi

V10V

V60V

50

15

V

V

72

5

1

60

17

211

2

1

2

1

===

=

=

=

Example :

Circuit with dependent current source

i21 2

3

I0

V 2

I S = 2 mA k10R3 =

V 1 k10R2 =

2 Ix

i1

Ix

k10R1 =

Find I0


12/42


# of essential nodes = N=3

Choose (3) to be reference node ( ground )

We have N-1 = 2 KCL equationat node(1) and( 2)

At node (1) , apply KCL

(1)A2V0.1V0.2

10kV

5kV

10kVV

10kVAm2

0R

VV

R

VAm2

0iiI

21

21211

2

21

1

1

21s

KK=

=+=

=

=


13/42


0II2i 0x2=

At node (2) apply KCL,

0II2R

VV

0I2Ii

0x2

21

0x2

=

=

(2)0V2V1

010kV

10kV2

10kVV

R

VI

R

ViIwhere

21

2121

3

20

1

11x

KK=+

=

=

==


14/42


Am0.4k10

4

R

VI

V4V

V8V

0

2

V2

V1

21

0.10.2

3

20

2

1

===

=

=

=


15/42


16/42


(1)16V18

1V0.201388

0

18

V

18

V

48

V

8

V16

018

VV48V

8V128

1

2111

2111

KK=

=+

=

KCL at (1)

(2)7V0.20555V18

1

0

10

70V

20

V

18

VV

21

2221

KK=

=

KCL at (2)


17/42


V60V

V96V

7

16

V

V

0.205518

118

10.201338

2

1

2

1

=

=

=

A1

10

70Vi

A42060

20Vi

A218

6096

18

VVi

A248

96

48

Vi

A48

96128

8

V128i

2e

2d

21c

1b

1a

=

=

===

=

=

=

===

===


18/42


A110

70Vi

A420

60

20

Vi

A2

18

6096

18

VVi

A24896

48Vi

A48

96128

8

V128i

2e

2d

21c

1b

1a

==

===

=

=

=

===

=

=

=

Special case:

What if a branch between two essential non-reference nodecontain a voltage source ?

This case is called super node" case.


19/42


Find I0

# of essential nodes = N = 3 Super node: is the voltage source and the two connecting nodes

# of equations = N-1-1 = 3-1-1=1

Reference node Super node

But we need (2) equations to find the two unknowns V1 and V2There is an equation that describe the super node.

1 2

3i2

V 2V 1

i1

i3

+ -

6 Vk6

Io

k6 k6k21k21

k6

supernode


20/42


Apply KCL at the super node :

(1)06k

V

6k

V

012k

V

12k

V

k12

V

k12

V0Iiii

21

2211

0321

KK

=+

=+++=+++

Am0.25

12k

3

12k

V2I

V3V

V3V

6

0

V2

V1

116k

1

6k

1)2(V6V

0

2

1

12

=

==

==

=

=+ KK

The super node is described by:


21/42


Example:

Circuits with dependent voltage sources

Find I0?

N = 3 N-1 = 2 equations

Io1 2

3

i2

V 2V 1 5

i1

i3

+-20 V

i4

22

02 01

8 Io


22/42


(1)10V5

1V

4

3

10V5

1V

5

1

20

1

2

1

21

21

KK=

=

++

5

V

5

V

20

V

2

V10

5

VV

20

V

2

V20

Iii

2111

2111

021

+=

+=

+=

KCL at node (1):


23/42


2

V

10

V

5

VV5

5

VV4

2

V

10

V

5

V

5

V

2

I8V

10

V

5

VV

iiI

2221

212221

02221

430

+=

+=

+=

+=

(2)0V

5

8V

0V2

1

10

11V

2

V

10

VVV

21

21

2221

KK=

=

++

+=


24/42


A1.25

6

5

VVI

V10V

V16V

0

10

V2

V1

581

5

1

4

3

210

2

1

==

=

=

=

=

Loop Analysis (Mesh)Mesh analysis : It is a technique in which KVL is used to

determine the current in all meshes

# of equations needed = # of meshes = be-(ne-1)Where :

be : # of essential branches

ne: # of essential nodes


25/42


Example :

(1)40i0i8i10

0)ii(82i40

321

211

KK=+=++KVL left loop :

6

+

-

+

- 20 V

42

8

40 V

6

+

-

Voi1i3

i2

Find V0?We have (3) meshes


26/42


A0.8i,A2i,A5.6i

20

0

40

i

i

i

1060

6208

0810

321

3

2

1

===

=

V28.8V

(3.6)82)(5.68)ii(8V

0

210

=

===

0i6i20i8

0)i(i6i6)i(i8

321

32212

=+

=++

20i10i6i0

04i)i(i620

321

323

=+

=++

KVL middle loop:

KVL right loop :


27/42


28/42


29/42


Special Case :

What happens if a current source is located between two meshes

Super mesh

i2 2

i1 i3+

-50 V

3

10

+

-

100 V

5 A

46

You dont know the voltage across the current source !!

Remove the whole branch that includes the current source


30/42


i2 2

i1i3

+

- 50 V

3

10

+

-

100 V

46

Apply KVL around the super mesh

(1)50i6i5i9

0i6i450)i(i2)i(i3100

321

132321

KK=+

=+++++


31/42


(2)0i2i15i30)i(i3)i(i2i10

321

12322

KK=+=++

(3)5ii0i

A5ii

321

13

KK=++

=

A6.75i

A1.25i

A1.75i

3

2

1

==

=

KVL around the upper loop:

We also know


32/42


Example:

Nodal Analysis :

001

+

-

128 V

501

256 V

+

-

005502

200 400

300

i

i50

Use nodal analysis and loop analysis to find power in the 300 ()

resistor ?


33/42


34/42


(1)1.7067V0.00333V0.01V0.0250 321 KK=

( )

(2)0.256V0.0118V0.004V0.0033

500

128

300

1

500

1

400

1

250

1V

250

V

300

V

0300

V1V

500

128V

400

V

150

VV

0iiii

321

321

33332

654

KK=+

=

++

=

=KCL at node (3):

You can notice that

(3)0V0.1667V1V0.166

V61V

61

300VV50V

i50V

321

1313

2

2

KK=+

=

=

=


35/42


V8V

V11.75V

V62.5V

0

0.256

1.7067

V3

V2

V1

0.166710.1667

0.01180.0040.0033

0.00330.010.0383

3

2

1

==

=

=

( ) ( ) W16.56753000.235RiP

A0.235i50Vi

22

300

2

===

==


36/42


37/42


(2)0i150i300i200

0)i(i200i50)i(i100

ii

0)i(i200i50)i(i100

521

12552

5

1252

KK=+

=+=

=++

KVL around loop (2):

5

543

4353

iisince

(3)0i200i400i6500i50)i(i400)i(i250

=

==+KK

(4)128i900i400

0)i(i400128i500

43

344

KK=+

=+

KVL around loop (3)

KVL around loop (4)


38/42


(5)0i650i250i100

0)i(i100)i(i250i300

532

25355

KK=+

=++

KVL around loop (5)

=

0

128

0

0

256

i

i

i

i

i

65002501000

090040000

20040065000

15000300200

000200350

5

4

3

2

1


39/42


W16.5675(300)(0.235)R)(iP

22

5300

===

A0.235i

A0.24i

A0.22iA0.9775i

A1.29i

5

4

3

2

1

=

=

==

=


40/42


Example:

Use the loop analysis to find V

10Ai3 =

(3) Meshes (3) equations

i3 1

i1 i2+

-

2

5 5

V2

10 A

+

-V

57 V


41/42


(1)55i5i7

07520i5i7

075)i(i5)i(i2

21

21

2131

KK=

=+

=+

KVL around loop (1):

( ) ( )

(2)0i3i2

i2i2ii552i

)i(i5V

5

V2i

21

21212

21

2

KK=

=

=

=

=

Equation of dependent source:


42/42

UAE University Department of Electrical Engineering Dr Hazem N Nounou

A10i

A15i0

55

i

i

32

57

2

1

2

1

=

=

=

V25V

V2510)(155

)i(i5V

21

=

==

=

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