EE100supplementarynoteweek03-2

8/4/2019 EE100supplementarynoteweek03-2

1/27

Node Voltage Method

This simplest among many circuit analysis methods is

applicable only for connected circuits N made of linear 2-

terminal resistors and current sources. The only variables in the

linear equations are the n-1 node voltages e1, e2 , , en-1 for an

n-node circuit.

Step 1. Choose an arbitrary datum node and label the

remaining nodes consecutively , , , , and

let e1, e2 , , en-1 be node-to-datum voltages.

Step 2. Express the current of each resistor Rj via Ohms

law in terms of 2 node-to-datum voltages:

1 2 n-1

ji

+ -jR

je

+

je

j

j

j

jei

R

e+ = (1)


2/27

5A2A

+

-

6v

6i

2e 3e

1e

+-3v

4v

+

-

5v+

-8 6

42+

-1v 2v

+

-

35

6i

e=

3i

11

2

2

ei

e=

12

3

4

ei

e=

24

8i e=

1

2 3

KCL at :1 31 2

( )( )52 4

e ee e

+ =1 (2)

KCL at : 1 2 2( )

22 8

e e e + = 2 (3)

KCL at : 1 3 3( ) 24 6

e e e + =3 (4)

3 1 1

1.EXAMPLE

R t E (2) (3)


3/27

3 1 1

4 2 41 5

0

2 81 50

4 1 2

2 0d e t

3 8 4

=

1(a) To solve for , replace colum n 1 of m atrix from (5) and calculatee

1

1 15

2 45

2 08

52 0

1 2

1 1 5d e t

9 6

=

C ramer's Ru le

11

115 2023

96 384Ve = = =

2(b) To solve for , replace colum n 2 of matrix from (5) and calculatee

22

76 2015.2

96 384e V

= = =

(c) To solve for replace colum n 3 of matrix from (5) and calculatee

2

3 1

54 41

2 021 5

24 1 2

7 6d e t

9 6

=

( 6 )

( 7 )

( 8 )

( 9 )

Solving eq. (5) by Cram er's Ru le (or any other m ethod) :

C ramer's Rule


4/27

{ }1 2 3O nce the node-to-datum voltages , , are found,e e eall resistor curren ts and voltages are found trivially via

K V L, K C L, and O hm 's law :

1 21 1

7.87.8 , 3.9

2v V i Ae e= = = =

1 32 2

4.44.4 , 1.1

4v V i Ae e= = = =

33 32 3.4 , 2v Ve e i A= = =

24 4

15.215.2 , 1.9

8v V ie A= = = =

35 518.618.6 , 3.1

6v V ie= = = =

16 623 , 5v V ie A= = =

V erification of Solution

T ellegen 's TheApp oly :rem

6

1 1 2 2 3 3 4 4

1

( ) ( ) ( ) ( )j jj

v i v i v i v i v i=

= + + +5 5 6 6( ) ( )v i v i+ +


5/27

Modified Node Voltage Method

Circuit N +v

1e1 e2

6V2A 3

4

i4

i2

i1

i3

_v3v4

+

-

+

_

+

_v2

+1

3

2

_

11

2

4

ei

e=

12

3

ie

=

KCL at : 1 1 2( ) 23 4e e e

+ =1 (1)

KCL at : 1 23

( )0

4i

e e + =2 (2)

For each voltage source , add an equation .j jjs j s

vev e+ =

2 6e = (3)

1 2When the circuit contains " " voltages , , , , uses s sv v v

1 2their associated currents , , , when applying KCL.s s si i i

Step 1.

Step 2.

Step 3.


6/27

Explicit matrix form of Node Voltage Equations

Assumption: Circuit N contains only linear resistors and

independent current sources which do not form cut sets.

Step 1. Delete all current sources from N and draw the

reduced digraph G of the remaining pure resistor circuit.

Assume Ghas n nodes and b branches.

Step 2. Pick a datum node and label the node-to-datum voltages

{ e1, e2 , , en-1 }, and derive the reduced-incidence matrix A.Define the branch admittance matrix Yb and independent

current source vector is as follow:

bY

1

2

1 1 1

2 2 2

0 0 0

0 0 0(1) , (2)

0 0 0

s

s

s

b b b s

ii Y v

ii Y v

i Y v i

=

i

i v1where , resistance of branch .j j

j

Y R jR

=

current= algebraic of all sources nodesu ,mms

i entering m


7/27

Deriving the Node Voltage Equation

The current sources can be

deleted since they can be trivially accounted

for by representing their net contribution at

each node m by the algebraic sum of all

current sources entering node m , m = 1, 2,

, n-1. The KCL equations therefore takes

the augmented form

Substituting (1) for i in (4), we obtain

Substituting KVL

s=A i i

b s=A Y v i

(4)

(5)

( )n s=Y e i


8/27

Writing Node-Admittance Matrix

Yn By Inspection

nY

(8)

:Node Votage Equation

1

2

1

11 12 1, 1

21 22 2, 1

1,1 1,2 1, 1

1

2

1n

sn

sn

n n n snn

iY Y Y

iY Y Y

Y eY

e

Y i

e

=

e si

Diagonal Elements of nY1

= sum of adm ittances of all resistorsm m j

j

Y Y

R

connected to node , 1, 2 , , -1m m n=

Off-Diagonal Elements of nY

1= sum of adm ittances of all resistors

jk j

j

Y YR

- ( connected across nod e and nod ej k )


9/27

5A2A

+

-

6v

6i

2e 3e

1e

+-3v

4v

+

-

5v+

-8 6

42+

-1v 2v

+

-

35

6

ei =

3i

1 21

2

e ei

=

1 32

4

e ei

=

24

8ei =

1

2 3

KCL at :1 31 2

( )( )

5 42 4

e ee e

+ = 1 (2)

KCL at : 1 2 2( )

4 2 32 8

e e e + = 2 (3)

KCL at : 1 3 3( ) 2 14 6

e e e + = +3 (4)

Matrix Form : 3 1 1

EXAMPLE

1A

4A

3A


10/27

Mesh Current Method

The next simplest among many circuit

analysis methods is applicable only for

connected planar circuit N (with a planardigraph) made of 2-terminal linear resistors

and voltage sources. The only variables in the

equations are l conceptual mesh currents

circulating in the l meshes in

a clockwise direction (by convention) :

1 2

, , ,lm m m

l l l

1R

10i

12i

7i4i

1i

2R

4R

5R

7R

8R

10R 12R

1i

2i3i


11/27

ji

jR

ci

ai

b

i

di


12/27

Mesh Current Method

The next simplest among many circuit

analysis methods is applicable only for connected

planar circuits made of 2-terminal linear resistors andvoltage sources. The only variables in the associated

mesh-current equations are l mesh current

which we define to be circulating in the l meshes in a clockwise direction (by convention).

Unlike node-to-datum voltages in the node voltage

method which are physical in the sense they can bemeasured by a volt meter, the mesh currents are

abstract variables introduced mathematically for

writing a set of equations whose solution can be usedto find each resistor current ij trivially via

2

, , lm ml l

1

,ml


13/27

1

2

3

6 0 2 2

0 14 8 22 8 10 5

i

i

i

=

(1)

Then calculate :

1 3 1

3 10

i i i= =

1 2 3

13 8 19 , ,20 20 20

i i i= = =

2 1

13

20

i i= =

3 1 2

5 20

i i i= =

4 3 2

11 20i i i= =

5 2

8i i= =

1 1

6210

v i= =

2 2

134

5

v i= =

3 2v =

4 4

22

8 5v i= =

5 5

126v i= =

Mesh

CurrentEquations

,

,

,

,

,

(2)

(3)


14/27

2V

+

-

6v

6i

+-3v

4v+

-

5v

+

-8 6

42

+

-

1v 2v+

-

5i3i

2i

4i

1i

3i1

i

2i

+

1 3 1 i i i=

2 1i i=

3 1 2 i i i=

4 3 2 i i i=

5 2i i=

( )1 3 1 2v i i=

2 14v i=

3 2v =

( )4 3 2 8v i i=

5 26v i=

6 5v =

,

,

,

,

,

,6 3

i i=

Loop equation around mesh 1:

1 2 3 3 1 1 0 2( ) 4 2 0v v v i i i + + = + =

1 3 6 2 2i i =

5V +-

(1)


15/27

5V

2V

+

-

6v

6i

+-3v

4v+

-

5v

+

-8 6

42

+

-

1v 2v+

-

5i3i

2i

4i

1i

3i1

i

2i

1 3 M esh 1 : 6 2 2i i =

2 3

M esh 2 : 14 8 2i i =

1 2 3 M esh 3 : 2 8 10 5i i i + =

1 1 3

1 2 S olving from (1)3 3

i i i =

2 2 3

4 1 S olving from (2)7 7

i i i =

S ub stituting (4) and (5) into (3)

(1)

(2)

(3)

(4)

(5)

+- +


16/27

1i

2i

4i

6i

3i5i

We can redraw this circuit so that there are

no intersecting branches.

1

2

3

4

1i

2i

3i5i

1 3

4


17/27

1i

2i

4i

6i

3i5i

We can redraw this circuit so that there are

no intersecting branches.

1

2

3

4

1i

2i

3i5i

1 3

2

4

8

65V2V

2

5V

2V 6

4

8

+

+

+

+-

+


18/27

All branch voltages and currents can be trivially

calculated from e1 and i3.

121 1 10 , 0

4

vv Ve i Ae= = = =

12

2 26 , 23

vv e V i A= = = =

23 36 , 0v V ie A= = =

14 46 , 2ev V i= = =

V erification of Solution T ellegen's T heorby :em

4

1 1 2 2 3 3 4 4

1

( ) ( ) ( ) ( )j jj

v i v i v i v i v i=

= + + +

(0)(0) (6)(2) (6)(0) ( 6)(2)= + + + ?

0=


19/27

Note:

The unknown variables in the

modified node voltage method consist

of the usual n-1 node-to-datum

voltages, plus the unknown currents

associated with the voltage sources.

Hence, if there are voltage

sources, the modified node voltage

method would consist of (n-1)+

independent linear equations involving

(n-1)+ unknown variables


20/27

Conservation of Electrical

Energy

The algebraic sum of

electrical energy flowing into all

devices in a connected circuit is

zero for all times .t>

Proof.

Tellegen's Theorem

( ) ( ) 0b

t

v t i t dt =


21/27

1 1

0 0 0 0 0 1 1 0 1

0 0 0 0 0 1 0 1 01 1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 01 0 0 1 0 0 0 0

0

00

00

0

0

0

0

0 0 0 0 1 0 0 0 0

0 1 0 0 0 4 0 0 0

0 0 1 0 0 0 3 0 0

0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0

6

0

t

2 1

de

=

A

1A


22/27

4 4

0 0 0 0 0 1 1 0 1

0 0 0 0 0 1 0 1 01 1 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0 00 1 0 1 0 0 0 0 0

1 0 0 0 1 0 0 0 0

0 0 1 0 0 4 0 0 0

0 0 0 0 0 0 3 0 0

0 0

0

00

00

0

0

0

60 1 0 0 0 0 0

0 0 0 0 0 0 0 0 12

det

= A

4A


23/27

9 9

0 0 0 0 0 0 1 1 1

0 0 0 0 0 0 1 0 01 1 1 0 0 0 0 0 0

1 0 0 1 0 0 0 0 00 1 0 0 1 0 0 0 0

1 0 0 0 0 1 0 0 0

0 0 1 0 0 0 4 0 0

0 0 0 1 0 0 0 3 0

0 0 0 0 1

0

00

0

0 0 0

0

0

0

0

6 0

0 0 0

det

0 0 120 0 0

=

A

9A


24/27

11 11 21 21 31 31 41 41 51 51

61 61 71 71 81 81 91 91 10,1 10,1

1 a A a A a A a A a Aa A a A a A a A a A

+ + + + = + + + + +

( )91 91 10,1 10,1 11 21 811 , because 0a A a A a a a= + = = =

( ) ( )10,191 91 10,1AA a a= +

11k 12k

11 126 2k k= +

11 1 12 1s sk v k i= +

11e

=


25/27

Instantaneous Power

of a 2-terminal device

D+

-

( )v t

( )i t1

2

( ) ( ) ( )p t v t i t Under Associated Reference Convention,

1( ) 0 at p t t T > =1means ( ) Watts of power ent ersp T

1.

( )flow ats into t T=D

2( ) 0 at p t t T < =

f l


26/27

Instantaneous Power

of ann-terminal device

1

1

( ) ( ) ( )n

j j

j

p t v t i t

=

=

1 2Energy entering from time to :T TD

D

k

n

1

2

n-1

1i

2i

ki

1ni

1v

2v kv

1nv


27/27

Tellegens Theorem has many deep

applications. For this course, it can be used to

check whether your answers in homework

problems, midterm and final exams are

correct.

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