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Node Voltage MethodThis 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 , , …, , andlet e1, e2 , …, en-1 be node-to-datum voltages.

Step 2. Express the current of each resistor Rj via Ohm’s law in terms of 2 node-to-datum voltages:

Step 3. Apply KCL to each node , , …, with each resistor current ij expressed in terms of and .

Step 4. Solve the (n-1) independent linear equations for e1

, e2 , …, en-1.

Step 5. Solve for the resistor currents via Eq.(1).

1 2 n-1

ji

+ -jR Ω

je+je−

jj

j

jei

Re+ −−

= (1)

1 2 n-1

je+je−

5A2A

+

-

6v

6i

2e 3e

1e

+- 3v

4v+

-5v +

-8Ω 6Ω

4Ω2Ω

+

-1v 2v

+

-

35 6i e

=3i

11

2

2ei e−

=1

23

4ei e−

=

24 8

i 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)

matrix form : 1

2

3

3 1 1

4 2 41 5

02 81 5

04 12

52

2

eee

− −

−

−

= −

(5)NodevoltageEquations

1.EXAMPLE

Recast Eqs. (2), (3), and (4) in

3 1 1

4 2 41 5

02 81 5

04 1 2

2 0d e t3 8 4

− −

−

−

∆ =

1(a) T o 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 t9 6

− −

−

∆ =

C ram er's R ule ⇒

11

115 20 2396 384

Ve ∆ = = = ∆ 2(b) T o solve for , replace colum n 2 of m atrix from (5) and calculatee

22

76 20 15.296 384

e V∆ = = = ∆

3(c) T o solve for , replace colum n 3 of m atrix from (5) and calculatee

33

93 20 18.696 384

e V∆ = = = ∆

2

3 15

4 41

2 021 5

24 1 2

7 6d e t9 6

−

− −

−

∆ =

3

3 15

4 21 5

22 81 2

04 1 2

9 3d e t9 6

−

− −

−

∆ =

( 6 )

( 7 )

( 8 )

( 9 )

(1 0 )

(1 1 )

Solving eq. (5) by C ram er's R ule (or any other m ethod) :

C ram er's R ule ⇒

C ram er's R ule ⇒

1 2 3O nce the node-to-datum voltages , , are found,e e eall resistor currents and voltages are found trivially via K V L, K C L, and O hm 's law :

1 21 17.8 7.8 , 3.92

v V i Ae e= − = = =

1 32 24.4 4.4 , 1.14

v V i Ae e= − = = =

33 32 3.4 , 2v Ve e i A= − = = −

24 415.2 15.2 , 1.9

8v V ie A= = = =

35 518.6 18.6 , 3.1

6v V ie A= = = =

16 6 23 , 5v V ie A= − = − =

V erification of SolutionT ellegen's T heApp oly :rem

6

1 1 2 2 3 3 4 41

( ) ( ) ( ) ( )j jj

v i v i v i v i v i=

= + + +∑

(7.8)(3.9) (4.4)(1.1) (3.4)( 2)= + + −

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

(15.2)(1.9) (18.6)(3.1) ( 23)(5)+ + + −

?0=

Modified Node Voltage MethodCircuit N + v1e1 e2

6V2A3Ω

4Ω

i4

i2

i1

i3

_v3v4 +-

+

_+

_v2

+1

3

2

_1

12

4ei e−

=

12 3

i e=

KCL at : 1 1 2( ) 23 4e e e−

+ =1 (1)

KCL at : 1 23

( ) 04

ie e−− + =2 (2)

For each voltage source , add an equation .j jjs j svev e+ −− =

2 6e = (3)

S u b stitu tin g (3 ) in to (1 ), w e o b ta in :

1 11

( 6) 2 63 4e e Ve−

+ = ⇒ = (4)

S u b stitu tin g (4 ) in to (2 ), w e o b ta in :

3 0i = (5)

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

αα

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

α

Step 1.

Step 2.

Step 3.1 21 2 1Solve the ( -1) + equations for , , , , , .n s s sn ie e e i i

αα −

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 “β” current sources from N and draw the reduced digraph G of the remaining pure resistor circuit. Assume G has “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:

Step 3. Form the Node voltage equation

bY

1

2

1 1 1

2 2 2

0 0 00 0 0

(1) , (2)

0 0 0

s

ss

b b b s

ii Y vii Y v

i Y v iβ

=

i

i v1where , resistance of branch .j j

j

Y R jR

=

current= algebraic of all sources node su , mmsi entering

1, 2, , -1.m n= …m

n s=Y e i (3)

where is called the .Tn b node - admittance matY AY A rix

Deriving the Node Voltage Equation

The “β ” 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

for v in (5), we obtain

s=A i i

b s=A Y v i

T=v A e

( )Tb s=A Y A e i

nY

(4)

(5)

(6)

(7)

( )n s=Y e i

Writing Node-Admittance Matrix Yn By Inspection

nY

(8)

:Node Votage Equation1

2

1

11 12 1, 1

21 22 2, 1

1,1 1,2 1, 1

1

2

1 n

sn

sn

n n n snn

iY Y YiY 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 YR

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

Off-Diagonal Elements of nY1 = sum of adm ittances of all resistorsjk j

j

Y YR

- (connected across node and node j k )

Symmetry Property:sym m etric m atri is a , i .e ., xn jk k jY Y=Y

P roof :

( )TT T T T Tn b b b n= = = =Y A Y A A Y A A Y A Y

S in ce in (1 ) is a d iag o n a l m atrix , Tb b b=Y Y Y

5A2A

+

-

6v

6i

2e 3e

1e

+- 3v

4v+

-5v +

-8Ω 6Ω

4Ω2Ω

+

-1v 2v

+

-

35 6

ei =3i

1 21 2

e ei −=

1 32 4

e ei −=

24 8

ei =

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 :

1

2

3

3 1 1

4 2 41 5

02 81 5

04 12

11

3

eee

− −

−

−

= −

(5)NodeEquations

EXAMPLE

1A

4A

3A

Mesh Current MethodThe next simplest among many circuit

analysis methods is applicable only for connected planar circuit N (with a planar digraph) 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

ˆ ˆ ˆ, , ,lm m ml l l

1R 10i

11i 13i

12i

3i

7i

2i

4i

5i

6i

9i8i

1i

2R

3R

4R

5R

6R

7R

8R

9R

10R

11R

12R

13R

1i2i

5i4i 6i

3i

ji

jR

ci

ai bi

di

ˆ ˆa bj i ii = −

Mesh Current MethodThe next simplest among many circuit

analysis methods is applicable only for connected planar circuits made of 2-terminal linear resistors and voltage 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 be measured by a volt meter, the “mesh” currents are abstract variables introduced mathematically for writing a set of equations whose solution can be used to find each resistor current ij trivially via

Where is the circulating current flowing through Rj in the same (resp., opposite) direction as the reference current ij.

2

ˆ ˆ, ,lm ml l

1

ˆ ,ml

ˆ ˆj a bi i i= −

ˆ ˆ (resp., )a bi i

1

2

3

ˆ6 0 2 2ˆ0 14 8 2ˆ2 8 10 5

iii

− − = − − −

(1)

Then calculate :

1 3 13ˆ ˆ

10i i i= − =

1 2 313 8 19ˆ ˆ ˆ, ,20 20 20

i i i= = =

2 113ˆ20

i i= =

3 1 25ˆ ˆ20

i i i= − =

4 3 211ˆ ˆ20

i i i= − =

5 28ˆ20

i i= =

6 319ˆ20

i i= =

1 162

10v i= =

2 21345

v i= =

3 2v = −

4 42285

v i= =

5 51265

v i= =

6 5v = −

MeshCurrent

Equations

Verification by Tellegen's Theorem :6

j=1j jv i =∑ ( ) ( )6 3 13 13 5 22 11 12 8 192 5

10 10 5 20 20 5 20 5 20 20 + + − + + + −

?0=

,

,

,

,

,

,

(2)

(3)

2V

+

-

6v

6i

+- 3v

4v+

-5v +

-8Ω 6Ω

4Ω2Ω

+

-1v 2v

+

-

5i3i

2i

4i

1i

3i1i

2i

+

1 3 1ˆ ˆi i i= −

2 1i i=

3 1 2ˆ ˆi i i= −

4 3 2ˆ ˆi i i= −

5 2i i=

( )1 3 1ˆ ˆ2v i i= −

2 14v i=

3 2v = −

( )4 3 2ˆ ˆ8v i i= −

5 26v i=

6 5v = −

,

,,

,,,

6 3i i=

Loop equation around mesh 1:1 2 3 3 1 1

ˆ ˆ ˆ0 2( ) 4 2 0v v v i i i− + + = ⇒ − − + − =

1 3ˆ ˆ6 2 2i i⇒ − =

Loop equation around mesh 2:3 5 4 2 3 2

ˆ ˆ ˆ0 ( 2) 6 8( ) 0v v v i i i− + − = ⇒ − − + − − =

2 3ˆ ˆ14 8 2i i⇒ − = −

5V +-

Loop equation around mesh 3:6 1 4 3 1 3 2

ˆ ˆ ˆ ˆ0 5 2( ) 8( ) 0v v v i i i i+ + = ⇒ − + − + − =

1 2 2ˆ ˆ ˆ2 8 10 5i i i⇒ − − + =

(1)

( 2 )

(3 )

5V2V

+

-

6v

6i

+- 3v

4v+

-5v +

-8Ω 6Ω

4Ω2Ω

+

-1v 2v

+

-

5i3i

2i

4i

1i

3i1i

2i

1 3ˆ ˆM esh 1 : 6 2 2i i− =

2 3ˆ ˆM esh 2 : 14 8 2i i− = −

1 2 3ˆ ˆ ˆM esh 3 : 2 8 10 5i i i− − + =

1 1 31 2ˆ ˆ ˆS olving from (1)3 3

i i i⇒ = −

2 2 34 1ˆ ˆ ˆS olving from (2)7 7

i i i⇒ = −

S ubstitu ting (4 ) and (5) in to (3) ⇒

(5 ) and (6) ⇒

319ˆ12

i A=

28ˆ

20i A=

113ˆ20

i A= (4 ) and (6) ⇒

(1 )

(2 )(3 )

(4 )

(5 )

(6 )

(7 )

(8 )

+- +

1i

2i

4i

6i

3i5i

We can redraw this circuit so that there are no intersecting branches.

1

2

3

4

1i

2i

4i

3i5i

1

2

3

46i

Hence the above circuit is planar and it is possible to formulate mesh current equations.

1i

2i

4i

6i

3i5i

We can redraw this circuit so that there are no intersecting branches.

1

2

3

4

1i

2i

4i

3i5i

1

2

3

46i

Hence the above circuit is planar and it is possible to formulate mesh current equations.

2Ω

4Ω

8Ω

6Ω5V2V

2Ω

5V

2V 6Ω

4Ω

8Ω

+

+

+

+-

+

All branch voltages and currents can be trivially calculated from e1 and i3.

121 1 1 0 , 0

4vv Ve i Ae= − = = =

12

2 2 6 , 23vv e V i A= = = =

23 3 6 , 0v V ie A= = =

14 4 6 , 2 Aev V i= − = − =

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

4

1 1 2 2 3 3 4 41

( ) ( ) ( ) ( )j jj

v i v i v i v i v i=

= + + +∑(0)(0) (6)(2) (6)(0) ( 6)(2)= + + + −

?0=

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 “α” voltage sources, the modified node voltage method would consist of (n-1)+αindependent linear equations involving (n-1)+α unknown variables

1 21 2 1, , , , , , .n s s sie e ie iα−

(n-1) node-to-datumvariables

α currentvariables

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 ⇒

1( ) ( ) 0

b t

j jj

v t i t dt−∞

=

=∑∫

for all .t

1 1

0 0 0 0 0 1 1 0 10 0 0 0 0 1 0 1 01 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 01 0 0 1 0 0 0 0

00000000

00 0 0 0 1 0 0 0 00 1 0 0 0 4 0 0 00 0 1 0 0 0 3 0 00 0 0 1 0 0 0 0 00 0 0 0 0 0 0

60

t

2 1

de

−−

−=

−

∆

−

A

1A

4 4

0 0 0 0 0 1 1 0 10 0 0 0 0 1 0 1 01 1 1 0 0 0 0 0 01 0 0 0 0 0 0 0 0

0 1 0 1 0 0 0 0 01 0 0 0 1 0 0 0 00 0 1 0 0 4 0 0 00 0 0 0 0 0 3 0 00 0

0000000060 1 0 0 0 0 0

0 0 0 0 0 0 0 0 12

det

−−

−−

−

−−

= ∆A

4A

9 9

0 0 0 0 0 0 1 1 10 0 0 0 0 0 1 0 01 1 1 0 0 0 0 0 01 0 0 1 0 0 0 0 0

0 1 0 0 1 0 0 0 01 0 0 0 0 1 0 0 00 0 1 0 0 0 4 0 00 0 0 1 0 0 0 3 00 0 0 0 1

0000

0 0 0

00006 0

0 0 0

det

0 0 120 0 0

−−

∆

−−

−−

=

−

A

9A

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= + = = =∆

( ) ( )10,19191 10,1

AA a a= +∆ ∆

11k 12k

11 126 2k k= • + •

11 1 12 1s sk v k i= • + •

11e ∆=∆

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 atp t t T> =

1means ( ) Watts of power ent ersp T1.( )flow at s into t T=D

2( ) 0 atp t t T< =

2means ( ) Watts of power lea vesp T2.flows out of( ) at t T=D

1 2Energy entering from time to :T TD2

1 21

( ) ( )T

T T TW v t i t dt− = ∫

Instantaneous Power of an n-terminal device

1

1( ) ( ) ( )

n

j jj

p t v t i t−

=

=∑

1 2Energy entering from time to :T TD

2

1 21

1

1( ) ( )

n T

T T j jTj

W v t i t dt−

−=

=∑∫

D

k

n

1

2

n-1

1i2i

ki

1ni −

1v2v kv

1nv −

Tellegen’s 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.