ECEN 301Discussion #7 – Node and Mesh Methods1 DateDayClass No. TitleChaptersHW Due date Lab Due...

Discussion #7 – Node and Mesh Methods

ECEN 301 1

Date Day ClassNo.

Title Chapters HWDue date

LabDue date

Exam

24 Sept Wed 7 Network Analysis 3.1 – 3.3

25 Sept Thu LAB 2

26 Sept Fri Recitation HW 3

27 Sept Sat

28 Sept Sun

29 Sept Mon 8 Network Analysis 3.4 – 3.5

NO LAB

30 Sept Tue

1 Oct Wed 9 Equivalent Circuits 3.6

Schedule…


ECEN 301 2

Learning is good only when humble2 Nephi 9: 28-2928 O that cunning plan of the evil one! O the vainness,

and the frailties, and the foolishness of men! When they are learned they think they are wise, and they hearken not unto the counsel of God, for they set it aside, supposing they know of themselves, wherefore, their wisdom is foolishness and it profiteth them not. And they shall perish.

29 But to be learned is good if they hearken unto the counsels of God.


ECEN 301 3

Lecture 7 – Network Analysis

Node Voltage and Mesh Current Methods


ECEN 301 4

Network Analysis

Determining the unknown branch currents and node voltagesImportant to clearly define all relevant variablesConstruct concise set of equations

• There are methods to follow in order to create these equations

• This is the subject of the next few lectures


ECEN 301 5

Network Analysis

Network Analysis Methods:Node voltage methodMesh current methodSuperpositionEquivalent circuits

• Source transformation• Thévenin equivalent• Norton equivalent


ECEN 301 6

Node Voltage Method

Network Analysis


ECEN 301 7

Node Voltage Method Identify all node and branch voltages

ia+v2

–

vs+_

+v4

–

+ v1 – + v3 –

R2

R1 R3

R4

a b c

d

ic

ib

Node Voltages Branch Voltages

va = vs vs = va - vd

= va

vb = v2 v1 = va - vb

vc = v4 v2 = vb - vd

= vb

vd = 0 v3 = vb - vc

v4 = vc - vd

= vc


ECEN 301 8

Node Voltage Method The most general method for electrical circuit analysis

Based on defining the voltage at each node One node is selected as a reference node (often ground)

• All other voltages given relative to reference node• n – 1 equations of n – 1 independent variables (node voltages)

Once node voltages are determined, Ohm’s law can determine branch currents

• Branch currents are expressed in terms of one or more node voltages

0

0

:KCL Using

321

321

R

v

R

v

R

v

iii

bdbcab

R

iva vb

R

vR

vvi

ab

ba

+R2

–

+ R1 – + R3 –va vd

vb

vc

i1

i2

i3


ECEN 301 9

Node Voltage Method1. Label all currents and voltages (choose arbitrary orientations

unless orientations are already given)2. Select a reference node (usually ground)

This node usually has most elements tied to it3. Define the remaining n – 1 node voltages as independent or

dependent variables Each of the m voltage sources is associated with a dependent variable If a node is not connected to a voltage source then its voltage is treated

as an independent variable4. Apply KCL at each node labeled as an independent variable

Express currents in terms of node voltages5. Solve the linear system of n – 1 – m unknowns


ECEN 301 10

Node Voltage Method

Example1: find expressions for each of the node voltages and the currents

+ R2 –

is

+R1

–

+R3

–


ECEN 301 11

Node Voltage Method Example1: find expressions for each of the node

voltages and the currents

is

i1 i3

+R1

–

+R3

–

+ R2 –

i2

Node a Node b

Node c

va

vb

vc

1. Label currents and voltages (polarities “arbitrarily” chosen)

2. Choose Node c (vc) as the reference node (vc = 0)

3. Define remaining n – 1 (2) voltages• va is independent since it is not

associated with a voltage source• vb is independent

4. Apply KCL at nodes a and b (node c is not independent) to find expressions for i1, i2, i3


ECEN 301 12



32

32 0

:b Nodeat KCL

ii

ii

is

i1 i3

+R1

–

+R3

–

+ R2 –

i2

Node a Node b

Node c

va

vb

vc

4. Apply KCL to find expressions for i1, i2, i3

0

:a Nodeat KCL

21 iiis

NB: whenever a node connects only 2 branches the same current flows in the 2 branches (EX: i2 = i3)


ECEN 301 13



11 R

vi ac

is

i1 i3

+R1

–

+R3

–

+ R2 –

i2

Node a Node b

Node c

va

vb

vc

22 R

vi ab

33 R

vi bc

0

:a Nodeat KCL

21

R

v

R

vi abas

32

:b Node at KCL

R

v

R

v bab

4. Express currents in terms of node voltages


ECEN 301 14



is

i1 i3

+R1

–

+R3

–

+ R2 –

i2

Node a Node b

Node c

va

vb

vc

5. Solve the n – 1 – m equations

221

111

Rv

RRvi bas

322

111

RRv

Rv ba


ECEN 301 15

Node Voltage Method Example2: solve for all unknown currents and voltages

I1 = 10mA, I2 = 50mA, R1 = 1kΩ, R2 = 2kΩ, R3 = 10kΩ, R4 = 2kΩ

I1 R1 R4

R2

R3

I2


ECEN 301 16



I1 R1 R4

R2

R3

I2

+I1

–

–I2

+

+R4

–

+R1

–

+ R2 –

+ R3 –i3

i1

i2

i4

Node aNode b

Node c


ECEN 301 17




2. Choose Node c (vc) as the reference node (vc = 0)

3. Define remaining n – 1 (2) voltages va is independent vb is independent

4. Apply KCL at nodes a and b

+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c

va

vb

vc


ECEN 301 18



+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c

4. Apply KCL at nodes a and b

va

vb

vc

0

0

:a Nodeat KCL

3211

3211

R

v

R

v

R

vI

iiiI

ababa 0

0

:b Nodeat KCL

2432

2432

IR

v

R

v

R

v

Iiii

babab


ECEN 301 19



+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c

4. Express currents in terms of voltages

va

vb

vc

323211

11111

RRv

RRRvI ba

432322

11111

RRRv

RRvI ba


ECEN 301 20



+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c


va

vb

vc

ba

ba

ba

ba

vv

vv

vv

RRv

RRRvI

6.06.110

)1.05.0()1.05.01(10

10105101051001.0

11111

44443

323211


ECEN 301 21



+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c


va

vb

vc

ba

ba

ba

ba

vv

vv

vv

RRRv

RRvI

1.16.050

)5.01.05.0()1.05.0(50

105101051010505.0

11111

44444

432322


ECEN 301 22



Vv

Vv

b

a

86.52

57.13

+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node aNode b

Node c


va

vb

vc

ba

ba

vv

vv

1.16.050

6.06.110


ECEN 301 23



mA

R

vi a

57.1310

57.133

11

Node a


+I1

–

+R1

–

+R4

–

+ R2 –

+ R3 –

–I2

+

i3

i2

i1i4

Node b

Node c

va

vb

vc

mA

R

vi ab

65.19102

86.5257.133

22

mA

R

vi b

43.26102

86.523

44

mA

R

vi ab

93.310

86.5257.134

33


ECEN 301 34

Node Voltage Method Example5: find all node voltages

vs = 2V, is = 3A, R1 = 1Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

R2 R4

R3

is

R1

+–

vs


ECEN 301 35


vs = 2V, is = 3A, R1 = 1Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+R2

–

+R4

–

+ R3 –

+is

–

i3

i2 i4

vavc

vd

+ R1 –

+–

i1

vs

vb 1. Label currents and voltages (polarities “arbitrarily” chosen)

2. Choose Node d (vd) as the reference node (vd = 0)

3. Define remaining n – 1 (3) voltages va is dependent (va = vs) vb is independent vc is independent

4. Apply KCL at nodes b, and c


ECEN 301 36


vs = 2V, is = 3A, R1 = 1Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+R2

–

+R4

–

+ R3 –

+is

–

i3

i2 i4

vavc

vd

+ R1 –

+–

i1

vs

vb

0

0

:b Nodeat KCL

321

321

R

v

R

v

R

v

iii

bcbsb 0

0

:c Nodeat KCL

43

43

R

v

R

vi

iii

cbcs

s

4. Apply KCL at nodes b, and c


ECEN 301 37


vs = 2V, is = 3A, R1 = 1Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+R2

–

+R4

–

+ R3 –

+is

–

i3

i2 i4

vavc

vd

+ R1 –

+–

i1

vs

vb 4. Express currents in terms of voltages

33211

1111

Rv

RRRv

R

vcb

s

433

111

RRv

Rvi cbs


ECEN 301 38


vs = 2V, is = 3A, R1 = 1Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+R2

–

+R4

–

+ R3 –

+is

–

i3

i2 i4

vavc

vd

+ R1 –

+–

i1

vs

vb

Vv

Vv

c

b

17.5

62.2


1853

827

cb

cb

vv

vv


ECEN 301 39

Node Voltage Method Example6: find the current iv

Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

Is R4R2

R1

R3

Vs

+ –

iv


ECEN 301 40

Node Voltage Method Example6: find the current iv

Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+Is

–

+R4

–

+R2

–

+ R1 –

+ R3 –

Vsi3

i1

i2 i4

+ –

iv

va vbvc

vd


2. Choose Node d (vd) as the reference node (vd = 0)

3. Define remaining n – 1 (3) voltages va is independent vb is dependent

(actually both vb and vc are dependent on each other so choose one to be dependent and one to be independent)(vb = vc + Vs)

vc is independent4. Apply KCL at nodes a, and c


ECEN 301 41

Node Voltage Method

Example6: find the current iv

Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+Is

–

+R4

–

+R2

–

+ R1 –

+ R3 –

Vsi3

i1

i2 i4

+ –

iv

va vbvc

vd

0

0

:a Nodeat KCL

31

31

R

v

R

vI

iiI

acabs

s

0

0

:c Nodeat KCL

43

43

R

vi

R

v

iii

cv

ac

v

4. Apply KCL at nodes a, and c

12

21 0

:b Nodeat BUT

R

v

R

vi

iii

abbv

v

0

0

:c Nodeat KCL

4123

43

R

v

R

v

R

v

R

v

R

vi

R

v

cabbac

cv

ac


ECEN 301 42

Node Voltage Method


Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+Is

–

+R4

–

+R2

–

+ R1 –

+ R3 –

Vsi3

i1

i2 i4

+ –

iv

va vbvc

vd

4. Express currents in terms of voltages

3131

1111

Rv

Rv

RRvI cbas

432131

1111110

RRv

RRv

RRv cba


ECEN 301 43

Node Voltage Method


Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+Is

–

+R4

–

+R2

–

+ R1 –

+ R3 –

Vsi3

i1

i2 i4

+ –

iv

va vbvc

vd

Vv

Vv

Vv

c

b

a

14.2

14.5

64.5


010912

42

cba

cba

vvv

vvv3 cb vv


ECEN 301 44

Node Voltage Method


Vs = 3V, Is = 2A, R1 = 2Ω, R2 = 4Ω, R3 = 2Ω, R4 = 3Ω

+Is

–

+R4

–

+R2

–

+ R1 –

+ R3 –

Vsi3

i1

i2 i4

+ –

iv

va vbvc

vd


A

R

v

R

vi abbv

32.22

14.564.5

4

14.512


ECEN 301 45

Mesh Current Method

Network Analysis


ECEN 301 46

Mesh Current Method Write n equations of n unknowns in terms of mesh

currents (where n is the number of meshes)Use of KVL to solve unknown currents Important to be consistent with current direction

+v_

R

i

NB: the direction of current defines the polarity of the voltage

Positive voltage

+v_

R

i

Negative voltage


ECEN 301 47



ia

+v2

–

vs+_

+v4

–

+ v1 – + v3 –

R2

R1 R3

R4ib

Two meshes n = 2


ECEN 301 48



ia

+v2

–

vs+_

+v4

–

+ v1 – + v3 –

R2

R1 R3

R4ib

Two meshes n = 2

i1 (current through R1) = ia but what about i2?According to the current directions of ia and ib, and the polarity of v2:i2 (current through R2) = ia – ib


ECEN 301 49



0)(

0

:bMesh around KVL

432

432

RiRiRii

vvv

bbba

ia

+v2

–

vs+_

+v4

–

+ v1 – + v3 –

R2

R1 R3

R4ib

0)(

0

:aMesh around KVL

21

21

RiiRiv

vvv

baas

s


ECEN 301 50



ia

+v2

–

vs+_

+v4

–

+ v1 – + v3 –

R2

R1 R3

R4ib

0)(

)(

:unknowns 2 equations, 2

4322

221

RRRiRi

vRiRRi

ba

sba


ECEN 301 51

Mesh Current Method1. Label each mesh current consistently

Current directions are chosen arbitrarily unless given2. Label the voltage polarity of each circuit element

Strategically (based on current direction) choose polarity unless already given

3. In a circuit with n meshes and m current sources n – m independent equations result

Each of the m current sources is associated with a dependent variable If a mesh is not connected to a current source then its voltage is treated

as an independent variable4. Apply KVL at each mesh associated with independent

variables Express voltages in terms of mesh currents

5. Solve the linear system of n – m unknowns


ECEN 301 57

Mesh Current Method Example8: find the voltages across the resistors

Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

R2

R4

vs2+_

R1

R3

vs1+_

R5


ECEN 301 58


Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

1. Mesh current directions given2. Voltage polarities chosen and labeled3. Identify n – m (3) mesh currents

ia is independent ia is independent ic is independent

4. Apply KVL around meshes a, b, and c

+R2

–

+ R4 –

vs2+_ ia

ib

ic

+R1

– +R3

–

vs1+_

– R5 +


ECEN 301 59


Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

+R2

–

+ R4 –

vs2+_ ia

ib

ic

+R1

– +R3

–

vs1+_

– R5 +

4. Apply KVL at nodes a, b, and c

0)()(

0

:cMesh around KVL

123

123

RiiRiiRi

vvv

cbcac

0)(

0

:aMesh around KVL

522

522

RiRiiv

vvv

acas

s

0)(

0

:bMesh around KVL

141

141

RiiRiv

vvv

cbbs

s


ECEN 301 60


Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

+R2

–

+ R4 –

vs2+_ ia

ib

ic

+R1

– +R3

–

vs1+_

– R5 +

0)( 32112 RRRiRiRi cba

2252 )( sca vRiRRi

1141 )( scb vRiRRi

4. Express voltages in terms of currents


ECEN 301 61


Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

+R2

–

+ R4 –

vs2+_ ia

ib

ic

+R1

– +R3

–

vs1+_

– R5 +

Ai

Ai

Ai

c

b

a

26.11

11.17

57.13

0711540

110153.16

110403.41

cba

cb

ca

iii

ii

ii

5. Solve the n – m equations


ECEN 301 62


Vs1 = Vs2 = 110V, R1 = 15Ω, R2 = 40Ω, R3 = 16Ω, R4 = R5 = 1.3Ω

+R2

–

+ R4 –

vs2+_ ia

ib

ic

+R1

– +R3

–

vs1+_

– R5 +


V

Riv c

16.180

16)26.11(33

V

Riiv cb

75.87

15)26.1111.17(

)( 11

V

Riiv ca

40.92

40)26.1157.13(

)( 22

V

Riv b

24.22

3.1)11.17(44

V

Riv a

64.17

3.1)57.13(55


ECEN 301 63

Mesh Current Method Example9: find the mesh currents

Vs = 6V, Is = 0.5A, R1 = 3Ω, R2 = 8Ω, R3 = 6Ω, R4 = 4Ω

R2

R1

R4

R3

Is vs–+ia ib

ic


ECEN 301 64


Vs = 6V, Is = 0.5A, R1 = 3Ω, R2 = 8Ω, R3 = 6Ω, R4 = 4Ω

+R2

–

+ R1–

+R4–

+R3 –

Is vs–+ia ib

ic

1. Mesh current directions given2. Voltage polarities chosen and labeled3. Identify n – m (3) mesh currents

ia is dependent (ia = is) ia is independent ic is independent

4. Apply KVL around meshes b and c


ECEN 301 65


Vs = 6V, Is = 0.5A, R1 = 3Ω, R2 = 8Ω, R3 = 6Ω, R4 = 4Ω

+R2

–

+ R1–

+R4–

+R3 –

Is vs–+ia ib

ic

4. Apply KVL at nodes b and c

0)()(

0

:bMesh around KVL

32

32

RiiRiiv

vvv

cbbss

s

0)()(

0

:cMesh around KVL

134

134

RiiRiiRi

vvv

cscbc


ECEN 301 66


Vs = 6V, Is = 0.5A, R1 = 3Ω, R2 = 8Ω, R3 = 6Ω, R4 = 4Ω

+R2

–

+ R1–

+R4–

+R3 –

Is vs–+ia ib

ic

2332 )( RivRiRRi sscb

14313 )( RiRRRiRi scb

4. Express voltages in terms of currents


ECEN 301 67


Vs = 6V, Is = 0.5A, R1 = 3Ω, R2 = 8Ω, R3 = 6Ω, R4 = 4Ω

+R2

–

+ R1–

+R4–

+R3 –

Is vs–+ia ib

ic

Ai

Ai

Ai

c

b

a

55.0

95.0

50.0

5.1136

10614

cb

cb

ii

ii



ECEN 301 68

Equation Solver Methods

CalculatorMatrix

Cramer’s RuleBrute Force (Substitution)


ECEN 301 69

TI-89 Equation Solver

1. Press F1 Select option number 8 (Clear Home)

2. Press APPS Select option number 1 (FlashApps) Select Simultaneous Eqn Solver

Select New…

3. In the new box enter n (the number of equations and unknowns)


ECEN 301 70


4. Input the equation coefficients For the set of 3 equations and 3 unknowns from

example 8:

Input the coefficients as follows:

41.3 0 -40 110

0 16.3 -15 110

-40 -15 71 0

0711540

110153.16

110403.41

cba

cb

ca

iii

ii

ii

1

2

3

a1 a2 a3 b1


ECEN 301 71


5. Press F5 (Solve)

26.11

11.17

57.13

3

2

1

x

x

x

ECEN 301Discussion #7 – Node and Mesh Methods1 DateDayClass No. TitleChaptersHW Due date Lab Due...

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Transcript of ECEN 301Discussion #7 – Node and Mesh Methods1 DateDayClass No. TitleChaptersHW Due date Lab Due...