Node and Mesh Equations

1
Node and Mesh Equations

• NETWORKS 1 ECE 09.201.02
• 10/09/07 Lecture 9
• ROWAN UNIVERSITY
• College of Engineering
• Prof. John Colton
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Fall 2007 - Quarter One

2
Some Administrative Items

• Test 2 10/15 Monday 800AM 915AM Room 131
• Node equations
• Mesh equations
• Problem Set 6 due Tuesday 10/16

3
Problem Set 6

• Dorf and Svoboda pp 186 - 193
• 5.2-1 through 5.2-3
• 5.3-1 through 5.3-3
• 5.4-1 through 5.4-3
• 5.5-1 through 5.5-3

4
Circuit Theorems

• Source Transformations
• Superposition
• Thévenins Theorem
• Nortons Theorem
• Maximum Power Transfer

5
Source Transformations

• Allows us to replace a voltage source plus
  series resistor
• with a current source plus a parallel resistor
  or vice versa
• Often helpful in simplifying a circuit
  configuration
• Replacement does not change the element current
  or
• voltage of any other element in the circuit

6
Equivalent Sources




A voltage source vs connected in series with a
resistance Rs , and a current source is
connected in parallel with a resistance Rp are
equivalent circuits, provided that Rp Rs and vs
Rs is
7
Equivalent Sources Example




For circuit (a), KVL says vs Rs i v or vs
/Rs i v /Rs For circuit (b), KCL says is i
v/Rp For these sources to appear equivalent to
the rest of the circuit,
is vs /Rs and Rs Rp
8
Summary Source Transformations




9
Source Transformations

• For equivalence, is vs /Rs 2A
• and Rp Rs 14 O
• Note polarities

• For equivalence, vs is Rs 24 V
• and Rs Rp 12 O
• Note polarities

10
Example Using Source Transformations




11
Example Using Source Transformations


12
Example Using Source Transformations


13
Example Using Source Transformations




14
Superposition

• The response of a linear circuit to several
  inputs working
• together is equal to the sum of the responses
  to each of
• the inputs working separately

15
Superposition

• The response of a linear circuit to several
  inputs working
• together is equal to the sum of the responses
  to each of
• the inputs working separately
• For example, a linear element satisfies
  superposition
• when it satisfies
• If i1 ? v1 and i2 ? v2 then i1 i2 ? v1 v2
  

16
Superposition

• The response of a linear circuit to several
  independent sources
• working together is equal to the sum of the
  responses to each
• of the independent sources working separately
• For example, a linear element satisfies
  superposition
• when it satisfies
• If i1 ? v1 and i2 ? v2 then i1 i2 ? v1 v2
  
• In a linear circuit containing independent
  sources, the
• voltage across or current through any element
  may be
• obtained by adding algebraically all the
  individual voltages
• or currents caused by each independent source
  acting
• alone with all other independent voltage
  sources replaced
• by short circuits and all other independent
  current sources
• replaced by open circuits.

17
Example Using Superposition



Calculate the current im in the ammeter

18
Example Using Superposition



Calculate the current im in the ammeter

19
Example Using Superposition



Calculate the current im in the ammeter

im i1 i2 6/9 2(3/9) 12/9 A
20
Superposition Theorem

• So, superposition results in more circuit
  configurations to
• evaluate, but each configuration is simpler
  that the original
• circuit configuration
• What about dependent sources? If a dependent
  source is
• present, it is never deactivated, and it must
  remain in each
• reduced circuit and active during each step of
  the
• superposition algebraic summing process

21
Example Using Superposition


22
Example Using Superposition


i2 - va /3 KCL 7 i2 (va - 3 i2 )/2 7
i2 (- 6 i2 )/2 -3 i2 i2 -7/4 A
Superposition i i1 i2 3 7/4 5/4 A
23
Example Using Superposition


Calculate current i


i2 - va /3 KCL 7 i2 (va - 3 i2 )/2 7
i2 (- 6 i2 )/2 -3 i2 i2 -7/4 A
KVL 24 i1 (3 2) 3i1 i1 3 A
Superposition i i1 i2 3 7/4 5/4 A
24
Example Using Superposition


Calculate current i


i2 - va /3 KCL 7 i2 (va - 3 i2 )/2 7
i2 (- 6 i2 )/2 -3 i2 i2 -7/4 A
KVL 24 i1 (3 2) 3i1 i1 3 A
Superposition i i1 i2 3 7/4 5/4 A
25
Thévenins Theorem

• Allows us to replace part of a circuit (Circuit
  A) with a
• voltage source plus a series resistance
• Doing so will not change the element current or
  voltage of
• any element in Circuit B

26
Thévenins Theorem Process

• In general, a Thévenin equivalent circuit
  involves calculation
• of three parameters (1) open circuit voltage
  voc , short
• circuit current isc , and Thévenin resistance
  Rt
• For Rt , circuit A is formed from circuit A by
  replacing all
• independent voltage sources with short circuits
  and all
• independent current sources with open circuits.
  Dependent
• current or voltage sources are not replaced.

27
Example Using Thévenins Theorem


28
Example Using Thévenins Theorem


Find i

voc -

29
Example Using Thévenins Theorem


Find i

voc -

30
Example Using Thévenins Theorem


Find i

voc -

31
Example Using Thévenins Theorem


Find i

voc -

i 40/(8 R)
32
Example Using Thévenins Theorem


Find the Thevenin equivalent circuit

33
Example Using Thévenins Theorem


Find the Thevenin equivalent circuit
Rt

KCL (vc 10)/10 vc /40 2 0 vc 8V
voc

Rt 4 8 12
34
Example Using Thévenins Theorem


Find the Thevenin equivalent circuit
Rt

voc
KCL (vc 10)/10 vc /40 2 0 vc 8V
voc

Rt 4 8 12
35
Example Using Thévenins Theorem


Find the Thevenin equivalent circuit
Rt

voc
KCL (vc 10)/10 vc /40 2 0 vc 8V
voc

Rt 4 8 12
36
Thévenins Theorem Process

• The open circuit voltage, Thevenin resistance,
  and short circuit
• current are related by the equation
• Rt voc /
  isc
• This approach is used to calculate the Thevenin
  resistance of a
• circuit if (1) there are known dependent
  sources, (2) there may be
• dependent sources, (3) the details of the
  circuit are unknown, or
• (4) the Thevenin equivalent is obtained by
  measurement.

37
Example Using Thévenins Theorem


38
Example Using Thévenins Theorem


39
Example Using Thévenins Theorem


isc calculation
Mesh i i1 i2 20 2(i1 i2 )
6i1 6(i1 i2) 6(i2 i1 ) 10i2
0 i2 120/136 A isc
40
Example Using Thévenins Theorem


isc calculation
Rt voc / isc 12 V/ (120/136 A) 13.6 O
Mesh i i1 i2 20 2(i1 i2 )
6i1 6(i1 i2) 6(i2 i1 ) 10i2
0 i2 120/136 A isc
41
Example Using Thévenins Theorem


isc calculation
Rt voc / isc 12 V/ (120/136 A) 13.6 O
Mesh i i1 i2 20 2(i1 i2 )
6i1 6(i1 i2) 6(i2 i1 ) 10i2
0 i2 120/136 A isc
13.6 O
12 V
42
Nortons Theorem

• Allows us to replace part of a circuit with a
  current source
• plus a parallel resistance
• Doing so will not change the element current or
  voltage of
• any other element in the circuit

43
Example Using Nortons Theorem


44
Example Using Nortons Theorem



Rn 6 KO // 12 KO 4 KO

45
Example Using Nortons Theorem



Rn 6 KO // 12 KO 4 KO

isc 15 V / 12 KO 1.25 mA
46
Example Using Nortons Theorem



Rn 6 KO // 12 KO 4 KO

isc 15 V / 12 KO 1.25 mA
1.25 A
4 KO
47
Example Using Nortons Theorem


48
Example Using Nortons Theorem


Rn 3O
49
Example Using Nortons Theorem


Rn 3O
(24 0)/4 3 isc 9A
50
Example Using Nortons Theorem


Rn 3O
(24 0)/4 3 isc 9A
51
Example Using Nortons Theorem


52
Example Using Nortons Theorem


vab 0 i 5/500 10 mA isc - 10i -100 mA
53
Example Using Nortons Theorem


vOC
vab vOC -250i Mesh 500(- vOC /250) vab
vab -5V
vab 0 i 5/500 10 mA isc - 10i -100 mA
54
Example Using Nortons Theorem


vOC
vab vOC -250i Mesh 500(- vOC /250) vab
vab -5V
Rt voc/isc -5/-0.1 50O
vab 0 i 5/500 10 mA isc - 10i -100 mA
55
Example Using Nortons Theorem


vOC
vab vOC -250i Mesh 500(- vOC /250) vab
vab -5V
Rt voc/isc -5/-0.1 50O
vab 0 i 5/500 10 mA isc - 10i -100 mA
56
Maximum Power Transfer Theorem

• Describes condition under which one circuit
  transfers as
• much power as possible to another circuit

57
Maximum Power Transfer Theorem

• Describes condition under which one circuit
  transfers as
• much power as possible to another circuit

58
Maximum Power Transfer Theorem

• Describes condition under which one circuit
  transfers as
• much power as possible to another circuit

p vs/(RL Rt) 2RL dp/dRL 0 when RL Rt d
2p/dRL2 lt 0 Max power in load when RL Rt