study of node

1. Nodal and Mesh Analysis
Boylestad: Art 8.9 , P:306; Alexender: Art:3.2, P:76,
Node: A node is a junction of two or more branches,
where a branch is any combination of series elements.
Boylestad P:290.
Mesh: A mesh is a loop ,no other loop inside in it.
Boylestad
Mesh: A mesh is a loop which does not contain any
other loops within it. Alex P:88,
Active Node and Reference Node:
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2. Determinants Methods For Solving Simultaneous
Equation
Boylestad: Appendix D, P:1128-1135
Simultaneous Equation Solution using
Calculator
“Youtube”
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3. Nodal Analysis
Boylestad: Art 8.9 , P:306; Alexender: Art:3.2, P:76,
Procedure
1. Determine the number of nodes within the network.
2. Pick a reference node and label each remaining node
with a subscripted value of voltage: V1, V2 , and so on.
3. Apply KCL at each node
4. Solve the resulting simultaneous equations for the nodal
voltages.
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4. Nodal Analysis
Alexender: Example:3.1, P:78
......(1
)
......(2
)
At node 1, applying KCL
At node 2, applying KCL
Solving (1)and(2)
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5. Nodal Analysis
Write the nodal equations and find the voltage across the 2Ω resistor for the network in Figure.
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6. Nodal Analysis
Alexender: Art:3.3, P:82
Supernode:A supernode is formed by enclosing a voltage source connected between two
nonreference nodes and any elements connected in parallel with it.
At node 1,
v1 =10V..........(1)
Now, the supernode condition
v2 - v3 = 5V..........(2)
At supernode (2&3), applying KCL
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18v1 15v2  2v3  0........(3)
v1 
v2 
v3 
After solving (1), (2) and (3)

7. Nodal Analysis
Alexender: Example:3.3, P:82
Now, the supernode condition
v2 - v1 = 2V..........(1)
At supernode (1&2), applying KCL
 7  0
2v1  v2  20........(2)

v2
2 10 4 10
5  0; 
v1
 5
v1

v2
2 4 2 4

v1  v2

v2

v2  v1
 2
v1
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8. Nodal Analysis
Supernode Practice
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9. Mesh Analysis
Boylestad: Art 8.7 , P:295, P:, Alexender: Art:3.4 P:87,
Steps of Mesh Analysis
1. Assign a distinct current in the clockwise/anticlockwise
direction to each mesh.
2. Indicate the polarities for each element in each mesh.
3. Apply KVL around each closed loop.
4. Solve the equations for the assumed mesh currents.
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10. Mesh Analysis
Alexender: Example:3.5, P:90
KVL in mesh-1
Solving above two equations
KVL in mesh-2
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11. Mesh Analysis
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12. Supermesh Analysis
Alexender: Art:3.5, P:92
A supermesh results when two meshes have a current source in common.
Now, the supermesh condition
i2 - i1 = 6A.........(1)
In supermesh (2&3), applying KVL
Solving above two equations
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13. Supermesh Analysis
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14. Network Theorems
Boylestad: Ch:9 , P:345, P:, Alexender: Chap:4 P:119,
1. Superposition Theorem
2. Thévenin’s Theorem
3. Norton’s Theorem
4. Maximum Power Transfer Theorem
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15. Superposition Theorem
Boylestad: Art:9.1 , P:345, Alexender: Art:4.3 P:123;
Statement: The current through, or voltage across, any element of a
linear network is equal to the algebraic sum of the currents or
voltages produced independently by each source.
Steps to Apply Superposition Principle:
1. Turn off all independent sources except one source. Find the
output (voltage or current) due to that active source using nodal
or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the
contributions due to the independent sources.
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16. Superposition Theorem
Alexender: Exam:4.3 P:123;
To obtain v1, we set the current
source to zero. Applying KVL to the loop.
To get v2, we set the voltage source to zero,Using
current division,
Or,
v2  (4 8) *3  8V
Therefore,
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17. Superposition Theorem
Alexender: Ex:4.5 P:126;
To get i1,
To get i2,
To get i3,
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18. Superposition Theorem
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19. Thévenin’s Theorem
Boylestad: Art:9.3 , P:353, Alexender: Art:4.5 P:132;
Statement: A linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source in series with a
resistor.
Steps to Apply:
1. Remove that portion of the network where the Thévenin
equivalent circuit is to be applied.
2. Mark the terminals of the remaining two-terminal network.
3. Calculate the open-circuit voltage (ETh) between the marked
terminals.
4. Caculate the seen resistance from the marked terminal.
5. Draw Thévenin equivalent circuit.
6. Complete the Thévenin equivalent circuit by connecting the
removed portion in step1
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20. Thévenin’s Theorem
Alexender: Ex:4.8 P:133;
At the top node, KCL gives
The Thevenin equivalent
circuit
Rth
Eth
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/Eth

21. Thévenin’s Theorem
Boylestad: Ex:9.10 P:359;
Eth
6 0.8 4
 6

Eth 10
 0
Eth

Eth
Eth
Eth=3V
Rth
Rth=2 kΩ
The Thevenin equivalent circuit
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22. Thévenin’s Theorem
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23. Norton’s Theorem
Boylestad: Art:9.3 , P:353, Alexender: Art:4.5 P:132;
Statement: A linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a current source in parallel with
a resistor.
Steps to Apply:
1. Remove that portion of the network where the Norton
equivalent circuit is to be applied.
2. Mark the terminals of the remaining two-terminal network.
3. Calculate the short circuit current (IN) between the marked
terminals.
4. Caculate the seen resistance from the marked terminal.
5. Draw Norton equivalent circuit.
6. Complete the Norton equivalent circuit by connecting the
removed portion in step1
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24. Norton’s Theorem
Alexender: Ex:4.11 P:138;
IN
= Rth
Eth
Rth
Eth=4V
Thevenin’s
circuit
Norton’s
circuit
RN
IN
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th
N
R
I 
Eth

25. Norton’s Theorem
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26. Maximum Power Transfer Theorem
Boylestad: Art:9.5 , P:367, Alexender: Art:4.8 P:142;
Statement:A load will receive maximum power from a network
when its resistance is exactly equal to the Thévenin resistance of
the network applied to the load.
The power delivered to the load
x=RL
u=RL
v=(RTh+RL)2
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27. Maximum Power Transfer Theorem
Alexender: Ex:4.13 P:143;
From,Thevenin
'stheorem
ETh VTh  22V
RTh 9
For maximum power to load
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28. Maximum Power Transfer Theorem
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29. AC Circuit
Analysis
Alexender: Sinusoids and Phasors, Chap:9 P:352;
Complex number:
Complex Algebra:
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30. AC Circuit
Analysis
Alexender: Sinusoids and Phasors, Chap:9 P:352;
A sinusoid is a signal that has the form of the sine or cosine function.
v1(t) = Vm sin ωt
v2(t) =Vm cos ωt =Vm sin (ωt+90°)
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v1(t) = Vm sin ωt =Vm cos (ωt-90°)
v2(t) =Vm cos ωt

31. AC Circuit
Analysis
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Alexender: Sinusoids and Phasors, Art:9.3 P:359;
A phasor is a complex number that represents the amplitude and phase of a
sinusoid.
Time Domain Represntation
v1(t) = Vm sin ωt
v2(t) =Vm sin (ωt+90°)
Time Domain Represntation
v2(t) =Vm cos ωt
v1(t) = Vm cos (ωt-90°)
Phasor Domain Represntation
V2  Vm0
V2 Vm90
V1 Vm0
V1  Vm90
Phasor Domain Represntation
In general,
v1(t) = Vm sin ωt
v2(t) =Vm sin (ωt±θ)
V1  Vm0
V2  Vm θ
2
Time Domain Represntation
v(t) = Vm sin ωt
Phasor Domain Represntation
V  Vm0 (Maximum Value )
V 
Vm
0(RMS Value )

32. AC Circuit
Analysis
Alexender: Art:9.4 P:359; Boylestad Art 14.3, P:589
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS: Resistor
Z=R
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I  Im ϕ
V  Vm 0
V  IR

33. AC Circuit
Analysis
3
3
Alexender: Art:9.4 P:359; Boylestad Art 14.3, P:589
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS: Inductor
Z=jωL=jXL=XL<90°
 jX L I L
VL  ωLI m sin( ωt  90 )
VL  ωLI m 90   jωLI L
V  jXLI
XL=ωL
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IL  Im 0
I  Im ϕ

34. AC Circuit
Analysis
Alexender: Art:9.4 P:359; Boylestad Art 14.3, P:589
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS: Capacitor
Z=-j/(ωC)=-jXC=XC<-90°
I C   j I C   jX C I C
VC 
iC  ωCV m sin( ωt  90 )
I C  ωCV m  90   jωCV C ( In phasor
jωC ωC
form )
1 1
V   jX C I
XC=1/(ωC)
I
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I   j I   jX
ωC
jωC
V 
I  ωCV
C
m
i  ωCVm cos(ωt  90)
In Phasor form
e j (ϕ90)
 jωCV
1
1
VC  Vm 0
V  Vm ϕ

35. AC Circuit
Analysis
Alexender: Art:9.4 P:359; Boylestad Art 14.3, P:589
PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS:
Ohm’s
Law: I
V
 Z  V  IZ
v I R cos( ωt  ϕ)
In time domain ,
ϕ* R
I  I ϕ;
Z  R
i  I m cos( ωt  ϕ)
In Phasor form
V  I Rϕ
V  IZ  I
m
m
m
m
v  ωLI cos( ωt  ϕ 90 )
In time domain ,
ϕ* jX
V  jωLI m ϕ ωLI m ϕ 90 
I  I ϕ;
V  IZ  I
Z  jX  jωL
L
i  Im cos( ωt  ϕ)
form
In Phasor
m
m L
m
i  ωCV cos( ωt  ϕ 90 )
  90 
V V ϕ
Z  jX
V  V cos( ωt  ϕ)
I 
 j
 X
Z   jX 
ωC
I  jωCV m ϕ ωCV m ϕ 90 
In time domain ,
m
C
 m
In Phasor form
V  Vm ϕ;
m
C
C
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36. AC Circuit
Analysis
Alex Art 9.7, P: 373;Boylestad Art 15.3, 15.7, P:643,664
Impedances in Series and Parallel :
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37. AC Circuit
Analysis
Alex Art 9.6, P: 372;Boylestad Art 15.3, 15.4, P:644,650
Kirchhoffs Voltage Law(KVL) :
Voltage Divider Rule(VDR) :
2
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2
1
2
1
2
1
1
* Z
* Z
Z  Z
E
V 
Z  Z
E
V 

38. AC Circuit
Analysis
Alex Art 9.7, P: 373;Boylestad Art 15.8,15.9, P:668,675
Kirchhoffs Current Law(KVL) :
Current Divider Rule(VDR) :
T
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IT Z 2

VP
) 
IT Z1 Z 2
T
P
I
Z 2
I
2
1
2
1
1
1
1 2
Z  Z
2
1
V  I (Z Z

Z Z  Z Z  Z


39. AC Circuit Analysis
ω1000rad /sec
Z1  R  30
Z2  jXL  jω
L  j*1000rad/sec*65mH j65
i(t)  i  0.24cos(1000t 38.13) A
 0.2438.13 A
Z 5053.13
I 
VS

InPhasor form
Vs 1215V
Z  Z1  Z2  Z3  30 j65 j25
 30 j40  5053.13
  j25
ω
C (1000rad/sec)*(40μ
F)

 j

 j
Z3   jXC
1215
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40. AC Circuit
Analysis
Z1  R  30
Z2  jXL  jω
L  j*1000rad/sec*65mH j65
  j25
ω
C (1000rad/sec)*(40μ
F)
InPhasor form
Vs 1215V
VR  IZ1  IR  0.2438.13*30 7.238.13V
VL  IZ2  I( jXL )  0.2438.13*( j65) 15.651.87V
VC  IZ3  I( jXC )  0.2438.13*( j25)  6128.13V

 j

 j
Z3   jXC
ω1000rad /sec
VS VR VL VC 1215V
 6128.13V
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( j25)*1215
Z
Z
15.651.87V
( j65)*1215
Z
Z
Z
38.13V
S
V 
T
C
V 
S
V 
T
L
V 
S
T
R
5053.13
5053.13
 7.2
5053.13

V  V
U sing KVL U sin gVDR
VS VR VL VC  0 Z 30*1215
3
2
1

41. AC Circuit
Analysis
Z1
Z2
Z3
Alex Ex9.11, P: 376;
ω 4rad /sec
Vo VS VR  2015(0.1774.04*60)
17.1515.66V
vo 17.15cos(4t 15.66)V
Z 116.6259.04
V 2015
  0.1774.04A
I  S
Z  Z1  Z2 Z3  60( j25) ( j20)
 60 j100 116.6259.04
 j
   j25
 j
Z2   jXC 
Z3  jXL  jω
L  j*4rad /sec*5H j20
InPhasor form
Vs  2015V
ω
C (4rad /sec)*(10mF)
Z1  R  60
17.1515.96V
Z Z  j25 j20
Vo  I3Z3 1715.96V
VDR(Z2 Z3  j100)
 0.68105.96A
I3  I  I2 (KCL)
 0.1774.040.68105.96
 0.8574.04 A
Z3I

j20*0.1774.04
I 
o
V 
Z1  Z2 Z3 60 j100
Z3)

2015* j100
VS *(Z2
2 3
2
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42. AC Circuit Analysis (Power)
Boylestad Art 14.5, P: 603;
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43. AC Circuit Analysis (Power)
Boylestad Ex 16.2, P: 715;
P  I 2
R  80 2
* 3  19200 Watt
R 1
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 VI cos(θ
V θI )  400 * 50 cos( 46.25  30)  19201 Watt
Psource
V  IZ  I2 Z2  I1Z1
 50 A136 .26 * ( j8)  400 46.25 V

44. AC Circuit Analysis (Power)
Boylestad Ex 16.4, P: 716;
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