Ena - K-17el43 (1).docx

MEHRAN UNIVERSITY OF ENGINEERING AND TECHNOLOGY SZAB CAMPUS KHAIRPUR MIR'S

ASSIGNMENT ROLL NO:- K-17EL43 SUBJECT:- ELECTRICAL NETWORK ANALYSIS

SUBMITTED TO:- ENGR. SHAKIR ALI

D E PA R T M E N T O F E L E C T R I C A L E N G I N E E R I N G

Electrical Network Analysis

Introduction of Network Analysis and its Different Topologies:1) Circuit Elements:  

The mathematical models of a two terminal electrical devices. Completely characterized by its voltage and current relationship. Can't be subdivided into other two terminal devices.

2) Node:A point at which two or more circuit elements have common connection. 3) Node Pair:Node Pair is simply two nodes which can be identified by specifying voltage variable. 4) Branch:A single path containing one circuit element, which connect one node to any other node. It is represented by line graph. 5) Path:A set of elements that maybe cross in order without passing through same node twice. 6) Loop:A closed selected path of a network is called loop. A path that maybe started from a particular node to other nodes through branches and comes to original starting point. This is also known as closed path or circuit. 7) Mesh:  

A loop that doesn't contain any other loops within it. Any mesh is a circuit/loop but any circuit/loop may not be mesh. A mesh is one which can't be further divided into loops.

8) Network:D E PA R T M E N T O F E L E C T R I C A L E N G I N E E R I N G

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Electrical Network Analysis

The interconnection of two or more circuit elements from an electrical network/complex is called network. 9) Circuit:The network that contains at least one closed path is called circuit. Every network is a circuit but every circuit is not a network.

Topology:It is defined as the graphical representation of complicated network in order to analyze the different parameters (current, voltage and power) in any easy way, it is simply the physical layout of network. It also deals with the properties of the network which are unaffected when the network is stretched, twisted, distorted the size and shape. 

Graph:A graph corresponding to a given network is obtained by replacing all circuit elements with line.



Connected Graph:A graph in which at least one path exists between any two nodes of the graph.



Oriented Graph:A graph that has all nodes and branches numbered and also directions are given to branches.



Subgraph:If line is removed from graph, we get subgraph.



TREE of connected graph:It is circuit less subgraph of N nodes and (n-1) branches. The tree of graph of N nodes has following properties: i. It contains all nodes of graph nodes are left and isolated portion. ii. It contains (n-1) branches. iii. There is no any closed path, when n = Number of nodes and (n-1) = number of tree branches.

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Electrical Network Analysis

Example:- Draw the graph, oriented graph, branches and link for the below circuit.

Circuit 1.1

a) Graph:-

b) Oriented Graph:-

Circuit 1.1(a)

Circuit 1.1(b)

c) Tree:Total Branches = n-1 where n = number of nodes, n-1=4-1=3

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Electrical Network Analysis

a)

b)

c)



d)

Link:Number of Links = Number of branches - Number of trees. L = B - (n-1) Link physically remains, and are open through dotted lines and can be connected that time if needed. So the links for above circuit is: L = 6 - (4 - 1) L=3

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Electrical Network Analysis

a)

c) 

b)

d)

Inductor:It is current dependent source, It stores the current point acting a short circuit when fully charged.



Capacitor:It is voltage dependent source, It will act as open circuit when it is fully charged.



Resistor:It can be represented by either link or branch.



Cutsets:Since a tree includes branches that connects all the nodes of graph without following closed path, therefore a particular tree, a cut set can be formed by cutting one tree branch and maximum number of links at a time. Total number of cutsets in a graph = Total number of branches of tree. Example:- Write the Cutset Equations for below figure (1.1).

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Electrical Network Analysis

Fig(1.1) Branches = 8 Nodes = 5 Links = B-(n-1) = 4

i. ii. iii. iv.

Cutset Equation 1:Cutset Equation 2:Cutset Equation 3:Cutset Equation 4:-

Ib1 + Ib5 - Ib4 = 0 -Ib1 + Ib2+Ib6 = 0 Ib3 + Ib7 - Ib2 = 0 -Ib3 + Ib4 +Ib8 = 0

When Current entering the node then negative sign is taken as symbol, when current is leaving the node then the positive sign is taken as symbol. 

Node Transformation Equation:It is used to find the node voltages of the network. The node voltage is difference between the two paths of the graph or difference of voltage between nodes with reference node. Example:- Write the equations of node for below figure (1.2) and then write these equations in matrix form.

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Electrical Network Analysis

Fig (1.2) 1) e1 = V1 - V2 5) e5 = V1

2) e2 = V2 - V3 6) e6 = V2

3) e3 = V3 - V4 7) e7 = V3

4) e4 = V4 - V1 8) e8 = V4

[ ][] [] 1 −1 0 0 0 1 −1 0 0 0 1 −1 −1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1



e e e e e e e e

1 2 3 4

v v v v

5

1

6

2

7

3

4

8

=

Tie Set Equation:Example:- Write the equation for below figure (1.3) and write these equations in matrix

form.

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Electrical Network Analysis

Fig (1.3) i. ii. iii. iv.

Equation 1:Equation 2:Equation 3:Equation 4:-

e1 + e6 - e5 = 0 e2 + e7 - e6 = 0 e3 + e8 - e7 = 0 e4 + e5 - e8 = 0

[] e e e e e e e e

1 2 3 4

[ 

1 0 0 0

0 1 0 0

0 0 1 0

0 −1 1 0 0 0 0 −1 1 0 0 0 0 −1 1 1 1 0 0 −1

]

5 6 7

=

8

Loop Transformation Equation:In loop transformation equation, we calculate loop currents of graph. Example:- Write the equations of loop for below figure (1.4) and then write these equations in matrix form.

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Electrical Network Analysis

Fig (1.4) 1) Ib1 = I1 5) Ib5 = I4-I1

2) Ib2 = I2 6) Ib6 = I1-I2

3) Ib3 = I3 7) Ib7 = I2-I3

4) Ib4 = I4 8) Ib8 = I3-I4

[ ][] 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 −1 0 0 1 1 −1 0 0 0 1 −1 0 0 0 1 −1

I I I I

1 2 3 4

= 0

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Electrical Network Analysis

Phasors:It provides a graphic mean for representing quantities that have both magnitude and direction(angle). Phasors are specially useful for representing sine wave in terms of their amplitude and phase angle and also for the analysis of reactive circuits.

Fig (2.1)

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Electrical Network Analysis

-) Phasor of 45°

-) Phasor of 180°

Fig 2.2(a) 

Fig 2.2(b)

Phasor Representation of Sine Wave :i. A full cycle can be represented by rotation of phasor through 0° - 360° ii. Sine wave shows how phasor traces out the sine wave as it goes from 0° - 360° iii. The length of phasor is equal to peak vale of sine wave (90 ° or 270°). The angle of phasor measured from 0° is corresponding (reference) angle point on sine iv.

wave. The drawn/below phasor is a phase shift of 45°.

Fig : 2.3 Phasor Representation of Sine Wave

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Electrical Network Analysis



Conversion of Complex Number from Rectangular form to Polar form :-

a) 8+j6 Sol:c = 8+j6 c = √ 82 +62 = √ 100 = 10 c = 10. tanϴ = P/B , ϴ = tan-1(6/8) = 36.87°. b) 10-j5 Sol:c = 10-j5 c = √ 102+ 52 = √ 125 = 11.8 c = 11.8. tanϴ = P/B , ϴ = tan-1(-5/10) = -26.56°. c) -5-j10 Sol:c = -5-j10 c = √ 52+ 102 = √ 125 = 11.8 c = 11.8. tanϴ = P/B , ϴ = tan-1(-10/-5) = 63.43°.



Conversion of Complex Number from Circular function to Polar form :1. 10 < 30° Sol:C = A + jB So, A = cCosϴ = 10Cos30° = 8.66 and B = cSinϴ = 10Sin30° = 5 then C = 8.66 + j5

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Electrical Network Analysis

2. 200 < -45° Sol:C = A + jB So, A = cCosϴ = 200Cos(-45°) = 141.4 and B = cSinϴ = 200Sin(-45°) = -141.4 then C = 141.4 - j141.4 

Angular Velocity :The periodic and frequency are related to velocity of rotation of phasor. The velocity of rotation is called angular velocity and denoted by ω. When phasor rotates through 360° (2π) are complete cycle is traced out, therefore time required for phasor to go through 2π radial is the period of sine wave. Because phasor rotates through 2π radial in time equal to period (T). The angular velocity can be expressed as: ω = 2π/T or ω = 2πf When phasor is rotated at velocity (ω) ; the ωt is angle through which phasor has passed at any instant by using following relationship: ϴ = ωt The relationship between angle and time, the equation for instantaneous value of sine wave voltage can be expressed as: V = VpSinωt Problem 1:- What is the value of sine wave voltage at 3µs from positive going zero crossing when Vp = 10V and frequency is 50KHz? Sol:V = VpSinωt = VpSin2πft V = 10Sin(2 x 3.14 x 50x103 x 3x10-6) V = 10Sin(0.942) V = 10 x 0.016 V = 0.16Volts.

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Electrical Network Analysis

R-L Series Circuit :In as series R-L circuit, the current is same through both resistor as well as inductor, thus the resistor voltage is in phase with current and inductor voltage leads current by 90°. There is a phase difference of 90° between resistor voltage and inductor voltage.

Circuit 2.1 then, VS = VR + jVL This equation can be written as: VS = √ V r 2+Vl 2

and

ϴ = tan-1 ( VL/ VR )

Phasor Diagram:-

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Electrical Network Analysis

Fig 2.4(a)

Fig 2.4(b)

Fig 2.4(c) Problem:- For a series R-L Circuit, determine the magnitude of total impedance and phase angle .

Circuit 2.2 Sol:Z=

√ R 2+ Xl2

and XL = 2πfL = 2 x 3.14 x 10x103 x 2x10-3

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Electrical Network Analysis

XL = 125.6 Ω Z=

√ 10002+125.6 2

= 1007.58Ω

ϴ = tan-1 ( XL/ R ) = tan-1 ( 125.6/ 1000 ) = 7.158° ϴ = 7.158° ZT = 1007.58 < 7.158°

Impedance of Series Circuit :The Impedance of total opposition to sinusoidal current in circuit and is express in ohms.

Circuit 2.3 The Impedance of R-L series circuit is determined by resistance and inductive reactance. Total impedance in phase is sum of R and jXL. Z = R + jXL Impedance Triangle:In A.C analysis, both R and XL are treated as phasor quantities, as shown in phasor diagram.

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Electrical Network Analysis

Fig 2.5(a) While XL is appearing at 90° angle with respect to R. This relationship comes from that fact that indicates voltage leads the current and thus the resistors voltage by 90°. Since Z is a phasor sum of R and XL , its phasor representation is,

Fig 2.5(b) and the Impedance trangle is:

Fig 2.5(c) The Length of each phase represents the magnitude of quantity and ϴ is phase angle.

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Electrical Network Analysis

By pythagoras theorem, we have:

√ R 2+ Xl2

Z=

and phase angle = ϴ = tan-1 ( XL/ R ) Combining both equations, we have: Z=

√ R 2+ Xl 2 <

tan-1 ( XL/ R )

Variation of Impedance and Phase Angle with Frequency:-

Fig 2.6 The above phasor diagram shows impedance triangle and the base of this triangle represents R. The resistance is independent of frequency, So if the frequency increases or decreases, the resistance remains constant. The formula for inductive reactance = 2πfL , So if the frequency is increased or decreased, the inductive reactance will also be increase and decrease. So, in series R-L circuit above points must be remembered. Problem :- The current I = 0.2 < 0° mA. Determine the source voltage and express it also in polar form.

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Electrical Network Analysis

Circuit 2.4(a) Sol:For Inductive Reactance, XL = 2πfL = 2 x 3.14 x 10x103 x 100x10-3 = 6.28 KΩ

Circuit 2.4(b) Z = R + jXL =

√ R 2+ Xl 2

=

√(10000)2+(6280)2

Z = 11.8 KΩ. ϴ = tan-1 ( XL/ R ) = tan-1 ( 6.28/ 10 ) ϴ = 32.18°. Z = 11.8 KΩ < 32.18° As we Know, V = IZ V = 0.2 < 0° mA x 11.8 < 32.18° KΩ V = 2.36 < 32.18°

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Electrical Network Analysis

-) Phasor Diagram:-

Fig 2.7

Parallel R-L Circuit:A basic parallel R-L circuit expression of total impedance is developed by:

Circuit 2.5 ZT = R < 0° + XL < 90°

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Electrical Network Analysis

ZT =

ZT =

R . X L <0 ° +90 ° X √ R2 + X L2 tan−1( RL ) R . X L <90 ° −tan −1 (

then Z =

√R + X 2

XL ) R

2 L

R . XL

and ϴ = 90° - tan-1 ( XL/ R )

√ R2 + X L2

or ϴ = tan-1 (R / XL )

Problem :- Find the total current of R-L circuit and also draw the phasor diagram.

. Circuit 2.6 Sol:IR =

Vs R

=

12 220

IR = 54.54 mA

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Electrical Network Analysis

and IL =

Vs XL

=

12 150

IL = 80 < -90° mA. I = I R + IL I = 54.5 + 80 < -90° I = 54.5 + ( 80Cos(-90°) + j80Sin(-90°) ) I = 54.5 + ( 0 + j80(-1) ) I = 54.5 - j80 So,

I=

√(54.5)2+(80)2

= 96.8 mA

and ϴ = tan-1 (IL / IR ) = tan-1 (-80 / 54.5 ) = -55.73° So,

IT = 96.8 < -55.73° mA. and ϴ = 90° - tan-1 (XL / R ) ϴ = 90° - tan-1 (150 / 220 ) = 90° - 34.286° ϴ = 55.7°

-) Phasor Diagram

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Electrical Network Analysis

Fig 2.8

Problem:- For each circuit determine magnitude and phase angle.

a) Circuit 2.7 Sol :-

As we know,

ZT =

R . X L <90 °−tan −1 (

√R + X 2

XL ) R

2 L

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Electrical Network Analysis

ZT =

100 .50< 90° −tan −1 (

50 ) 100

√ 1002+ 502

ZT =

−1 1 5000< 90 °−tan ( ) 2 111.8

ZT =

5000< 90 °−26.56 ° 111.8

ZT = 44.7 < 63.43 °

b) Circuit 2.8 Sol :-

As we know,

ZT =

ZT =

R . X L <90 °−tan −1 (

√R + X 2

XL ) R

2 L

1 2000 .1000< 90° −tan −1 ( ) 2 2 2 √ 2000 + 1000

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Electrical Network Analysis

2000000< 90 °−26.56 ° 2236

ZT =

ZT = 894.5 < 63.5 °



Susceptance and Admittance :Susceptance is the imaginary part of admittance where the real part is conductance.

Susceptance is measure in siemens. Y = G +jB where Y = admittance , G = conductance and B = susceptance.

Admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as reciprocal of Impedance. It's unit is siemens. 1 Y

R = Z and Z =

XL =

1 BL

or Y =

or BL =

1 Z

1 XL

Susceptance is reciprocal of reactance. G=

1 R

or

R=

1 G

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Electrical Network Analysis

where G = conductance For a parallel R-L circuits, BL =

and Admittance is Y =

1 X L <90 °

1 Z <±θ

or

BL = BL < -90 °

= -jBL.

or Y = Y < ∓ θ .

The admittance phasor diagram is the sum of conductance and susceptance. Y = G - jBL or

Y=

√G +B 2

2 L

.

Problem :- Determine admittance diagram of following circuit.

Circuit 2.9 Sol:G =

1 R

=

1 330 Ω

= 0.0033 Siemens.

XL = 2πfL = 2 x 3.14 x 1000 x 100x10-3 XL = 628Ω.

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Electrical Network Analysis

So, BL =

1 XL

=

1 628

= 0.00159 Siemens

Y = G - jBL = 0.0033 - j0.00159 Y=

√ 0.00332+ 0.001592

= 0.00345 Siemens.

ϴ = tan-1 (BL / G ) = tan-1 (-0.00159 / 0.0033 ) ϴ = -27.9 ° So, YT = 0.00345 < -27.9 °

Siemens.

-) Phasor Diagram

Figure 2.9

Problem:- Determine the total current and the phase angle of the given circuit.

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Electrical Network Analysis

Circuit 2.10 Sol:XL = 2πfL = 2 x 3.14 x 1.5x103 x 150x10-3 XL = 1413.9Ω.

As we know,

ZT =

ZT =

R . X L <90 ° −tan −1 (

√R + X 2

XL ) R

2 L

2200 x 1413.9< 90 °−tan−1(

1413.9 ) 2200

√ 22002 +1413.92 ZT = 1189.44Ω < 57 ° Y=

1 Z

=

1 1189.44

= 8.407 x 10-4

IT = VY = 10 x 8.407 x 10-4 = 8.407 mA.

-) Phasor Diagram

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Electrical Network Analysis

Figure 2.10

Problem:- In Circuit of figure, determine the values:

Circuit 2.11(a) a) ZT

b) IT

c) ϴ

Sol:-

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Electrical Network Analysis

a)

For Z1 ,

XL1 = 2πfL1

= 2 x 3.14 x 5x103 x 250x10-3

XL1 = 7.85KΩ. Z1 = R1 + jXL1 Z1 = (4.7 + j7.85)KΩ.

For Z2 , XL2 = 2πfL2

(1)

= 2 x 3.14 x 5x103 x 100x10-3

XL2 = 3.14KΩ. R2 = 3.3KΩ.

then G2 =

1 R2

1 3300

= 1 XL 2

and BL2 =

=

= 303 µSiemens. 1 3140

= 318 µSiemens.

BL2 = -j318µS = 318< -90 ° µS. Y2 = G2 - jBL2 = 303 - j318µS. Y2 =

√ 3032+ 3182

tan −1 (

<

−318 ) 303

Y2 = 439.6 < -46.43 ° µS. Then,

Z2 =

1 Y2

=

1 439.6<−46.43 ° µS .

Z2 = 2.28 < 46.43 ° KΩ. Convert it into rectangular form. Z2 = 2.28 ( Cos46.43 °

+ jSin46.43 ° )

Z2 = ( 1.57 + j1.65 ) KΩ. Now,

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Electrical Network Analysis

Circuit 2.11(b) ZT = Z1 + Z2 ZT = ( 4.7 + j7.85 ) + ( 1.57 + j1.65 ) ZT = ( 6.27 + j9.5) KΩ.

√ 6.272 +9.52

ZT =

<

−1

tan (

9.5 ) 6.27

ZT = 11.38 < 56.58 ° KΩ. V ZT

b)

IT =

=

c)

ϴ = 56.58 °

10< 0° 11.38 <56.58 ° K Ω

= 0.878 < -56.58 ° mA.

Problem:- Determine the voltage across each element? Sketch Voltage phasor diagram.

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Electrical Network Analysis

Circuit 2.12(a) Sol:XL1 = 2πfL1 XL1 = 628Ω

= 2 x 3.14 x 2x106 x 50x10-6 & R1 = 330Ω.

then Z1 = 330 + j628Ω Z1 =

√ 3302+ 6282

Z1 = 709.4 < 62.2 ° and XL2 = 2πfL2

−1

tan (

<

628 ) 330

Ω.

= 2 x 3.14 x 2x106 x 100x10-6

XL2 = 1.256KΩ

& R2 = 1KΩ.

then Z2 = 1000 + j1256 Ω Z2 =

√ 10002+1256 2

<

tan −1 (

1256 ) 1000

Z2 = 1600 < 51.3 ° Ω.

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Electrical Network Analysis

Now, Circuit 2.12(b) I1 = Vs / Z1 = 10 / 709.4 < 62.2 °

Ω. = 14.09 < -62.2 °

I2 = Vs / Z2 = 10 / 1600 < 51.3 ° Ω

= 6.25 < -51.3 °

mA. mA.

Now Calculate the Voltage Drop Across Each Element.    

VR1 = I1 x R1 = 14.09 < -62.2 ° mA x 330Ω = 4.64 < -62.2 ° Volts. VL1 = I1 x XL1 = 14.09 < -62.2 ° mA x 628 < 90 ° Ω = 8.85 < 27.8 ° Volts. VR2 = I2 x R2 = 6.25 < -51.3 ° mA x 1000Ω = 6.25 < -51.3 ° Volts. VL2 = I2 x XL2 = 6.25 < -51.3 ° mA x 1256 < 90 ° Ω = 7.85 < 38.9 ° Volts.

-) Phasor Diagram:-

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Electrical Network Analysis

Figure 2.11

R-C Circuit Sinusoidal Response:When sinusoidal voltage is applied to R-C Circuit, each resulting voltage drop and current in circuit is also sinusoidal with same frequency as the applied voltage.

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Electrical Network Analysis

Circuit 2.13 As shown in above circuit, resistors voltage, capacitor voltage and current are all sine waves with frequency of source. Sinusoidal response with general phase relationship phase of V R , VC and I relative to source voltage. VR and I are in phase, VR leads VC , VC lags VS , and VR and VC are at 90 ° means out of phase, and phase shift is introduced because of capacitance. The amplitude of phase relationship of voltage and current depend on the ohmic values of resistance and capacitive reactance. When circuit is purely resistive phase angle between V and I is zero. When circuit is purely capacitive, the phase angle between V and I is 90 ° with current leading the voltage. When there is combination of resistance and capacitive reactance in circuit, the phase angle between applied voltage and IT is same where between zero and 90 ° , depending on the relative values of resistance and reactance.

Circuit 2.14

Fig 2.12

Impedance of RC Circuit:Impedance is total opposition to sinusoidal current expressed in units is ohm. In purely capacitive circuit, the impedance is the total capacitive reactance. D E PA R T M E N T O F E L E C T R I C A L E N G I N E E R I N G

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Electrical Network Analysis

Circuit 2.15 XC = -jXC The total impedance of the circuit is: Z = R - jXC

.

Fig 2.13(a) Z=

Fig 2.13(b)

√R + X 2

2 C

. < - tan-1 (XC / R)

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Electrical Network Analysis

Problem:- For series RC circuit determine the magnitude of total Impedance and phase angle.

Circuit 2.16 Sol:XC = Now,

1 2 πf C

=

1 2 x 3.14 x 10 x 103 x 0.01 x 10−6

ZT =

√R + X

ZT =

√ 10002+15922

2

2 C

= 1592Ω.

. < - tan-1 (XC / R) . < - tan-1 (1592 / 1000)

ZT = 1880 < -57.7 ° .

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Electrical Network Analysis

Problem:- If the current expressed in polar form is I = 0.2 < 0 °

mA. Determine source

voltage and also express in polar form.

Circuit 2.17 Sol:XC =

1 2 πf C

=

1 2 x 3.14 x 103 x 0.01 x 10−6

= 15.92 KΩ, and R = 10

KΩ then Z =

√ 100002+15920 2

= 18.8 KΩ.

VS = IZ = 0.2 < 0 ° mA x 18.8 KΩ = 0.2 x 18.8 < 0 ° Volts. Vs = 3.76 < 0 ° Volts.

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Electrical Network Analysis

R-C Parallel Circuit:A basic parallel R-C Circuit is shown in fig, the expression for total impedance is given by:

Circuit 2.18 Z=

Z=

R . Xc √ R2 + Xc 2

R< 0 °+ Xc< 90° R− jXc < 90 ° +tan

-1

( XC / R)

Problem:- Determine the magnitude of total impedance and phase angle.

Circuit 2.19 Sol:-

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Electrical Network Analysis

Z=

100 .50 √1002 +502

then Z = 44.72Ω.

ϴ = 90 ° +tan

-1

( XC / R)

ϴ = 90 ° +tan

-1

( 50 / 100)

ϴ = 90 ° +tan

-1

( 0.5 )

ϴ = 90 ° +26.5° ϴ = 116.5 ° Then ZT = 44.72 < 116.5 ° Ω. 

Conductance, Susceptance and Admittance:G=

1 R< 0 °

=G<0 °

The parallel capacitive susceptance (B C) is reciprocal of capacitive reactance is expressed as: BC =

1 Xc<−90 °

= BC < 90 °

BC = jBC Admittance is the reciprocal of Impedance , Y =

1 Z <±θ

or Y = Y < ∓ θ .

Y = G + jBC

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Electrical Network Analysis

Problem:- Determine the admittance of circuit given, Sketch admittance phasor diagram.

Circuit 2.20 Sol:1 R

G=

XC =

=

1 330

1 2 π fC

= 0.0030 Siemens.

then BC = 2 π fC

and BC = 2 x 3.14 x 103 x 0.2x 10-6 BC = 0.001256 Siemens. Y = G + jBC Y = 0.0030 + j0.001256 Y=

√ 0.00302+ 0.0012562

Y = 0.00325 < 22.78 ° Now,

Z=

1 Y

=

< tan

-1

( 0.001256 / 0.0030 )

Siemens.

1 0.00325< 22.78°

Z = 307.69 < -22.78 ° Ω. Ans.

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Electrical Network Analysis



Admittance Diagram:-

Fig 2.14

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Electrical Network Analysis

Problem:- Determine the total current of fig. Sketch the general phasor diagram.

Circuit 2.21 Sol:As , Y = 0.00325 < 22.78 °

Siemens.

and IT = VY IT = 10 < 0 °

x 0.00325 < 22.78 °

IT = 0.0325 < 22.78 ° A 

Phasor Diagram:-

Fig 2.15

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Electrical Network Analysis

Problem:- Find the total impedance, current and the phase angle by which IT leads VS.

Circuit 2.22 Sol:For ZT ,

XC1 =

1 2 π f C1

then XC1 = 318.3Ω.

XC2 =

1 2 π f C2

then XC2 = 636.6Ω.

As R1 and C1 are in series, then: Z1 = R - jXC1 Z1 = 1000 - j318.3 Ω And R2 and C2 are in parallel, then: G2 =

1 R2

BC2 =

1 XC2

1 680

=

=

1 636.6

= 0.01478 Siemens.

= 0.001578 Siemens.

Y2 = G + jBC2 Y2 = 0.01478 + j0.001578 Y2 =

√ 0.001472 +0.001572

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Electrical Network Analysis

Y2 = 0.00215 < 46.8 ° Z2 =

1 Y2

=

1 0.00215< 46.8 °

= 465.8 < -46.8 °

Converting it into rectangular form: Z2 = 465.8 ( Cos(-46.8 ° ¿

+ jSin(-46.8 ° ¿

)

Z2 = 318 - j339.5 Ω ZT = Z1 + Z2 ZT = 1000 - j318.3 + 318 - j339.5 ZT = 1318 - j657.8 Ω ZT = 1473 > -26.5 ° IT = VY =

V ZT

IT = 6.78 < 26.5 °

=

10 1473>−26.5 ° mA

The total current leads the Capacitor applied voltage by 26.53 °

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Electrical Network Analysis

R-L-C Series Circuit:A series RLC circuit shown in below fig that contains resistance, inductance and capacitance.

Circuit 2.23 XC has the opposite effect that causes the current to leads the voltage. XL causes the total current to lag the applied voltage, Hence XL and XC tends to offset each other. XT = | XL - XC |

(1)

Eq (1) means the absolute value of difference of two reactance's.  

XL > XC XC > XL

( Predominantly inductive reactance) ( Predominantly capacitive reactance) Z = R + j(XL - XC) Z = √ R 2+ X T 2 < tan-1 ( XT / R )

Problem:- Find the total impedance of series RLC circuit.

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Electrical Network Analysis

Circuit 2.24 Sol:XL = 2 π f L = 2 x 3.14 x 100 x 10x10-3 XL = 6.28Ω and XC =

1 2 π fC

=

1 −6 2 x 3.14 x 100 x 500 x 10

XC = 3.18Ω and R = 5.6Ω 2 then Z = √ R + X T 2 < tan-1 ( XT / R ) Z = √ 5.62+ 3.12 < tan-1 ( 3.1 / 5.6 ) Z = 6.4 < 28.9 °

Ω

Problem:- Find the total impedance of the circuit and drop across each element.

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Electrical Network Analysis

Circuit 2.25 Sol:XT = XL - XC = 25Ω - 60Ω = -35Ω Z = R + jXT = 75 - j35 Ω Z=

√R + X

Z=

√ 752+ 352

2

2

< tan-1 ( XT / R )

T

< tan-1 ( -35 / 75 )

Z = 82.76 < -24.9 ° I =

Vs ZT

10 82.76<−24.9°

=

I = 0.12 < 24.9 ° a) VR = IZ = 0.12 < 24.9 ° VR = 9 < 24.9 ° b)

A x 75

Volts.

VL = IXL = 0.12 < 24.9 ° VL = 3 < 115 °

VC = 7.2 < -65 ° 

x 25 < 90 °

Volts.

VC = IXC = 0.12 < 24.9 °

c)

Ω

x 60 < -90 °

Volts.

Impedance of Parallel R-L-C Circuit:-

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Electrical Network Analysis

Circuit 2.26 A parallel R-L-C circuit as shown in fig, The total Impedance can be calculated by using the sum of reciprocals method. 1 Z Z=

=

1 R< 0 °

=

1 X L <90 °

1 1 1 1 + + R<0 ° X L < 90 ° X C <−90 °

=

1 X C <−90°

or Conductance, Susceptance and Admittance. 1 0° -) G = R< 0 ° = G < 1 -) BC = X <−90° = jBC C 1 -) BL = = -jBL X L <90 ° 1 and Y = or Y = Y < ∓ θ . Z <±θ or Y = G + jBC - jBL

Problem:- Find the total impedance in polar form.

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Electrical Network Analysis

Circuit 2.27 Sol:1 X L <90 °

1 Z

=

1 R< 0 °

1 Z

=

1 100

1 Z

= 0.01 - 0.01j + 0.02j

1 Z

= 0.01 + 0.01j

=

=

1 100<90 °

1 √0.01 +0.012

Z

=

Z

= 70.7 < - tan-1 ( 1 )

Z

= 70.7 < - 45 °

2

=

=

1 X C <−90° 1 50<−90 °

< - tan-1 ( 0.01 / 0.01 )

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Electrical Network Analysis

Circuit 2.28 Sol:G=

1 10< 0 °

BL =

1 5< 90 °

BC =

1 10<−90 °

= 0.1 < 0 ° Siemens.

= -j0.2 Siemens.

= j0.1 Siemens

Y = G + jBC - jBL Y = 0.1 + j0.1 - j0.2 Y = 0.1 - j0.1 Converting rectangular form into polar form, we get: Y = 0.1414 < - 45 °

Siemens.

Problem:- Find the total current and current across each element.

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Electrical Network Analysis

Circuit 2.29 Sol:IR =

VS Z

=

5 2.2

IXL =

VS XL

=

5 10< 90 °

IXC =

VS XC

=

= 2.27 A

5 5 ←90°

= 0.5 < - 90 ° = -j0.5 A

= 1 < 90 ° = 1j A

Now Total Current: IT = IR + IXL + IXC IT = 2.27 -j0.5 + 1j IT = 2.27 + j0.5.

Problem:- For the Series R-L-C Circuit, Determine:

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Electrical Network Analysis

a) Total Impedance b) Voltage Drop across each element c) Phasor Diagram d) Prove the total sum of voltages as source voltage.

Circuit 2.30(a) Sol:a) For Total Impedance:-

RT = 6Ω + 4Ω = 10Ω , LT = 0.05H + 0.05H = 0.1H and CT =

1 200 µ F

+

1 200 µ F

= 100µF

then XL = 2πfL = ωL = 377 x 0.1 = 37.7Ω XL = 37.7Ω and XC =

1 2 πfC

XC = 26.53Ω

=

1 ωC

=

1 377 x 100 x 10−6

= 26.53Ω

then:

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Electrical Network Analysis

Circuit 2.30(b) ZT = R < 0° + XL < 90° + XC < -90° ZT = 10Ω + j37.7Ω - j26.53Ω ZT = 10Ω + j11.17Ω ZT =

√ 102+ 11.172

< tan-1 ( 11.17 / 10 )

ZT = 15Ω < 48.16° b) For Voltage Drop Across Each Element:IT =

E ZT

=

20V 15 Ω< 48.16 °

IT = 1.33A < - 48.16° 

VR = IR = 1.33A < - 48.16° x 10Ω VR = 13.30 V < - 48.16



VL = IXL = 1.33A < - 48.16° x 37.7Ω < 90° VL = 50.14 V < 41.84°



VC = IXC = 1.33A < - 48.16° x 26.53Ω < - 90° VC = 35.28V < -138.16°

c) Phasor Diagram:-

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Electrical Network Analysis

Fig 2.16 d) Prove the total sum of voltages as source voltage:E = √ V R2 +(V L −V C )2 E = √ 13.32+(14.86)2 E = 20V

Problem:- For the Parallel R-L-C Circuit, Determine: a) Total Admittance and Total Impedance b) Admittance Diagram c) Total Voltage and Current through each element and their total sum. d) The equivalent series circuit as far as the terminal characteristics of the network are ???????????concerned.

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Electrical Network Analysis

d) Power Delivered to the Network.

Circuit 2.31(a) Sol:a) Total Impedance and Total Admittance RT = 10Ω || 40Ω = 8Ω LT = 6mH || 12mH = 4mH CT = 80µF + 20µF = 100µF XL = ωL = (1000) x (4mH) = 4Ω XC =

1 ωC

=

1 −6 1000 x 100 x 10

= 10Ω

Circuit 2.31(b)

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Electrical Network Analysis

YT = YR + YL + YC = G < 0° + BL < -90° + BC < 90° YT =

1 8Ω

+

1 4Ω

¿−90 ° +

1 10 Ω

¿ 90 °

YT = 0.125S < 0° + 0.25S < -90° + 0.1S < 90° YT = 0.125S - j0.25S + j0.1S YT = 0.125S - j0.15S YT =

√ 0.1252+ 0.152

< tan-1 ( -0.15 / 0.125)

YT = 0.195S < - 50.194° and

ZT =

1 YT

=

1 0.195 S<−50.194 °

ZT = 5.13Ω < 50.194°

b) Admittance Diagram

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Electrical Network Analysis

Fig 2.17 c) Total Voltage and Current through each element and their total sum. E = IZT = (12A < 0°) x (5.13Ω < 50.194°) E = 61.56 < 50.194°

IL =

V XL

=

61.56 <50.194 ° 4 Ω<90 °

IL = 15.39A < - 39.81° IC =

V XC

=

61.56 <50.194 ° 10 Ω<−90°

IC = 6.153A < 140.19°

IR =

V R

=

61.56 <50.194 ° 8 Ω<0 °

IR = 7.7A < 0° IT =

√I

2 R

+( I L −I C )2

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Electrical Network Analysis

IT =

√ 7.72+(9.237)2

IT = 12A. d) The equivalent series circuit As,

ZT = 5.13Ω < 50.194°

, then:

ZT = 5.13Ω ( Cos(50.1°) + jSin(50.1°) ) ZT = 3.28Ω ZT =

R

then R = 3.28Ω

and

+ j3.94 Ω + jXL XL = 3.94 Ω ,

So:

XL = ωL or L = XL / ω = 3.94Ω / 1000 L = 3.94mH

Circuit 2.31(c) e) Power Delivered to the Network P = I2R = 122 x 3.28Ω P = 472.32 W

RESONANCE:-

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Electrical Network Analysis

In an electrical circuit, the condition that exists when the inductive reactance and the capacitive reactance are of equal magnitude, causing electrical energy to oscillate between the magnetic field of the inductor and the electric field of the capacitor. Here XL = XC or XL - XC = 0 as

Z=

√ R +( X 2

L

−X C )2

then Z =

√ R 2+ 0

= R

The circuit, when XL = XC and Z = R, is said to be resonance and V = VR 

Resonant Frequency:At resonance XL = XC , then 1 2πfrL = 2 πfrC 1 fr2 = or (2 π )2 LC For parallel circuit,



1 2π

√

(fr is the resonant frequency) fr = 1 R − LC L2

1 2 π √ LC

( for series circuit)

2

Effects of Series Resonance:1) When a series in R-L-C Circuit attains resonance X L = XC i.e the net reactance of circuit is zero. 2) Z = R, i.e the impedance of circuit is minimum. 3) Since Z is minimum, I = V / Z will be maximum. 4) Since I is maximum, the power dissipated would be maximum P = I2R. 5) Since V L = VC, V = VR , i.e the supply voltage is in phase with the supply current.



Q-Factor in R-L-C Series Circuit:In a case of R-L-C series circuit Q-Factor is defined as the voltage magnification of circuit at resonance. Q - Factor =

1 2 π f r CR

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Electrical Network Analysis



Comparison of Series and Parallel Resonance:Table 2.1

Description Impedance at Resonance

Series Circuit

Parallel Circuit

Minimum given by Z = R

Maximum given by Z = L/CR

Current at Resonance Resonant Frequency When f < fr

Maximum I = V/R f r=

1 2 π √ LC

Circuit is Capacitive (as the net reactance is

When f > fr Power factor at resonance Q - factor It magnifies at resonance



Minimum I =

V L /CR

√

1 1 R2 − 2 2 π LC L Circuit is Inductive ( as the

net reactance is negative)

negative) Circuit is Inductive ( as the Circuit is Capacitive (as the net reactance is positive) Unity XL / R Voltage

net reactance is positive) Unity XL / R Current

AC Power:AC Impedance is a complex quantity made up of real resistance and imaginary reactance. Z = R + jX [Ω] AC Apparent power is a complex quantity made up of real active power and





imaginary reactive power. S = P + jQ [VA] AC Active/Real Power (P):The active power is the power that is dissipated in the resistance of the load. It uses the same formula used for DC. P = I2R = V2/R [Watt] AC Reactive/Imaginary Power (Q):The reactive power is the power that is exchanged between reactive components (Inductor and capacitor). Q = I2X = V2/X Unit:- Volts-Amp-Reactive [VAR] Q is negative for capacitor and is positive for inductor.

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Electrical Network Analysis

 

AC Apparent Power (S):The apparent power is the power that is “appears” to flow to the load. S = VI = I2Z = V2Z [VA] Power Traingle:The power triangle graphically shows the relationship between real(P), reactive(Q) and apparent power(S).

POLYPHASE CIRCUITS 

Power Factor:-

The cosine angle between the voltage and current is in A.C circuit is

known as the power factor. Power factor involves the relationship between two types of power, working power and reactive power. Electrical energy is almost exclusively generated, transmitted and distributed in the form of Alternating current, therefore the question of power factor comes immediately into picture. Most of the loads (80%) in electrical distribution systems are inductive in nature and hence they have low power factor which is highly undesirable. Low power factor causes an increase in reactive current, resulting and additional loss of active power in all elements of power systems. It is important to have power factor as close as to unity as possible.

KW KVA Three Phase System:- The system which has three phases, i.e., the current will pass Cos Ф =



through the three wires, and there will be one neutral wire for passing the fault current to the earth is known as the three phase system. In other words, the system which uses three wires for generation, transmission and distribution is known as the three phase system. The three phase system is also used as a single phase system if one of their phase and the

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Electrical Network Analysis

neutral wire is taken out from it. The sum of the line currents in the 3-phase system is equal to zero, and their phases are differentiated at an angle of 120º. The three-phase system has four wire, i.e., the three current carrying conductors and the one neutral. The cross section area of the neutral conductor is half of the live wire. The current in the neutral wire is equal to the sum of the line current of the three wires and consequently equal to √3 times the zero phase sequence components of current. The three-phase system has several advantages like it requires fewer conductors as compared to the single phase system. It also gives the continuous supply to the load. The three-phase system has higher efficiency and minimum losses.

Star or Wye(Y) Connection:In this method of interconnection, the similar ends say, 'star' ends of three coils are joined together at point N as shown in below figure (3.1).

Fig 3.1 The point N is known as star point or neutral point. The three conductors meeting at point N are replaced by a single conductor known as neutral conductor, and this above system is four wire - 3 phase system as shown in figure (a). And these conductors are balanced and having exactly equal magnitude but are 120° out of phase with each other. Hence their vector sum is zero.

i.e IR + IY + IB = 0 .

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Electrical Network Analysis

These three currents (IR, IY, IB) in three phase system, At no instant will all the three currents flow in the same direction either outwards or inwards.



Fig 3.2 Voltages and Currents in Star Connection:The voltage induced in each winding is called the phase voltage (V ph) and current in each phase is known as phase current (I ph) , and the voltage available between any pair of terminals is called line voltage (VL) and current flowing in each line is called line current (IL).

Fig 3.3 Line Voltage ( VRY) = ER - EY VYB = EY - EB VBR = EB - ER If ER = EY = EB = Eph , then : 60 ° ¿ VRY = 2 x Eph x Cos( 2 √3 VRY = 2 x Eph x 2

(L1 & L2) (L2 & L3) (L3 & L1)

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VRY =

√ 3 Eph

Now VRY = VYB = VBR = VL Hence VL = √ 3 Eph

It is noted from figure (3.3) that : 1) Line voltages are 120 ° apart. 2) Line voltages are 30 ° ahead of their respective phase voltages. 3) Angle between line currents and corresponding line voltages is ( 30 °

+Ф)



with current lagging. Line currents and phase currents:a. Current in Line 1 = IR b. Current in Line 2 = IY c. Current in Line 3 = IB IR = IY = IB = Iph So IL = Iph Power:Total active power = 3 x phase power. P = 3Vph Iph CosФ VL As , Vph = and Iph = IL √3 VL Then P = 3 IL CosФ √3 So , P = √ 3 VL IL CosФ



Problem:- A balance star connected load of (8+j6) Ω per phase is connected to a



balanced 3 phase 400V supply. Find the line current, power factor, power and total voltampere. Sol:Z = 8+j6 Ω then Zph =

√ 82 +62

and Zph = 10Ω

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Electrical Network Analysis

VL √3 V ph Z ph

400 = 231 Volts. √3 231 Iph = = = 23.1 A and Iph = IL = 23.1 A 10 R 8 Power Factor = = = 0.8 Z 10 P = √ 3 VL IL CosФ = √ 3 x 400 x 23.1 x 0.8 = 12800W = 12.8 kW. S = √ 3 VL IL = √ 3 x 400 x 23.1 = 16000 VA = 16kVA. Vph =



=

Advantages of 3 Phase :1) It is more economical as it requires less conducting material as compared to single phase. 2) 3 phase motors give more output as compare to single phase motors. 3) 3 phase motors have uniform torques while single phase motors have pulsating torques. 4) Domestic power and industrial/commercial power can supplied from same. 5) Voltage regulation is better in 3 phase system. 6) 3 phase system has less ripples in its waveform and can be easily converted to smooth DC as compared to single phase. 7) Mostly industrial load is 3 phase, so it has variety of applications.

Delta (∆) or Mesh Connection:In this form of interconnection, the dissimilar ends of the three phase winding are joined together i.e the ‘starting’ end of one phase is joined to the ‘finishing’ end of the other phase and so on as shown in figure(3.4) below:

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Electrical Network Analysis

Fig 3.4 In other words, the three windings are joined in series to form a closed mesh. If the system is balanced, then sum of three voltages round the closed mesh is zero. This type of connection is 3 – phase 3 – wire system because it doesn’t have neutral conductor. 

Line Voltages and Phase Voltages:In Delta Connection, the voltage between any pair of lines is equal to the phase voltage of phase winding connected between the two lines considered. Since phase sequence is R Y B , the voltage having its positive direction from R to Y leads by 120 °

on that having its positive direction from Y to B. Calling the voltage between lines

1 and 2 as VRY and between lines 2 and 3 as V YB , we find that VRY leads VYB by 120 ° , Similarly VYB leads VBR by 120 ° . Let VRY = VYB = VBR = VL, Then the VL = Vph



Line Currents and Phase Currents:-

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Electrical Network Analysis

Fig 3.5 It will be seen from above figure (b) that current in each line is the vector difference of the two phase currents flowing through that line. a. Current in Line 1 is : I1 = IR - IB b. Current in Line 2 is : I2 = IY – IR c. Current in Line 3 is : I3 = IB – IY If IR = IY = Iph then current in line no 1 is, I1 = 2 Iph Cos (60 ° / 2) I1 = 2 Iph √ 3 / 2 I1 = √ 3 Iph . Similarly, I2 = IY – IR = √ 3 Iph I3 = IB – IY = √ 3 Iph So IL = √ 3 Iph It is noted from figure (b) that: 1) Line currents are 120 ° apart. 2) Line currents are 30 ° behind the respective phase currents. 3) The angle between the line currents and the corresponding line voltages is ( 30 ° + Ф ) with the current lagging. 

Power:Phase power = Vph Iph CosФ. Total Power = 3Vph Iph CosФ. As Vph = VL and Iph = IL / √ 3 Then P = 3VL x IL / √ 3 x CosФ P = √ 3 VLILCosФ

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Electrical Network Analysis

Delta – Star Transformation:-) For Delta Connection the Resistance between (1) and (2) is: R 12( R23 + R31 ) R12 + R23 + R31 -) For Star Connection the Resistance between (1) and (2) is: R1 + R 2 Since resistances are equivalent, then: R 1 + R2 =

R 12( R23 + R31 ) R12 + R23 + R31

R 2 + R3 =

R 23( R 12+ R31 ) R12 + R23 + R31

R 3 + R1 =

R 31( R12 + R23 ) R12 + R23 + R31

(1) Similarly, between terminals (2) and (3) is:

(2) Similarly, between terminals (3) and (1) is:

(3) Subtract equation (2) from (1) and add it in equation (3). R12 . R31 R23 . R31 R1 – R3 = R 12+ R23 + R31 R 12+ R23 + R31 R12 . R31 R23 . R31 R1 + R 3 = + R 12+ R23 + R31 R 12+ R23 + R31 2 R12 . R31 2R1 = R 12+ R23 + R31

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and So,

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Electrical Network Analysis

So,

R1 =

R12 . R31 R 12+ R23 + R31

R2 =

R23 . R12 R 12+ R23 + R31

R3 =

R23 . R31 R 12+ R23 + R31

(A)

(B)

(C)

Star – Delta Transformation:Dividing equation(A) by equation(B) we get, R 1 R31 R1 = or R31 =R23 ( ) R 2 R23 R2 Now divide equation(A) by equation(C) we get, R1 R 1 R12 = or R12=R23 R R 3 R23 3

( )

Put R12 and R31 in equation(A) we get,

R 1= R23

R1 R . R23 1 R2 R3

( ) ( ) ( ) ( )

R 23

R1 R + R23+ R 23 1 R3 R2

R 12 R . R2 . R 3 R 1= R 1 . R 2+ R 2 . R3 + R 1 . R 3 R 2 . R3 2 23

R23=

R1 . R 2 + R2 . R 3 + R 1 . R3 R1

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Electrical Network Analysis

So,

R12=

R 1 . R2 + R 1+ R 2 R3

(D)

R 2 . R3 + R3 R1

(E)

R 1 . R3 + R 1+ R 3 R2

(F)

R23=R2 +

R31=

TWO - PORT NETWORK 

Port:- Two associated terminals are given, the terminal pair or part suggesting port of



entry into network. Two Port:- The network have two ports of entry (or driving part) is called two-port



network. Examples:- Transmission Line, Transformer, Extension, Transistor etc. Parameters:1) Open Circuit Parameter / Z – Parameter. 2) Short Circuit Parameter / Y – Parameter. 3) Transmission Line Parameter. 4) Inverse Transmission Line Parameter. 5) Hybrid Parameter or H – Parameter. 6) Inverse Hybrid Parameter.

1) Z – Parameter:-

V = ZI V1 = Z V2

[] []

[] I1 I2

[

V1 Z 11 Z 12 = V2 Z 21 Z 22 V1 = Z11I1 + Z12I2 V2 = Z21I1 + Z22I2

] [] I1 I2

(1) (2)

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Electrical Network Analysis

From equation (1) V1 = Z11I1 + Z12I2 V 1 −Z 12 I 2 Z11 = I2 = 0 I1 V1 Z11 = I1 From equation (2) V 2 −Z 22 I 2 Z21 = I2 = 0 I1 V2 Z21 = I1 From equation (1) V 1 −Z 11 I 1 Z12 = I1 = 0 I2 V1 Z12 = I2 From equation (2) V 2 −Z 21 I 1 Z22 = I1 = 0 I2 V2 Z22 = I2

Z=

[ ] V1 I1 V2 I1

V1 I2 V2 I2

2) Y – Parameter:-

V = IZ or I = V / Z or I = VY I1 Y 11 Y 12 V1 = I2 Y 21 Y 22 V2 I1 = Y11V1 + Y12V2 I2 = Y21V1 + Y22V2 From equation (1) I 1−Y 12 V 2 Y11 = V2 = 0 V1 I1 Y11 = V1

[]

[

] [ ]

(1) (2)

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Electrical Network Analysis

From equation (2) I 2−Y 22 V 2 Y21 = V2 = 0 V1 I2 Y21 = V1 From equation (1) I 1−Y 11 V 1 Y12 = V1 = 0 V2 I1 Y12 = V2 From equation (2) I 2−Y 21 V 1 Y22 = V1 = 0 V2 I2 Y22 = V2

Y=

[ ] I1 V1 I2 V1

I1 V2 I2 V2

3) Transmission Line Parameter:-

[] V1 I1

[ ] [] [ ] [ ]

= T

V2 −I 2

V1 A B = I1 C D V1 = AV2 - BI2 I1 = CV2 - DI2 From equation (1) V 1 + B I2 A= V2 V1 A= V2

V2 −I 2

(1) (2) I2 = 0

From equation (1)

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Electrical Network Analysis

V 1 −A V 2 −I 2 −V 1 B= I2 From equation (2) I 1 +D I 2 C= V2 I1 C= V2 From equation (2) I 1−C V 2 D= −I 2 −I 1 D= I2 B=

T=

V2 = 0

I2 = 0

V2 = 0

[ ] V1 V2 I1 V2

−V 1 I2 −I 1 I2

4) Inverse Transmission Line Parameter:-

[ ] [] V2 I2

=

T

[ ] V1 −I 1

'

[

] [ ]

V2 V1 A' B' = ' ' I2 −I 1 C D ' ' V 2 = A V 1 - B I1 I2 = C' V1 – D' I1 From equation (1) ' V 2+ B I 1 ' I1 = 0 A = V1 V2 A' = V1 From equation (1) V2 = A ' V1 - B ' I1 V2 B' = I1 From equation (2)

(1) (2)

And V1 = 0. So,

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Electrical Network Analysis '

'

C = C' =

I 2+ D I 1 V1 I2 V1

I1 = 0

From equation (2) I2 = C' V1 – D' I1 −I 2 D' = I1 T'

=

And V1 = 0. So,

[ ] V2 V1 I2 V1

V2 I1 −I 2 I1

5) Hybrid Parameter:-

[] [] [] [ ] [] V1 I2

= h

V1 I2

=

I1 V2

h11 h12 h21 h22 V1 = h11I1 + h12V2 I2 = h21I1 + h22V2 V1 Then : h11 = I1 I2 And h21 = I1

So,

h=

I1 V2

and and

(1) (2) V1 h12 = V2 I2 h22 = V2

[ ] V1 I1 I2 I1

V1 V2 I2 V2

6) Inverse Hybrid Parameter:-

[] I1 V2

= g

[] V1 I2

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Electrical Network Analysis

[] I1 V2

[

] []

g 11 g 12 V1 g 21 g 22 I2 I1 = g11V1 + g12I2 V2 = g21V1 + g22I2 I1 Then : g11 = and V1 V2 And g21 = and V1

So,

=

h=

(1) (2) I1 g12 = I2 V2 g22 = I2

[ ] I1 V1 V2 V1

I1 I2 V2 I2

Interconnection of Two – Port Network:a) Cascaded Algorithm b) Series Algorithm c) Parallel Algorithm a) Cascaded Algorithm:-

-) Transmission Line Parameter (1) V 11 = I 11

[

A 1 B1 V 21 C1 D1 −I 21

-) Transmission Line Parameter (2) V 12 = I 12

[

A 2 B2 V 22 C2 D2 −I 22

[ ]

][ ]

(1)

[ ]

][ ]

(2)

[ ] [ ] [ ] [ ][ ]

Now Now further,

V 21 −I 21

V 21 −I 21

=

=

V 12 I 12

A 2 B2 V 22 C2 D2 −I 22

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Electrical Network Analysis

Then from equation (1) V 11 A 1 B1 = I 11 C1 D1 Hence, V 11 = T1 x T2 I 11 So, Or

[ [ [ [

V 1n I1 n

] [ ][ ][ ] ] [ ] ] [ ] ] [ ] A 2 B2 V 22 C 2 D2 −I 22

V 22 −I 22

V 2n −I 2 n

= T1 x T2 ……………..Tn

V 1n I1 n

= Teq

V 2n −I 2 n

-) Z – Parameter:-

Z=

[

][

Z 11 Z 12 Z +Z Z3 = 1 3 Z 21 Z 22 Z3 Z 1 +Z 2

]

b) Series Algorithm:-

[] [ ] [ ] V1 V2

=

V 11 V 21

+

V 12 V 22

+ ………… +

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[ ] V 1n V 2n

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Electrical Network Analysis

[] [ ] [ ] [] [ ] [ ] [ ] [] []

V1 I 11 I 12 = Z1 + Z2 + ………… + Zn V2 I 21 I 22 For the series network, I1 I 11 I 12 I 1n = = = ………….. I2 I 21 I 22 I 2n So,

V1 V2

= ( Z1 + Z2 + Z3 + ……… + Zn )

V1 V2

= Zeq

[ ] I 1n I 2n

[ ] [] I1 I2

I1 I2

c) Parallel Algorithm:-

[ ] [ ] [] [] [] [] V1 V2

Y So, Then,

V1 V2

[] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [] [ ] I1 I2

= =

V 11 V 21

I 11 I 21

=

I1 I2

=

I1 I2

= Y1

I 12 I 22

I1 I2

= (Y1+Y2+Y3)

I1 I2

= Yeq

+

V 11 V 21

V 12 V 22

+ Y2

+

V 12 V 22

V 13 V 23

=

= ………..

I 13 I 23

+ Y3

V 13 V 23

V1 V2

V1 V2

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Electrical Network Analysis



Problem:- Reduce the network shown in figure, calculate it’s all parameters.

Sol:For S – 1 Z11 = 3Ω + 2Ω = 5Ω Z12 = Z21 = 2Ω

Zeq1 =

[ ] 5 2 2 5

and ∆Zeq1 = 21

Z22 = 5Ω Using Algorithm, find Teq1 Z 11 ∆ Z 5 21 Z 21 Z 21 2 2 Teq1 = = Z 22 1 5 1 2 2 Z 21 Z 21

[ ] [ ]

For S – 2 1 0 1 0 Teq2 = = Y 1 2 1 For S – 3 1 Z 1 2 Teq3 = = 0 1 0 1 Now, Teq = Teq1 x Teq2 x Teq3 5 21 2 2 1 0 Teq = x 1 5 2 1 2 2

[ ] [ ] [ ] [ ]

[ ]

[ ] [ ] x

1 2 0 1

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Electrical Network Analysis

[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]

Teq =

5 2 1 2

21 2 5 2

x

[ ] 1 2 2 5

47 115 2 2 Teq = 11 27 2 2 Using Table, Calculate it’s Zeq A ∆T 47 C C 11 Zeq = = 1 D 2 C C 11 Now Yeq, D −∆ T 27 B B 115 Yeq = = −1 A −2 B B 115 Now Heq, finally B ∆T 115 D D 27 Heq = = −1 C −2 D D 27



and ∆Teq = 1

2 11 27 11

−2 115 47 115

2 27 11 27

Problem:- Solve the network and find it’s Teq.

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Electrical Network Analysis

Sol:For S – 1 (Delta Connection) Yeq1 =

[

Y A +Y C −Y C −Y C Y B +Y C

]

=

[ ] 11 30 −1 5

−1 5 9 20

1 . 8 Using Table, Converting into Zeq Y 22 −Y 12 18 8 ∆Y ∆Y 5 5 Zeq1 = = −Y 21 Y 11 8 44 5 15 ∆Y ∆Y For S – 2 (Vertical) 1 0 1 0 Teq2 = = 1 and ∆Teq2 =1. 1 Y 1 3 Using Table, we have: A ∆T C C 3 3 Zeq2 = = 1 D 3 3 C C and ∆Yeq1 =

[ ] [ ] [ ]

[ ]

[ ] [ ] [ ]

Now, Zeq1 + Zeq2 =

18 5 8 5

33 23 5 5 Zeq12 = 23 89 5 15 Now using table,

8 5 44 15

[ ]

+

[ ] 3 3 3 3

and ∆Zeq12 =

1350 75

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Electrical Network Analysis

Teq12 =

[ ] [ ] Z 11 Z 21 1 Z 21

∆Z Z 21 Z 22 Z 21

For S – 3 (Vertical) 1 0 Teq3 = Y 1

=

[ ] [ =

1 0 1 /2 1



Teq =

90 23 89 69

]

[ ] [ ] 33 23 5 23

Now, Teq12 x Teq3 =

.

33 23 5 23

78 23 119 138

90 23 89 69

and ∆Teq3 = 1

x

[

1 0 1 /2 1

]

90 23 89 69

Problem:- Solve the network and find its all parameters for the given below circuit.

Sol:For S-1 (Delta Connection)

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Electrical Network Analysis

[

Yeq1 =

[ ] ] [ ] [ ] [ ] [ ]

Y A +Y C −Y C −Y C Y B +Y C

]

=

3 2

−1

−1

4 3

For S-2 (Delta Connection) Yeq2 =

[

Y A +Y C −Y C −Y C Y B +Y C

Now, Yeq = Yeq1 x Yeq2 3 −1 2 Yeq = 4 −1 3

=

x

2 3 −1 2

2 3 −1 2

−1 2 3 3

−1 2 3 3

3 2 −4 3

−3 2 Yeq = 3 2 1 and ∆Yeq = . 4 Now, we find Zeq Y 22 −Y 12 ∆Y ∆Y Zeq = −Y 21 Y 11 ∆Y ∆Y For T – Parameter, −Y 22 −1 Y 21 Y 21 Teq = −∆ Y −Y 11 Y 21 Y 21

[ ]

=

[ ] 6 16 3

6 6

[ ] [ ]

Now H-Parameter, 1 −Y 12 Y 11 Y 11 Heq = Y 21 ∆ Y Y 11 Y 11

=

9 8 3 16

3 4 9 8

[ ] [ ] =

2 3 −8 9

1 1 6

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Electrical Network Analysis



Problem:- Calculate the Yeq of the below network.

Sol:For S – 1 (Horizontal) T1= 1 Z = 1 4 0 1 0 1

[ ][ ]

For S – 2 (Vertical) T2= 1 0 = 1 0 Y 1 4 1

[ ][ ] [ ][ ]

Now Teq12 =

1 4 . 1 0 0 1 4 1

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Electrical Network Analysis

Teq12 =

[

17 4

4 1

]

and ∆Teq12 = 1

Using Table, Convert into Y-Parameter D −∆ T 1 −1 B = 4 4 Y eq12= B −1 A −1 17 B B 4 4

[ ][ ] [ ][ ] [ ] [ ][ ] [ ] [ ] For S – 3 (Delta Connection) 5 −1 Y 11 Y 12 12 4 Y eq3 = = −1 7 Y 21 Y 22 4 12 5 Y eq3 = 12 −1 4

−1 4 7 12

1 4 Y eq =Y eq12 +Y eq = −1 4

−1 5 4 12 + 17 −1 4 4

2 Y eq = 3 −2 4

2 3 −1 2

−2 4 58 12

=

−1 4 7 12

−1 2 29 6

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Electrical Network Analysis



Problem:- Calculate it’s H-Parameter.

Sol:For S – 1 (Star Connection) Z Z 12 Z eq 1= 11 =6 4 4 10 Z 21 Z 22

[

][ ]

and ∆Zeq1 = 60-16 = 44. Using Table, we have: Z 22 −Z 12 10 −4 ∆ Z = 44 44 Y eq1= ∆ Z −Z21 Z 11 −4 6 44 44 ∆Z ∆Z

[ ][ ]

[ ]

5 22 Y eq1= −1 11

−1 11 3 22

For S – 2 (Delta Connection) Y Y 12 Y eq2= 11 = 3 −1 Y 21 Y 22 −1 5

[

][

]

and ∆Yeq2 = 14. Now, Yeq = Yeq1 + Yeq2

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Electrical Network Analysis

[ ][ [ ]

5 Y eq = 22 −1 11

71 22 Y eq = −12 11

−1 11 + 3 −1 3 −1 5 22

]

−12 11 113 22

and ∆Yeq =

7447 . 484

Now converting it in to Heq, using table. −Y 12 Y 11

[ ]

1 Y H eq = 11 Y 21 Y 11

∆Y Y 11

[ ]

22 71 H eq = −24 71

24 71 677 142

FOURIER SERIES A Fourier series is expansion of periodic functions in term of an infinite sum of Sine and Cosine.

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Electrical Network Analysis ∞

Expression: f(t) =

∞

1 a + ∑ a cos ( nx ) dx+ ∑ bn sin ( nx ) dx 2 0 n=1 n n =1 π

1 Where a0 = ∫ f ( x ) dx , 2 π −π π

an =

1 ∫ f ( x ) cos (nx )dx π −π π

1 And bn = ∫ f ( x ) sin (nx) dx π −π an and bn are well known as coefficient of Cosine and Sine wave. Basically fourier series is used to represent periodic signal in terms of cosine and sine 

wave. Periodic Signals:- Periodic signal is just a signal that repeats its pattern at same period.



Sine waves and Cosine waves depends on frequency and amplitude. Background:- Fourier Series is basically used to calculate heat equation and introduced by Jean Baptiste Joseph Fourier. Through Fourier series research, the fact was established that continuous



functions can be represented by trigonometric series. Applications:i. MP3 format uses audio compression. ii. Most audio and image like JPEG and MP3. iii. Partial Differential Equation, we use to solve higher order differential equations iv.



by method of separation of variables. Advanced noise cancellation and call phone network signal processing (Best

application). v. Control Theory finds dynamic response of function. Formulas:T

∫ Cosn ω0 tdt=0 0 T

∫ Sinnω 0 tdt =0 0 T

∫ Cosn ω0 t . Sinnω0 tdt=0 0

Now,

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Electrical Network Analysis π

1 a0 = ∫ f (t ) dt 2 π −π (+ c ) 0

π

∫ (−c ) +∫ ¿ 0

−π

1 a 0= ¿ 2π a0 =0 π

an =

1 ∫ f ( t ) cos (n ωt )dt π −π

bn =

1 ∫ f ( t ) sin( n ωt) dt π −π

π



Procedure to Calculate Fourier Coefficient:i. Multiply equation (1) by suitable factor on both sides. ii. Integrate the resulting expression terms by over the interval, if term iii.

varies from 0 to T. Make the use of different trigonometric formulas to evaluate Fourier coefficients. For T

a0 =? T

∞

∫ f (t ) dt=∫ a 0 dt +∑ 0 T

0

n=1

[

T

T

∫ an Cosn ω0 tdt +∫ b n Sinnω 0 tdt 0

0

]

∞

∫ f (t ) dt=a 0 [t ]T0 +∑ [ 0+0 ] 0 T

n=1

∫ f (t ) dt =a 0 [T −0 ] 0

T

Then , a0 =

1 ∫ f (t ) dt T 0

Now, an = ? Multiply equation by C osn ω 0 t T

T

∞

[

T

T

∫ f (t ) C osn ω 0 t dt=∫ a 0 C osn ω0 t dt +∑ ∫ an Cosn ω 0 t .C osn ω0 tdt +∫ bn Sinn ω0 t . C o 0

n=1

0

T

T

0

0

0

0

∫ f (t ) Cosnω 0 tdt=a 0 [0 ]T0 +[an ( 2t ) + bn ( 0 )T0 ]

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Electrical Network Analysis T

∫ f (t ) Cosnω0 tdt=a n [ 2t ] 0

T

Or

2 an = ∫ f (t ) Cosnω 0 tdt T 0

And bn = ? Multiply equation by sin n ω0 t T

T

∞

[

T

T

∫ f (t ) sin n ω0 t dt=∫ a0 sin n ω0 t dt +∑ ∫ a n Cosn ω0 t . sin n ω 0 tdt +∫ b n Sinn ω0 t .sin n ω 0

n=1

0

T T

∫ f (t ) sin n ω0 t dt =a0 [0]0 +[0+b n 0

0

0

t T ] 2 0

()

T

∫ f (t ) sin n ω0 t dt=bn [ 2t ] 0

T

Or



bn =

2 ∫ f ( t ) sin n ω0 t dt T 0

Problem:- For the waveform shown in the figure, find the fourier series?

Sol:V(t) = +V V(t) = -V V(t) = +V

0 ≤ t ≤ t/4 t/4 ≤ t ≤ 3t/4 3t/4 ≤ t ≤ t

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So,

a0 =? a n=? b n=? a0 =0 and bn =0 (because there is cosine wave) For an , T 2 an = ∫ f (t ) Cosnω 0 tdt T 0 vCosn ω 0 tdt t/4

3 t /4

t

∫ vCosn ω0 tdt + ∫ −vCosn ω0 tdt + ∫ ¿ 0

t /4

3t/4

2 an= ¿ T Cosn ω0 tdt t/4

3t/4

t

∫ Cosn ω 0 tdt + ∫ −Cosn ω0 tdt+ ∫ ¿ 0

t/4

a n= Sinn ω0 t n ω0 ¿ ¿ Sinn ω0 t n ω0 ¿ ¿ Sinn ω0 t n ω0 ¿ ¿ ¿ 2V an = ¿ T

Sinn ω0

3t /4

2V ¿ T

]

( 4t )−( Sinnω ( 3t4 )−Sinn ω ( 4t ))+( Sinnω t−Sinnω ( 34t )) 0

0

0

0

2V an= ¿ n ω0 T 2π ω 0=2 πf = As, T π 3π π 3π Sinn − Sinn −Sinn + Sinn ( 2 π )−Sinn 2 2 2 2 V an= ¿ nπ

()( ( )

( )) (

]

( ))

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Electrical Network Analysis

N = 1,3,5,……. (Odd)

V [1−(−1−1 ) + ( 0+1 ) ] π 4V a1 = π a1=

Similarly,

a3 =

−4 V 3π

and

Hence, v(t) = a1 cos ω 0 t v(t) =

a5 =

+ a3 cos 3 ω 0 t

4V cos ω0 t π

-

4V 5π

+ a5 cos 5 ω 0 t

4V cos 3 ω0 t 3π

+

4V 4V cos 5 ω0 t + …+ Cosn ω 0 t 5π nπ cos ω 0 t v(t) = 4 V ¿ π

-

1 cos 3 ω 0 t 3

+

1 cos 5 ω 0 t ¿ 5

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Electrical Network Analysis



Problem:- Find the response of the circuit. E = 0.318Em + 0.500Em sin ωt – 0.012Em cos 2 ωt

– 0.422Em cos 4 ωt

Sol:Here we consider only first three of them. E = 0.318Em + 0.500Em sin ωt – 0.422Em cos 4 ωt E = 63.60 +100 sin ωt – 42.4 sin(2 ωt +90 ° ) Using phasor notation circuit below,

By applying superposition theorem, we can solve this so firstly consider E0.

XL = 2πfL = 2 x 3.14 x 0 x 0.01 = 0

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Electrical Network Analysis

XL = 0 ZT = R = 6< 0 ° E 0 63.6 I0 = = =10.6 A R 6 VR0 = I0.R = 10.6 x 6 = 63.6 VR0 = 63.6 VL0 = 0 Now Average Power, P0 = I02.R = (10.6)2 x 6 P0 = 674.2 Watts. Now,

XL = ωL = 377 x 0.1 = 37.7Ω ZT = R + jXL = 6 + j37.7 ZT = 38.17 < 80.9 ° E1 70.7 = I1 = Z T 38.17<80.9 ° I1 = 1.85 < - 80.9 ° A VR1 = I1.R = 1.85 < - 80.9 ° x 6 VR1 = 11.11 < - 80.9 ° Volts. VL1 = I1.XL = 1.85 < - 80.9 ° x 37.7 < 90 ° VL1 = 69.74 < 9.10 ° Volts. P1 = I12.R = 20.5 Watts. Now,

XL = ωL = 754 x 0.1 = 75.4Ω ZT = R + jXL = 6 + j75.4 ZT = 75.6 < 85.45 °

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E2 29.9<−90 ° = Z T 75.6<85.45 ° I2 = 0.39 < - 175.45 ° A I2 =

VR2 = I2.R = 0.39 < - 175.45 ° x 6 VR2 = 2.34 < - 175.45 ° Volts. VL2 = I2.XL = 0.39 < - 175.45 ° x 75.4 < 90 ° VL2 = 29.4 < -85.45 ° Volts P2 = I22.R = 0.912 Watts. So the equations of fourier series are, I ( t )=10.6+1.85 √ 2 sin ( 377 t −80.9° ) +0.39 √ 2sin ( 754 t−175.45 ° ) +… V R ( t )=63.6+ 11.1 √2 sin ( 377 t−80.9 ° )+ 2.34 √ 2sin ( 754 t−175.45 ° ) +. . V L (t )=0+69.74 √ 2 sin ( 377 t +9.05 ° ) +29.93 √2 sin ( 754 t−85.45 ° ) +…

Now Effective value of current, voltage and power. a) Ieff = √ I 02 + I 12 + I 22 = √ 10.62+ 1.852+ 0.392 Ieff = 10.76 A b) VReff = √ 63.62 +11.12 +2.342 VReff = 64.6 V c) VLeff = √ 02 +69.742 +29.92 VLeff = 75.89 V d) Peff = Ieff2.R = 10.762 x 6 = 694.4 Watts.

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

Problem:- For the waveform determine the trigonometric fourier series.

Sol:T 1 a0 = ∫ f (t ) dt T 0 0 d ( ωt ) π

2π

∫ vSinωt d ( ωt )+∫ ¿ 0

π

1 a0= ¿ 2π π 1 a0 = [∫ vSin ωt d ( ωt ) ] 2π 0 v v π a0 = [−cos ωt ]0 = 2π π v a0 = π Now an, T 2 an = ∫ f (t)Cosn ωt dt T 0 π

∫ vSinωt .Cosn ωt d ω t 0

2 an= ¿ 2π

]

π

∫ 12 {sin ( ωt +nωt ) +sin ( ωt −nωt ) } d ωt 0

v an = ¿ π

]

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π

∫ sin ωt ( n+1 ) d ωt+∫ sin ωt ( 1−n ) d ωt 0

v an = 2π

[(

0 ] v a n= ¿ 2π −cos ( n+1 ) ωt π −cos ( 1−n ) ωt + n+ 1 1−n 0

) (

[

)] π

0

]

v −cos ( n+1 ) π 1 cos ( n−1 ) π 1 + + − 2π n+1 n+1 n−1 n−1 For n = even, Cos(n+1) π = -1 and Cos(n-1) π = -1 v 1 1 1 1 an = + − − 2 π n+1 n+ 1 n−1 n−1 v 2 2 an = − 2 π n+1 n−1 v 2 ( n−1 ) −2(n+1) an = 2 π ( n+1)(n−1) −2 v an = π (n2−1) −2 v −2 v −2 v a2 = a 4= a6 = So, , and 3π 15 π 35 π For n = odd, Cos(n+1) π = 1 and Cos(n-1) π = 1 v −1 1 1 1 a1 = + + − 2 π n+1 n+1 n−1 n−1 a1=0 , a3 =0 a5 =0 and an =

[ [

]

]

[

]

[

]

Now finally bn = ? T

bn =

2 ∫ f (t) Sinn ωt dt T 0

π

∫ vSinωt . Sinnωt d ω t 0

2 bn = ¿ 2π

]

π

∫ 12 {cos ( ωt−nωt )−cos ( ωt +nωt ) } d ωt 0

v bn= ¿ π π

π

∫ cos ωt ( 1−n ) d ωt −∫ cos ωt ( 1+ n ) d ωt 0

]

0

v bn = ¿ 2π

]

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Electrical Network Analysis

bn = bn =

v 2π

[(

sin (1−n ) ωt π sin (1+ n ) ωt − 1−n 1+ n 0

) (

[

)] π 0

]

v sin ( 1−n ) π sin (1+ n ) π − 2π 1−n 1+ n For n = 1, equation becomes infinite For n = 2,3,4,5,………. Sin(1-n) π = 0 and Sin(1+n) π Hence bn = 0 For n = 1,

=0

π

∫ sin ωt . Sinnωt d ω t 0

v b 1= ¿ π

]

π

∫ sin2 ω td ω t

] v b 1= ¿ π π ∫ 1−cos2 2 ωt d ω t 0 ] v b1= ¿ π v ω t sin2 ωt π b1 = [ − ] π 2 4 0 v b1= 2 0

Hence the fourier series, v ( t )=

V 2 vCos 2 ωt 2 vCos 4 ωt 2 vCos 6 ωt vSinωt − − − +… π 3π 15 π 35 π 2

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Electrical Network Analysis



Problem:- For the waveform find fourier series of trigonometric functions.

Sol:Since the line passes through origin, Y= mx , y = v(t) V m = perp/base = and x = ωt π V ¿ ωt Now, v(t) = ( 0 ≤ ω t ≤ π π π ≤ ω t ≤ 2π v(t) = 0

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For a0 = ? T 1 a0 = ∫ f (t ) dt T 0 2π

1 a0 = ∫ f ( t ) dt 2π 0 π

∫ ( Vπ )ωt d ω t

2π

+ ∫ 0d ωt ] 1 π a 0= ¿ 2π π ∫ ( Vπ )ωt d ω t 0 ] 1 a 0= ¿ 2π 2 π v (ω t) a0 = 2 [ ] ] 2 0 2π v a0 = 4 0

For an = ? T

2 an = ∫ v (t)Cosn ωt dt T 0 π

∫( 0

2π

V ωt . Cosnωt d ω t+ ∫ 0.Cosn ωt d ω t π π ] 2 an = ¿ 2π

)

π

∫ ωt . Cosnωt d ω t 0

a n=

V ¿ 2 π

]

Apply Integration by Parts, an =

π V d ωt [ωt Cosn ωt d ω t− [ Cosn ωt d ω t ]] ∫ ∫ d ωt ∫ π2 0

an =

V Sinnωt Sinn ωt d ω t [ωt . −∫ ] 2 n n π 0

π

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an =

V Cosn ωt [ωt . Sinn ωt+ ] 2 n nπ 0

an =

V Cosn π 1 [ωt . Sinn π + − 0+ ] 2 n n nπ

an =

V Cosn π 1 [ − ] 2 n n nπ

an =

−2 V n2 π2

( )

()

for n=1,2,3,…..

When n is even, Cosn π =1

Sinn π =0

and an =0

When n is odd, an =

−2 V n2 π2

a1=

,

−2V , π2

a3 =

−2 V 9 π2

and

a5 =

−2 V 25 π 2

For bn = ? T

2 bn = ∫ v (t)Sinn ωt dt T 0 π

∫( 0

2π

V ωt . Sinn ωt d ω t +∫ 0. Sinn ωt d ω t π π ] 2 b n= ¿ 2π

)

π

∫ ωt . Sinn ωt d ω t 0

V b n= 2 ¿ π

]

Apply Integration by Parts, bn =

V −Cosn ωt −Cosn ωt d ω t π [ωt . − ] ∫ n n π2 0

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bn =

−V Sinnωt [ωt . Cosn ωt− ] 2 n nπ 0

bn =

−V Sinnπ [πCosnπ− −(0−0)] 2 n nπ Sinn π =0

For n=1,2,3,…… bn =

−V [πCosnπ ] nπ 2

bn =

−V Cosnπ nπ

So,

b1=

V π

,

b2=

−V 2π

and b3 =

V 3π

Now, Fourier Series:

v 2v 2v 2v V V V v ( t )= − 2 cos ωt− cos 3 ωt− cos 5 ωt+ sin ωt− sin 2 ωt + sin 3 ωt 2 2 4 π π 2π 3π (3 π ) (5 π )

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

Problem:- For the waveform, determine the fourier series of trigonometric functions.

Sol:Y= mx

V ¿ ωt π Here a0 = 0, and an = 0, we will find bn because the waveform is sinusoidal.

V(t) = ( Now,

T

2 bn = ∫ v (t)Sinn ωt dt T 0 π

∫ ( Vπ ) ωt . Sinnωt d ω t −π

2 b n= ¿ 2π

]

π

∫ ωt . Sinn ωt d ω t −π

V b n= 2 ¿ π

]

Apply Integration by Parts, π

bn =

V −Cosn ωt −Cosn ωt d ω t [ωt . −∫ ] 2 n n π −π

bn =

V Cosn ωt Sinn ωt [−ωt . + ] 2 2 n π n −π

π

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−n πCosnπ + Sinnπ− ( nπCosn (−π ) + Sinn (−π ) ] V bn = 2 2 ¿ n π bn =

V [ −n πCosnπ + Sinnπ−nπCosnπ−Sinnπ ] n2 π2

bn =

V [ −2 n πCosnπ ] 2 2 n π

bn =

−2 V Cosnπ nπ

So,

b1=

2V , π

b2=

−2V , 2π

b3 =

2V 3π

and b 4=

−2 V 4π

Hence, v ( t )=

2v 2V 2V sin ωt− sin 2 ωt+ sin 3 ωt +… π 2π 3π

v ( t )=

2v 1 1 [sin ωt− sin 2 ωt + sin 3 ωt +…] π 2 3

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