Electrical Engineering Fundamentals_V. Del Toro

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Scilab Textbook Companion for

Electrical Engineering Fundamentals

by V. Del Toro1

Created byAditi Pohekar

Electrical EngineeringElectrical Engineering

IIT BombayCollege Teacher

Madhu N. BelurCross-Checked by

Mukul R. Kulkarni

August 10, 2013

1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the Textbook Companion Projectsection at the website http://scilab.in


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Book Description

Title: Electrical Engineering Fundamentals

Author: V. Del Toro

Publisher: Prentice - Hall International

Edition: 2

Year: 2009

ISBN: 9780132475525

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Scilab numbering policy used in this document and the relation to theabove book.

Exa Example (Solved example)

Eqn Equation (Particular equation of the above book)

AP Appendix to Example(Scilab Code that is an Appednix to a particularExample of the above book)

For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa scilab code whose theory is explained in Section 2.3 of the book.

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Contents

List of Scilab Codes 5

1 The Fundamental laws of Electrical Engineering 8

2 The circuit elements 9

3 Elementary network theory 16

4 Circuit differential equations Forms and Solutions 22

5 Circuit dynamics and forced responses 23

6 The laplace transform method of finding circuit solutions 25

7 Sinusoidal steady state response of circuits 26

9 Semiconductor electronic devices 34

11 Binary logic Theory and Implimentation 37

12 Simplifying logical functions 39

15 Magnetic circuit computations 40

16 Transformers 44

18 The three phase Induction motor 50

19 Computations of Synchronous Motor Performance 53

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20 DC machines 55

23 Principles of Automatic Control 59

24 Dynamic behaviour of Control systems 60

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List of Scilab Codes

Exa 1.1 force between two like charges in free space . . . . . . 8Exa 2.1.a Determine the current flow and voltage drop across the

resistor . . . . . . . . . . . . . . . . . . . . . . . . . . 9Exa 2.1.b Determine the current flow and voltage drop across each

resistor . . . . . . . . . . . . . . . . . . . . . . . . . . 9Exa 2.1.c Repeat parts A and B with the voltage source replaced

by a current source of 1A . . . . . . . . . . . . . . . . 10Exa 2.2 From the given list of resistors choose a suitable resistor

which can carry a current of 300mA . . . . . . . . . . 10Exa 2.3 Find the resistance of the round copper conductor hav-

ing the given specifications . . . . . . . . . . . . . . . 11Exa 2.4 Find the resistance of the round copper conductor hav-

ing the given specifications . . . . . . . . . . . . . . . 11

Exa 2.5 Find the time variation of the voltage drop appearingacross the inductor terminals . . . . . . . . . . . . . . 12

Exa 2.6 Find the time variation of the capacitor voltage . . . . 13Exa 2.7 Find A actual value of the voltage gain of the opamp

circuit B ideal value of the voltage gain C percent error 14Exa 2.8 Design a non inverting opamp circuit of voltage gain 4 14Exa 2.9 Find the input resistance of an inverting opamp circuit

with voltage gain of 4 . . . . . . . . . . . . . . . . . . 15Exa 3.1 for the given circuit calculate the current flowing from

the voltage source . . . . . . . . . . . . . . . . . . . . 16Exa 3.2 Calculate the potential difference across terminals bc . 16

Exa 3.3 Determine the equivalent series circuit . . . . . . . . . 17Exa 3.4 Value of E for which power dissipation in R5 is 15W R5

is 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

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Exa 3.8 Find current flowing through all the branches of the

circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Exa 3.12 Find A current flowing through Rl B valye of Rl forwhich the power transfer is maximum and the maximumpower . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Exa 3.13 Find the current flowing through R2 by using Nortonscurrent source equivalent circuit . . . . . . . . . . . . 21

Exa 4.3 Determine the operational driving point impedances ap-pearing at terminals ad and dg . . . . . . . . . . . . . 22

Exa 5.2 Find the expression for the current flowing through thecircuit and the total energy dissipated in the resistor . 23

Exa 6.1 Find the laplace transform of the given pulse . . . . . 25

Exa 7.1 Find the average value of the given periodic function . 26Exa 7.2 Determine the power factor and average power delivered

to the circuit . . . . . . . . . . . . . . . . . . . . . . . 26Exa 7.3 Find the expression for the sum of i1 and i2 . . . . . . 27Exa 7.4 Find the effective value of the resultant current . . . . 27Exa 7.5 Find the time expression for the resultant current. . . 28Exa 7.6 Find the value of the given expression . . . . . . . . . 29Exa 7.7 Find A value of steady state current and the relative

phase angle C magnitude and phase of voltage dropsappearing across each element D average power E power

factor . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Exa 7.8 Find the equivalent impedance appearing between pointsa and c . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Exa 7.9 Find the current which flows through branch Z3 . . . 31Exa 7.10 Find the current in the Z3 branchby using the Nodal

method . . . . . . . . . . . . . . . . . . . . . . . . . . 32Exa 7.11 Find the current flowing through Z3 by using Thevinins

theoram. . . . . . . . . . . . . . . . . . . . . . . . . . 33Exa 9.2 Find the values of self bais source resistance and drain

load resistance at Q point . . . . . . . . . . . . . . . . 34Exa 9.3 Find A midband frequency current gain of the first stage

B bandwidth of the first stage amplifier . . . . . . . . 34Exa 11.1 Determine th decimal equivalents of the binary numbers

A 101 B 11011 . . . . . . . . . . . . . . . . . . . . . . 37Exa 11.2 Determine the decimal equivalent of A octal number 432

B hexadecimal number C4F . . . . . . . . . . . . . . . 37

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Exa 11.3 Find the binary and octal equivalents of 247. . . . . . 38

Exa 15.1 Find A magneto motive force B current C relative per-miability and reluctance of each material . . . . . . . 40Exa 15.3 Find the mmf produced by the coil . . . . . . . . . . . 41Exa 15.5 B Find the magnetic force exerted on the plunger . . . 43Exa 16.1 Find A equivalent resistance and reactance referred to

both the sides B voltage drops across these in Volts andin per cent of the rated winding voltage C Repeat B forthe low voltage side D equivalent leakage impedancesreferred to both the sides . . . . . . . . . . . . . . . . 44

Exa 16.2 Compute the 6 parameters of the equivalent circuit re-ferred to the high and low sides . . . . . . . . . . . . . 46

Exa 16.3 For the transformer compute A efficiency B voltage reg-ulation . . . . . . . . . . . . . . . . . . . . . . . . . . 48

Exa 18.1 Find A input line current and power factor B developedelectromagnetic torque C horse power output D efficiency 50

Exa 19.1 Find A induced excitation voltage per phase B line cur-rent C power factor . . . . . . . . . . . . . . . . . . . 53

Exa 20.2 Caculate A electromagnetic torque B flux per pole Crotational losses D efficiency E shaft laod . . . . . . . 55

Exa 20.3 Determine the new operating speed. . . . . . . . . . . 56Exa 20.4 find A motor speed B required pulse frequency C repeat

part A and B for the given ON time to cycle time ratio 57Exa 23.1 Determine the new transfer gain and feedback factor . 59Exa 24.2 find A dynamic response of the system B position lag

error C change in amplifier gain D damping ratio andmaximum percent overshoot E output gain factor formaximum overshoot equal to 25percent . . . . . . . . 60

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Chapter 1

The Fundamental laws of

Electrical Engineering

Scilab code Exa 1.1 force between two like charges in free space

1 E 0 = 1 /( 3 6* % p i * 10 ^9 ) ; // p e r m i t i v i t y i n f r e e s pa ce2 k = 4* %pi *E0 ;

3 q 1 = 1; / / c ha r ge on t he f i r s t p a r t i c l e i n c ou lo mb s4 q 2 = 1; // c ha rg e on t he s ec on d p a r t i c l e i n co ul om bs5 d = 1 ; // d i s t a n ce b et we e n t he p a r t i c l e s i n m e te r6 F = ( q 1 * q2 ) / ( k* d ^ 2) ; / / f o r c e b et we en t h e two

p a r t i c l e s i n n ew t ons7

8 disp (F , f o r c e i n f r e e s pa ce b et w ee n t he twop a r t i c l e s i s i n Newtons i s : )

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Chapter 2

The circuit elements

Scilab code Exa 2.1.a Determine the current flow and voltage drop acrossthe resistor

1 V = 1 ; // v o l t a g e s up pl y2 R = 10; // r e s i s t a n c e i n ohms3 I = V / R / / c u r r e nt f l o w i n g t hr ou gh R4 disp ( a ) )5 disp (V , v o l t ag e a c r o s s t he r e s i s t o r ( i n v o l t s )= )

6 disp (I , c u r r e nt f l o wi n g t hr ou gh t he r e s i s t o r ( i namps) = )

Scilab code Exa 2.1.b Determine the current flow and voltage drop acrosseach resistor

1 V = 1 ; // v o l t a g e s up pl y2 R 1 = 10; // f i r s t r e s i s t a n c e i n ohms3 R 2 = 5; // r e s i s t a n c e o f t h e s ec o nd r e s i s t o r4 Vr1 = V * ( R1 /( R1 + R2 )) ; / / v o l t a g e a c r o s s R15 Vr2 = V - Vr1 ; / / v o l t a g e a c r o s s R26 I r = V r1 / R1 ; / / c u r r e nt f l o w i n g t hr ou gh R

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7

8 disp ( V r 1 , v o l t a g e a c r o s s t he f i r s t r e s i s t o r ( i nv o l t s )= )9 disp ( V r 2 , v o l t a ge a c r o s s t he s ec on d r e s i s t o r ( i n

v o l t s )= )10 disp ( I r , c u r r e nt f l o w i n g t hr ou gh t he r e s i s t o r ( i n

amps) = )

Scilab code Exa 2.1.c Repeat parts A and B with the voltage source re-

placed by a current source of 1A

1 // c a2 R 1 = 10; // f i r s t r e s i s t a n c e i n ohms3 I = 1 ; // c u r r e nt s o ur c e4 V = I *R1 ; // v o l t a ge a c r o s s R5 disp ( c a ) )6 disp (V , v o l t ag e a c r o s s t he r e s i s t o r ( i n v o l t s )= )7 disp (I , c u r r e nt f l o wi n g t hr ou gh t he r e s i s t o r ( i n

amps) = )8

9 // c b10 Vr1 = I * R1 ; // v o l t a ge a c r o s s R111 Vr2 = I * R2 ; / / v o l t a g e a c r o s s R212 disp ( c b ) )13 disp ( V r , v o l t a g e a c r o s s t he r e s i s t o r ( i n v o l t s )= )14 disp (I , c u r r e nt f l o wi n g t hr ou gh t he r e s i s t o r ( i n

amps) = )

Scilab code Exa 2.2 From the given list of resistors choose a suitable re-sistor which can carry a current of 300mA

1 R = 100; // r e s i s t a n c e i n ohms2 I = 0.3; // c u r r e nt i n amps

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3 P = I ^2 * R ; / / p ow er

4 // power s p e c i f i c a t i o n o f t h e r e s i s t o r s a v a i l a b l e i nt he s t o ck5 P a = 5;

6 P b = 7. 5;

7 P c = 10;

8

9 if Pa > P then

10 disp ( we s ho u ld s e l e c t r e s i s t o r a )11 end

12 if Pb > P then

13 disp ( we s h o u ld s e l e c t r e s i s t o r b )

14 end15 if Pc > P then

16 disp ( we s ho u ld s e l e c t r e s i s t o r c )17 end

Scilab code Exa 2.3 Find the resistance of the round copper conductorhaving the given specifications

1 L = 1 ; // l e ng t h o f t he c op pe r w ir e i n m et er s2 A = 1 * 10^ -4; // c r o s s s e c t i o n a l a re a o f t h e w i re

i n m et er s q ua r e3 rho = 1 .7 24 * 10^ - 8; / / r e s i s t i v i t y o f c o p p e r i n ohm

me t e r4 R = rho *L / A ; // r e s i s t a n c e o f t he w ir e i n ohm5

6 disp (R , r e s i s t a n c e o f t he w i re ( i n ohms )= )

Scilab code Exa 2.4 Find the resistance of the round copper conductorhaving the given specifications

1 / / 1 i n c h e s = 0 . 0 2 5 4 m e te r s

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2 // 1 f o o t = 0 . 3 04 8 m et er s

3 d = 0 . 1* 0 .0 2 54 ; // d ia me te r o f t he w ir e i n m et er s4 L = 1 0* 0. 30 48 ; // l e ng t h o f t he w ir e i n m et er s5 r ho = 1 .7 24 *1 0^ - 8 ; / / r e s i s t i v i t y o f t h e w i r e i n ohm

me t e r6 A = % pi * ( d /2 ) ^ 2; // c r o s s s e c t i o n a l a re a o f t h e w i re7 R = r ho * L /A ; // r e s i s t a n c e o f t he w ir e i n ohm8 disp (R , r e s i s t a n c e o f t he w i re ( i n ohm )= )

Scilab code Exa 2.5 Find the time variation of the voltage drop appearingacross the inductor terminals

1 L = 0.1; // i nd uc ta nc e o f t h e c o i l i n he n r y2 t 1 = [ 0 :0 . 00 1 :0 . 1] ;

3 t 2 = [ 0 .1 0 1: 0 .0 0 1: 0 .3 ] ;

4 t 3 = [ 0 .3 0 1: 0 .0 0 1: 0 .6 ] ;

5 t 4 = [ 0 .6 0 1: 0 .0 0 1: 0 .7 ] ;

6 t 5 = [ 0 .7 0 1: 0 .0 0 1: 0 .9 ]

7 // c u r r e n t v a r i a t i o n a s a f u nc t i o n o f t i m e8 i 1 = 1 00 * t1 ;

9 i 2 = ( -50* t2 ) + 15;

10 i 3 = -100* sin ( % p i * ( t 3 - 0 . 3 ) / 0 . 3 ) ;

11 i 4 = ( 10 0* t4 ) - 60;

12 i 5 = ( -50* t5 ) + 45;

13

14 t = [ t1 , t 2 , t3 , t 4 , t5 ] ;

15 i = [ i1 , i 2 , i3 , i 4 , i5 ] ;

16 plot (t , i )

1718 d t = 0 .0 01 ;

19 di = diff ( i ) ;

20 V = L *di / dt ; // v o l t a g e d ro p a p pe a ri n g a c r o s s t hei n d uc t or t e r mi n a l s

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21 T v = [ 0 :0 . 00 1 :0 . 89 9 ];

22 plot ( T v , V , g r e e n )

Scilab code Exa 2.6 Find the time variation of the capacitor voltage

1 C = 0. 01 ; // c a pa c it a nc e o f t h e c a p ac i t o r i n F a r ad s2 t 1 = [ 0 :0 . 00 1 :0 . 1] ;

3 t 2 = [ 0 .1 0 1: 0 .0 0 1: 0 .3 ] ;4 t 3 = [ 0 .3 0 1: 0 .0 0 1: 0 .6 ] ;

5 t 4 = [ 0 .6 0 1: 0 .0 0 1: 0 .7 ] ;

6 t 5 = [ 0 .7 0 1: 0 .0 0 1: 0 .9 ]

7 // c u r r e n t v a r i a t i o n a s a f u nc t i o n o f t i m e8 i 1 = 1 00 * t1 ;

9 i 2 = ( -50* t2 ) + 15;

10 i 3 = -100* sin ( % p i * ( t 3 - 0 . 3 ) / 0 . 3 ) ;

11 i 4 = ( 10 0* t4 ) - 60;

12 i 5 = ( -50* t5 ) + 45;

13

14 t = [ t1 , t 2 , t3 , t 4 , t5 ] ;15 i = [ i1 , i 2 , i3 , i 4 , i5 ] ;

16 plot (t , i )

17

18 / / v o l t a g e a c r o ss t h e c a p a c i t o r a s a f u n c t i o n o f t i m e

19 V 1 = (1 /C )* i n t e g r a t e ( 10 0t , t , 0 , t 1 ) ;20 V 2 = (1 /C )* i n t e g r a t e ( (50 t )+15 , t , 0 . 1 01 , t 2 ) ;21 V 3 = (1 /C )* i n t e g r a t e ( 100s i n ( %pi ( t 0 . 3 ) / 0 . 3 ) , t

, 0 . 3 0 1 , t 3 ) ;

22 V 4 = (1 /C )* i n t e g r a t e ( ( 100

t )

60 , t , 0 . 6 01 , t 4 ) ;23 V 5 = (1 /C )* i n t e g r a t e ( (50 t ) + 4 5 , t , 0 . 7 01 , t 5 ) ;24 V = [ V 1 , V 2 , V 3 , V 4 , V 5 ] ;

25

26 plot (t , V , g r e e n )

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Scilab code Exa 2.7 Find A actual value of the voltage gain of the opampcircuit B ideal value of the voltage gain C percent error

1 // a2 R i = 1;

3 R f = 39;

4 A = 10 ^5 ; // open l oo p g ai n o f t he opamp5 G = A /( 1 + ( A *Ri / ( Ri + Rf ) )) ; // a c t ua l v o l t a g e g ai n o f

t h e c i r c u i t6 disp ( a )7 disp (G , a c t u a l v o l t a ge o f t h e c i r c u i t = )8

9 // b10 G1 = 1 + ( Rf /Ri ); // v o l t a g e g ai n o f t h e c i r c u i t

w i t h i n f i n i t e o pe n l o o p g a i n11 disp ( b )12 disp ( G 1 , f o r i d e a l c as e t h e v o l t a ge g ai n = )13

14 // c15 e r = (( G 1 - G ) /G ) *1 00 ; / / p e r c e n t e r r o r16 disp ( c )17 disp ( e r , p e rc e nt e r r o r o f t he i d e a l v al ue compared

t o t he a c t u a l v a lu e= )

Scilab code Exa 2.8 Design a non inverting opamp circuit of voltage gain4

1 G = 4 ; // v o l t a g e g ai n o f t h e c i r c u i t2 r = G -1; // r a t i o o f t h e r e s i s t a n c e s i n t h e non

i n v e r t i n g opamp c i r c u i t3 disp (r , R f /R i = )

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4 // R e su l t :

5 //A s u i t a b l e c h o i ce f o r R1 i s 10K, Hence Rf = 30K

Scilab code Exa 2.9 Find the input resistance of an inverting opamp cir-cuit with voltage gain of 4

1 G = 4 ;

2 r = G ; // r a t i o o f t h e r e s i s t a n c e s i n t h e i n v e r t i n gopamp c i r c u i t

3 disp (r , R f / R i )4 // R e su l t ;5 / /A s u i t a b l e c h o i c e f o r Rf =30K and R1 =7 .5K6 // t h e r e f o r e i np ut r e s i s t a n c e R1 = 7 . 5K

This code can be downloaded from the website wwww.scilab.in

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Chapter 3

Elementary network theory

Scilab code Exa 3.1 for the given circuit calculate the current flowing fromthe voltage source

1 V = 100; // v o l a ta g e s up pl y i n v o l t s2 R s = 40; // r e s i s t a n c e i n s e r i e s i n ohms3 // p a r a l l e l r e s i s t a n c e s i n ohms4 R p1 = 3 3. 33 ;

5 R p2 = 50;

6 R p3 = 20;7 R p i n v = ( 1 / R p1 ) + ( 1 / R p 2 ) + (1 / R p 3 ) ; / / r e c i p r o c a l o f

e q u i v a l e n t r e s i s t a n c e i n p a r a l l e l8 Req = Rs + (1/ Rpinv ) ;

9 I = V / R e q ; // c u r r e n t f l o wi n g from t he v o l t ag e s o ur c ei n amps

10 disp (I , c u r r e nt f l o w i n g fr om t he v o l t a g e s o u r ce ( i na m p s ) = )

Scilab code Exa 3.2 Calculate the potential difference across terminals bc

1 V = 100; // v o l a ta g e s up pl y i n v o l t s

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2 R s = 40; // r e s i s t a n c e i n s e r i e s i n ohms

3 // p a r a l l e l r e s i s t a n c e s i n ohms4 R p1 = 3 3. 33 ;5 R p2 = 50;

6 R p3 = 20;

7 R p i n v = ( 1 / R p1 ) + ( 1 / R p 2 ) + (1 / R p 3 ) ; / / r e c i p r o c a l o f e q u i v a l e n t r e s i s t a n c e i n p a r a l l e l

8 R p = 1/ R pi nv ; // e q u i v a l e n t e s i s t a n c e i n p a r a l l e l9 V b c = V *( Rp / ( Rs + Rp ) ); // p o t e n t i a l d i f f e r e n c e

a c r o s s bc10 disp ( V b c , p o t e n t i a l d i f f e r e n c e a c r o ss bc = )

Scilab code Exa 3.3 Determine the equivalent series circuit

1 / / r e s i s t a n c e s i n ohms2 R 1 = 25;

3 R 2 = 30 0;

4 R 3 = 80;

5 R 4 = 30;

6 R 5 = 60;

78 R c d = R5 * R4 / ( R5 + R4 ) ;

9 Rbd1 = Rcd + R3 ;

10 R bd = R bd 1 * R2 / ( Rb d1 + R 2) ;

11 Req = Rbd + R1 ; // e q u i v a l e nt r e s i s t a n c e12 disp ( R e q , e q ui v a l e n t r e s i s t a n c e = )

Scilab code Exa 3.4 Value of E for which power dissipation in R5 is 15W

R5 is 15

1 / / r e s i s t a n c e s i n ohms2 R 1 = 25;

3 R 2 = 30 0;

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4 R 3 = 80;

5 R 4 = 30;6 R 5 = 60;

7

8 P 5 = 15; // p ower d i s s i p a t e d i n R5 ( i n wa tt )9

10 I5 = sqrt ( P 5 / R 5 ) ; / / c u r r e n t f l o w i n g t h ro u gh R511 V5 = R5 *I5 ; / / v o l t a g e a c r o s s R512 Vcd = V5 ; / / v o l t a g e a c r o s s cd13

14 I 4 = V cd / R4 ; / / c u r r e n t f l o w i n g t h ro u gh R415 Icd = I5 + I4 ; / / c u r r e nt f l o w i n g t hr ou gh cd

1617 V bd = ( I cd * R3 ) + Vc d ; / / v o l t a g e a c r o s s bd18 I b d = ( V bd / R 2 ) + Ic d ; / / c u r r e n t t h ro u g h bd19

20 V 1 = R1 * Ib d; / / v o l t a g e a c r o s s R121

22 E = V1 + Vbd ;

23 disp (E , E = )24

25 / / R e s ul t : V al ue o f E f o r which power d i s s i p a t i o n i n

R i s 1 5W = 2 00V

This code can be downloaded from the website wwww.scilab.in This

code can be downloaded from the website wwww.scilab.in This code can be

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downloaded from the website wwww.scilab.in

Scilab code Exa 3.8 Find current flowing through all the branches of thecircuit

1 / / mesh e q u a t i o n s :2 / / 6 0I 1 20I 2 = 203 //20 I 1 + 80I 2 = 654

5 R = [60 -20; -20 8 0] ;

6 E = [ 12 0; - 6 5] ;7 I = inv ( R ) * E ;

8 I 1 = I (1 , :) ; / / c u r r e n t f l o w i n g i n f i r s t mesh9 I 2 = I (2 , :) ; / / c u r r e nt f l o w i n g i n s ec on d mesh

10

11 Ibd = I1 - I2 ; / / c u r r e n t f l o w i n g t h ro u gh b ra nc h bd12 Iab = I1 ; // c u r r e n t f l o w i n g t h ro ug h b ra nc h ab13 I cb = - I2 ; // c u r r e n t f l o w i n g t hr o ug h b ra nc h cb14

15 disp ( I b d , c u r r e n t f l o w i n g t hr ou gh b ra nc h bd = )16 disp ( I a b , c u r r e nt f l o w i n g t hr ou gh b ra nc h ab = )

17 disp ( I c b , c u r r e nt f l o w i n g t hr ou gh b ra nc h cb = )


code can be downloaded from the website wwww.scilab.in This code can be

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downloaded from the website wwww.scilab.in

Scilab code Exa 3.12 Find A current flowing through Rl B valye of Rl forwhich the power transfer is maximum and the maximum power

1 // a2 // c i r c u i t p ar am et er s3 E 1 = 12 0;

4 R 1 = 40;

5 R 2 = 20;

6 R 3 = 60;7

8 V o c = E1 * R2 / ( R2 + R1 ) ; // open c i r c u i t v o l t a gea p pe ar i n g a t t e rm i na l 1

9 R i = R3 + ( R1 * R2 /( R1 + R2 )) ; // e q u i v a l e n t r e s i s t a n c el o o k i n g i n t o th e ne t w o r k from t e rm in al p ai r

0110

11 function I = Il ( Rl )

12 I = Voc /( Ri + Rl ) / / c u r r e n t t h ro u g h Rl13 e n d f u n c t i o n

14

15 I l 1 = Il ( 10 ) ; / / R l = 1 0 o h m16 I l 2 = Il ( 50 ) ; / / R l = 5 0 o h m17 I l3 = Il ( 20 0) ; / / R l = 2 0 0 ohm18

19 disp ( a )20 disp ( I l 1 , I l ( R l = 10 ohm ) = )21 disp ( I l 2 , I l ( R l = 50 ohm ) = )22 disp ( I l 3 , I l ( R l = 2 00 ohm ) = )23

24 // b25 // f o r maximum pow er R l = R i26 R l = Ri ;

27 P lm ax = ( V oc / (2 * Ri ) ) ^2 * Ri ; //maximum power to Rl28 disp ( b )

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29 disp ( P l m a x , maximum p o we r t o R l ( i n Watt ) = )

Scilab code Exa 3.13 Find the current flowing through R2 by using Nor-tons current source equivalent circuit

1 // c i r c u i t p a ra m et er s2 // v o l t a g e s o u r c e s3 E 1 = 12 0;

4 E 2 = 65;

5 // r e s i s t a n c e s6 R 1 = 40;

7 R 2 = 11;

8 R 3 = 60;

9

10 I = ( E1 / R1 ) + ( E2 / R3 ); / / n o rt on s c u r r e n t s o u r c e11 R e q = R1 * R3 / ( R1 + R3 ) ; // e q u i v a l e n t r e s i s t a n c e12

13 I2 = I * Req /( Req + R2 ); / / c u r r e n t f l o w i n g t h ro u gh R214

15 disp ( I 2 , c u r r en t f l o w i n g t hr ou gh R2 = )

This code can be downloaded from the website wwww.scilab.in This code

can be downloaded from the website wwww.scilab.in

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Chapter 4

Circuit differential equations

Forms and Solutions

Scilab code Exa 4.3 Determine the operational driving point impedancesappearing at terminals ad and dg

1 //ad2 Z ab = c om p le x ( 1 , - 0. 5) ; // i m pe da nc e a p p e ar i n g a c r o s s

t e r m i n a l s ab3 Z b g = c om p le x ( 1) ; // i m pe da nc e a p p e ar i n g a c r o s s

t e r m i n a l s bg4 Z bc d = c o mp le x ( 2 +1 , 2 ) ; // i m pe da nc e a p p e ar i n g a c r o s s

t e r m i n a l s bcd5 Z ad = Z ab + ( Z bg * Z bc d /( Z bg + Z bc d )) ; / / i m p e d a n c e

a p pe ar i n g a c r o s s t e r mi n a l s ad6 disp ( Z a d , i mp ed an ce a p pe a ri n g a c r o s s t e r m i n a l s ad =

)7

8 //dg

9 Z d g = Z bg + ( Z ab * Z b cd / ( Z ab + Z b cd ) ) ; / / i m p e d a n c ea p pe ar i n g a c r o s s t e r m a i n al s dg10 disp ( Z d g , i mp ed an ce a p pe a ri n g a c r o s s t e r m i n a l s dg =

)

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Chapter 5

Circuit dynamics and forced

responses

This code can be downloaded from the website wwww.scilab.in

Scilab code Exa 5.2 Find the expression for the current flowing throughthe circuit and the total energy dissipated in the resistor

1 C = 10*10^ - 6 ; / / c a p a c i t a n c e ( i n f a r a d s )2 R = 0 .2 *1 0^ 6; // r e s i s t a n c e ( i n ohms )3 V i = 40; // i n i t i a l v o l t a g e o f t h e c a p a c i t o r ( i n

v o l t s )4 W c = ( 1/ 2) * C * Vi ^ 2 ; / / e ne rg y s t o r e d i n t he c a p a c i t o r5 / / c ur r e nt f l o w i n g i n c i r c u i t a s a f u n c t i o n o f t i m e i

( t ) = 2104 exp( t /2 )6 / / power d i s s i p a t e d i n t he r e s i s t o r = Ri 27 Wr = i n t e g r a t e (R4108 exp( t ) , t ,0,100)

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8 disp ( W c , e ne rg y s t o r e d i n t he c a p a c i t or ( i n J o u l e s ) =

)9 disp ( W r , e ne rg y d i s s i p a t e d i n t he r e s i s t o r ( i n J o ul e s) = )


code can be downloaded from the website wwww.scilab.in This code can

be downloaded from the website wwww.scilab.in This code can be down-

loaded from the website wwww.scilab.in This code can be downloaded from

the website wwww.scilab.in This code can be downloaded from the website

wwww.scilab.in

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Chapter 6

The laplace transform method

of finding circuit solutions

Scilab code Exa 6.1 Find the laplace transform of the given pulse

1 function F = l ap la ce ( s, T1 , T2 )

2 / / p u l s e :3 // f = u ( t T1 ) u ( t T2 )4 F = i n t e g r a t e ( e xp ( s t ) , t , T 1 , T 2 ) ; // l a p l a c e

t r a ns f o r m o f t he p u l se5 e n d f u n c t i o n

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Chapter 7

Sinusoidal steady state

response of circuits

Scilab code Exa 7.1 Find the average value of the given periodic function

1 V m = 2; / / a s s u mp t i on2 // a v e ra ge v al ue o f t he f u n c t i o n3 // v ( t ) = Vm a l ph a / ( %pi / 3 ) f o r 0


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4

5 V m = 17 0; // p ea k v o l t a g e6 I m = 1 4. 14 ; // p ea k c u r r e n t7

8 P a v = V m * Im * p f /2 ; / / a v er a ge power d e l i v e r e d t o t hec i r c u i t

9 disp ( P a v , a v e ra ge power d e l i v e r e d t o t he c i r c u i t = )

Scilab code Exa 7.3 Find the expression for the sum of i1 and i2

1 / / l e t s assume t ha t i 1 and i 2 a re s t a t i o n a r y and t h ec o or d i n a t e s y s t e m i s r o t a t i n g w i t h an a ng ul ar

f r qu e n cy o f w . And i 1 l i e s on t he xa x i s ( i . e .making an a n gl e o f 0 d e g r ee w i th t he xa x i s )

2 t h e ta = % pi / 3 ; // p ha se d i f f e r e n c e b et w ee n i 1 and i 2 ;3 I 1 = 10* sqrt ( 2 ) ; // peak v al ue o f i 14 I 2 = 20* sqrt ( 2 ) ; // peak v al ue o f i 25 I = sqrt ( I1 ^ 2 + I2 ^ 2 + 2* I 1* I2 * cos ( t h e t a ) ) ; / / p e a k

v al ue o f t h e r e s u l t a n t c u r r e n t

67 p hi = atan ( I 2 * sin ( t h et a ) /( I 1 + I 2 *cos ( t h e t a ) ) ) ; //

p ha se d i f f e r e n c e b et we e n t he r e s u l t a n t and i 1 ( i nr a d i a n s )

8 disp (I , peak v al ue o f t he r e s u l t a n t c u rr e n t = )9 disp ( p h i , p ha se d i f f e r e n c e bet w ee n t he r e s u l t a n t and

i 1 = )10 // r e s u l t : i = I s i n ( wt + ph i )

Scilab code Exa 7.4 Find the effective value of the resultant current

1 I 1 = 10; // p eak v al ue o f i 12 I 2 = 20; // p eak v al ue o f i 2

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3 t h e ta = % pi / 3 ; // p ha se d i f f e r e n c e b et w ee n i 1 and i 2

4 / / compl ex r e p r e s e n t a t i o n o f t he two c u r r en t s5 i 1 = c o mp l ex ( 1 0) ;6 i 2 = c o mp l ex ( 2 0* cos (%pi /3) ,20* sin ( % p i / 3 ) ) ;

7

8 i = i 1 + i2 ; / / r e s u l t a n t c u r r e n t9 I = sqrt ( real ( i) ^2 + imag ( i ) ^ 2 ) ; // c a l c u l a t i n g t he

peak v al ue o f t h e r e s u l t a n t c u r r e n t by u si ng i t sr e a l and i m ag i na r y p a r t s

10 p hi = atan ( imag ( i ) / real ( i ) ) ; / / c a l c u l a t i g t he p ha seo f t h e r e s u l t a n t c ur r e nt by u si ng i t s r e a l andi m a gi n a ry p a r t s

11 disp (i , r e s u l t a n t c u r r e n t = )12 disp (I , peak v al ue o f t he r e s u l t a n t c u rr e n t = )13 disp ( p h i , p h a s e o f t h e r e s u l t a n t c u r r e n t = )14 // r e s u l t : i = I s i n ( wt + p hi )

Scilab code Exa 7.5 Find the time expression for the resultant current

1 I 1 = 3; // peak v al ue o f i 1

2 I 2 = 5; // peak v al ue o f i 23 I 3 = 6; // peak v al ue o f i 34 t h e ta 1 = % pi / 6 ; // p ha se d i f f e r e n c e be tw e en i 2 and i 15 t he t a2 = - 2* % pi / 3 ; // p ha se d i f f e r e n c e be t we en i 3 and

i 16 // compl ex r e p r e s e n t a t i o n o f t he c u r r e n t s7 i 1 = c o mp l ex ( 3 ) ;

8 i 2 = c o mp l ex ( 5 * cos ( % p i / 6 ) , 5 * sin ( % p i / 6 ) ) ;

9 i 3 = c o mp l ex ( 6 * cos ( - 2 * % p i / 3 ) , 6 * sin ( - 2 * % p i / 3 ) ) ;

10

11 i = i1 + i2 + i3 ; / / r e s u l t a n t c u r r e nt12 I = sqrt ( real ( i) ^2 + imag ( i ) ^ 2 ) ; // c a l c u l a t i n g t hepeak v al ue o f t h e r e s u l t a n t c u r r e n t by u si ng i t sr e a l and i m ag i na r y p a r t s

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13 p hi = atan ( imag ( i ) / real ( i ) ) ; / / c a l c u l a t i g t he p ha se

o f t h e r e s u l t a n t c ur r e nt by u si ng i t s r e a l andi m a gi n a ry p a r t s14 disp (I , peak v al ue o f t he r e s u l t a n t c u rr e n t = )15 disp ( p h i , p h a s e o f t h e r e s u l t a n t c u r r e n t = )16 // r e s u l t : i = I s i n ( wt + p hi )

Scilab code Exa 7.6 Find the value of the given expression

1 / / f i n d VZ1/Z22 V = c om p le x ( 4 5* sqrt ( 3) , - 45 ) ;

3 Z 1 = c o mp l ex ( 2 . 5* sqrt ( 2) , 2 .5 * sqrt ( 2 ) ) ;

4 Z 2 = c o mp l ex ( 7 .5 , 7 .5 * sqrt ( 3 ) ) ;

5 / / we h ave t o f i n d VZ1/Z26 Z = V *Z1 / Z2 ;

7 disp (Z , VZ 1 / Z 2 = )

Scilab code Exa 7.7 Find A value of steady state current and the relativephase angle C magnitude and phase of voltage drops appearing across eachelement D average power E power factor

1 // a2 f = 60; / / f r eq u en c y o f t he v o l a tg e s o ur c e3 V = c o mp le x ( 1 41 ) ; / / v o l t a g e s u p p l y V = 1 41 s i n ( wt )4 R = 3 ; // r e s i s t a n c e o f t h e c i r c u i t5 L = 0 .0 10 6; // i nd uc ta nc e o f t h e c i r c u i t6 Z = c o mp le x ( R , 2* % p i *f * L ); // i mpeda nce o f t he c i r c u i t

= R + j w L7 i = V / Z ; / / c u r r e n t8 I = sqrt ( real ( i) ^2 + imag ( i ) ^ 2 ) ; // c a l c u l a t i n g t he

peak v al ue o f t h e c u r r e n t by u si ng i t s r e a l andi m a gi n a ry p a r t s

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9 p hi = atan ( imag ( i ) / real ( i ) ) ; / / c a l c u l a t i g t he p ha se

o f t h e r e s u l t a n t c ur r e nt by u si ng i t s r e a l andi m a gi n a ry p a r t s10 disp ( a )11 disp (I , e f f e c t i v e v al ue o f t he s te ad y s t a t e c u r r e n t

= )12 disp ( p h i , r e l a t i v e p ha se a n gl e = )13

14 // b15 / / e x p r e ss i o n f o r t h e i n st a nt a ne o us c u rr e n t can be

w r i t t e n a s16 / / i = I s i n ( wt + p h i )

1718 // c19 R = c om pl ex ( 3) ;

20 vr = V *R /Z ; // v o l t a g e a c ro s s t h e r e s i s t o r21 Vr = sqrt ( real ( vr ) ^2 + imag ( v r ) ^ 2 ) ; // p ea k v a l ue o f

t h e v o l t a g e a c r o s s t h e r e s i s t o r22 p h i1 = atan ( imag ( v r ) / real ( v r ) ) ; // p h as e o f t he

v o l t a g e a c r o s s t h e r e s i s t o r23

24 v l = V - v r ; / / v o l t a g e a c r o s s t he i n d uc t or25 Vl = sqrt ( real ( vl ) ^2 + imag ( v l ) ^ 2 ) ;

// p ea k v a l ue o f t h e v o l t a ge a c r o s s t he i n d u ct o r26 p h i2 = atan ( imag ( v l ) / real ( v l ) ) ; // p h as e o f t he

v o l t a ge a c r o s s t he i n d u ct o r27 disp ( c )28 disp ( V r , e f f e c t i v e v al ue o f t h e v o l t a ge dr op a c r o s s

t h e r e s i s t o r = )29 disp ( p h i 1 , p h a s e o f t he v o l t a ge dr op a c r o s s t h e

r e s i s t o r = )30 disp ( V l , e f f e c t i v e v al ue o f t h e v o l t a ge dr op a c r o s s

t he i n d uc t or = )

31 disp ( p h i 2 , p h a s e o f t he v o l t a ge dr op a c r o s s t h ei n d u c t o r = )

32

33 // d

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34 P av = V *I *cos ( p h i ) ; / / a v er a ge power d i s s i p a t e d by

t h e c i r c u i t35 disp ( d )36 disp ( P a v , a v e ra ge power d i s s i p a t e d by t h e c i r c u i t =

)37

38 // e39 pf = cos ( p h i ) ; // p ower f a c t o r40 disp ( e )41 disp ( p f , power f a c t o r = )

Scilab code Exa 7.8 Find the equivalent impedance appearing betweenpoints a and c

1 // i mp ed an ce s i n t he c i r c u i t2 Z 1 = c o mp l ex ( 1 0 , 10 ) ;

3 Z 2 = c o mp l ex ( 1 5 , 20 ) ;

4 Z 3 = c o mp l ex ( 3 , - 4) ;

5 Z 4 = c o mp l ex ( 8 , 6) ;

6

7 Y b c = ( 1/ Z 2 ) + ( 1 / Z 3 ) + (1 / Z 4 ) ; / / a d mi t ta n ce o f t h ep a r a l l e l c om bi na ti on

8 Z bc = ( 1/ Y bc ) ; // i mpeda nce o f t he p a r a l l e lc o m b i n a t i o n

9

10 Z = Z1 + Zbc ; // e q u i v al e n t i mpe danc e o f t he c i r c u i t11

12 disp (Z , e q u i v a l e n t i mp eda nce o f t he c i r c u i t = )

Scilab code Exa 7.9 Find the current which flows through branch Z3

1 V 1 = c o mp l ex ( 1 0) ;

2 V 2 = c o mp l ex ( 1 0* cos (-%pi /3) ,10* sin ( - % p i / 3 ) ) ;

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3 Z 1 = c o mp l ex ( 1 , 1) ;

4 Z 2 = c o mp l ex ( 1 , - 1) ;5 Z 3 = c o mp l ex ( 1 , 2) ;

6

7 // by mesh a n a l y s i s we g e t t he f o l l o w i n g e q ua t i o n s :8 // I1 Z11 I 2 Z 12 = V 19 //I 1 Z 2 1 + I 2 Z 2 2 = V2 ; where I 1 and I 2 a r e t he

c u r r r e n t s f l o w i n g i n t he f i r s t and s ec on d m es hesr e s p e c t i v e l y

10 Z11 = Z1 + Z1 ;

11 Z12 = Z1 + Z2 ;

12 Z 21 = Z12 ;

13 Z22 = Z2 + Z2 ;14

15 / / t he mesh e q ua t i o n s can be r e p r e s e nt e d i n t hem at ri x form a s I Z = V

16 Z = [ Z11 , - Z12 ; - Z21 , Z22 ]; / / i m p ed a nc e m a t r i x17 V = [ V 1 ; - V 2 ] ; / / v o l t a g e m a tr i x18 I = inv ( Z ) * V ; / / c u r r e n t m a t r i x = [ I 1 ; I 2 ]19

20 I 1 = I (1 , :) ; // I 1 = f i r s t row o f I m at ri x21 I 2 = I (2 , :) ; // I 1 = se co nd row o f I m at ri x22

23 Ibr = I1 - I2 ; / / c u r r e n t f l o w i n g t h ro u gh Z324

25 disp ( I b r , c u r r e nt f l o w i n g t hr ou gh Z3 = )

Scilab code Exa 7.10 Find the current in the Z3 branchby using the Nodalmethod

1 V 1 = c o mp l ex ( 1 0) ;2 V 2 = c o mp l ex ( 1 0* cos (-%pi /3) ,10* sin ( - % p i / 3 ) ) ;

3 Z 1 = c o mp l ex ( 1 , 1) ;

4 Z 2 = c o mp l ex ( 1 , - 1) ;

5 Z 3 = c o mp l ex ( 1 , 2) ;

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6 //By a p p l in g t he n od al a n a l y s i s we g et t he f o l l o w i n g

e q u a t i o n :7 / / Va ( ( 1 / Z1 ) + (1 / Z2 ) + (1 / Z3 ) ) = ( V1 / Z1 ) + ( V2 / Z2 )8

9 Y = ( 1/ Z 1 ) + (1 / Z 2) + ( 1/ Z 3 ) ;

10 V a = ( 1/ Y ) *( ( V1 / Z1 ) + ( V2 / Z2 ) ); / / v o l t a g e o f node a11

12 I b r = Va / Z3 ; / / c u r r e nt f l o w i n g t hr ou gh Z313


Scilab code Exa 7.11 Find the current flowing through Z3 by using Thevininstheoram

1 V 1 = c o mp l ex ( 1 0) ;

2 V 2 = c o mp l ex ( 1 0* cos (-%pi /3) ,10* sin ( - % p i / 3 ) ) ;

3 Z 1 = c o mp l ex ( 1 , 1) ;

4 Z 2 = c o mp l ex ( 1 , - 1) ;

5 Z 3 = c o mp l ex ( 1 , 2) ;

6

7 Z th = Z3 + ( Z1 * Z2 / ( Z1 + Z2 ) ); // t h e v i ni n r e s i s t a n c e8

9 I = ( V 1 - V 2 ) / ( Z 1 + Z 2 ) ; // c u r r e n t f l o wi n g t hr ou ght he c i r c u i t when R3 i s n ot c on ne ct ed

10 Vth = V1 - I *Z1 ; // t h e v i n i n v o l t a g e11

12 I br = V th / Zt h ; / / c u r r e n t f l o w i n g t h ro u gh Z313


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Chapter 9

Semiconductor electronic

devices

Scilab code Exa 9.2 Find the values of self bais source resistance and drainload resistance at Q point

1 / / Q u ie s ce n t p o i n t2 I d q = 0 .0 0 34 ; // d r ai n c u r r e n t3 V dq = 15; // d r ai n v o l t a g e4 V gq = 1; // g at e v o l t a g e5

6 V dd = 24; // d r a in s u pp ly v o l t a g e7

8 R s = V gq / Id q ;

9 disp ( R s , The v al ue o f s e l f b a i s s o ur c e r e s i s t a n c e i s( i n ohm ) : )

10

11 Rd = ( Vdd - Vdq )/ Idq ;

12 disp ( R d , The v al ue o f d r a i n l oa d r e s i s t a n c e i s ( i n

ohm) : )

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Scilab code Exa 9.3 Find A midband frequency current gain of the first

stage B bandwidth of the first stage amplifier

1 // a2 // t r a n s i s t o r p a ra m et er s3 R2 = 0 . 62 5;

4 h ie = 1 . 67 ;

5 Rb = 4 .1 6;

6 Rl = 2 .4 ;

7 R oe = 1 50 ;

8

9 Cc = 25 * 10^ -6;

10 r BB = 0 . 29 ;11 r BE = 1 . 37 5;

12 Cd = 6 90 0 * 10 ^ -1 2;

13 Ct = 4 0 * 10^ -12;

14 gm = 0 . 03 2;

15

16 R eq = ( R l* R oe ) /( R l + Ro e) ;

17 hfe = 4 4;

18 a = 1 + ( R2 / Req );

19 b = 1 + ( hie /Rb );

20 A im = - h fe / ( a *b ) ; // mid band f r e q u e n c y g a i n21 disp ( a )

22 disp ( A i m , The mid band f r e q u e n c y g a i n o f t h e f i r s ts t a g e o f t h e c i r c u i t i s : )

23

24 // b25 T l = 2 * %p i *( R eq + R 2 )* C c * (1 0 ^3 ) ;

26 Fl = 1/ Tl ;

27

28 R p = ( R eq * R2 ) /( R eq + R 2 );

29 C = C d + C t *(1 + gm * Rp * 10 ^3 ) ;

30 d = Rb + hie ;31 e = rBE * ( Rb + rBB )* 1 0^3 * C ;

32 F h = d / (2 * % pi * e ) ;

33

34 BW = Fh - Fl ;

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35 disp ( b )

36 disp ( B W , The b an dw id th o f t h e f i r s t s t a g ea m p l i f i e r i n Hz i s : )

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Chapter 11

Binary logic Theory and

Implimentation

Scilab code Exa 11.1 Determine th decimal equivalents of the binary num-bers A 101 B 11011

1 // a2 N2 = 101 ; / / b i n a r y o r d e r ed s e q ue n c e3 N = b i n2 de c ( N 2 ); / / d ec i ma l e q u i v a l e n t o f N24 disp ( a )5 disp (N , d ec im al e q u i v al e n t o f 101 = )6

7 // b8 N2 = 1101 1 ; / / b i n a r y o r d e r ed s e q ue n c e9 N = b i n2 de c ( N 2 ); / / d ec i ma l e q u i v a l e n t o f N2

10 disp ( b )11 disp (N , d ec im al e q u i v al e n t o f 1 10 11 = )

Scilab code Exa 11.2 Determine the decimal equivalent of A octal number432 B hexadecimal number C4F

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1 // a

2 N8 = 432 ; / / o c t a l n um be r3 N = o c t2 de c ( N 8 ); // d ec i ma l r e p r e s e n t a t i o n o f N84 disp ( a )5 disp (N , d e ci ma l e q u i v a l e n t o f 432 = )6

7 // b8 N 16 = C4F ; / / h e x a d e c i m a l n um be r9 N = hex2dec ( N 1 6 ) ; / / d e ci m al r e p r e s e n t a t i o n o f N16

10 disp ( b )11 disp (N , d e c im a l e q u i v a l e n t o f C4F = )

Scilab code Exa 11.3 Find the binary and octal equivalents of 247

1 N = 247;

2 N 2 = d e c2 b in ( N ) ; / / b i n ar y e q u i v a l e n t o f N3 N 8 = d e c2 o ct ( N ) ; // o c t a l e q u i v a l e n t o f N4 disp ( N 2 , b in ar y e q u i v a l e n t o f 247 = )5 disp ( N 8 , o c t a l e q ui v a l e n t o f 247 = )

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Chapter 12

Simplifying logical functions

This code can be downloaded from the website wwww.scilab.in This code

can be downloaded from the website wwww.scilab.in This code can be down-

loaded from the website wwww.scilab.in

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Chapter 15

Magnetic circuit computations

Scilab code Exa 15.1 Find A magneto motive force B current C relativepermiability and reluctance of each material

1 // a2 p h i = 6 *1 0^ - 4 ; / / g i v e n m a g n et i c f l u x ( i n Wb)3 A = 0 .0 01 ; // c r o s s s e c t i o n a l a r e a ( i n me te r s qu ar e )4 B = phi /A ; //5 H a = 10; // m a gn et ic f i e l d i n t e n s i t y o f m a t e r i a l a

n ee de d t o e s t a b l i s h t h e g i v en m ag n e t i c f l u x6 H b = 77; // m ag ne ti c f i e l d i n t e n s i t y o f m a t e r i a l b7 H c = 27 0; // m ag ne t ic f i e l d i n t e n s i t y o f m a t e r i a l c8 L a = 0. 3; // a r c l e ng t h o f m a t e r i a l a ( i n m et er s )9 L b = 0. 2; / / a r c l e n g th o f m a t e r i a l b ( i n m et er s )

10 L c = 0. 1; // a r c l e ng t h o f m a t e r i a l c ( i n m et er s )11

12 F = Ha *La + Hb *Lb + Hc *Lc ; // m a g ne to mo t iv e f o r c e13 disp ( a )14 disp (F , m ag ne to mo ti ve f o r c e n ee de d t o e s t a b l i s h a

f l u x o f 6

10

4 ( i n At ) = )1516 // b17 N = 100; / / no . o f t u r ns18 I = F / N ; / / c u r r e n t i n amps

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19 disp ( b )

20 disp (I , c u r r e n t t h a t must b e made t o f l o w t hr o ug ht he c o i l ( i n amps ) = )21

22 // c23 M U0 = 4 * %p i * 10 ^ - 7;

24 MUa = B / Ha ; // p e r me a b i l i t y o f m a t e r i a l a25 MUb = B / Hb ; // p e r me a b i l i t y o f m a t e r i a l b26 MUc = B / Hc ; // p e r me a b i l i t y o f m a t e r i a l c27

28 M U ra = M Ua / M U0 ; / / r e l a t i v e p e r m e ab i l i t y o f m a te r ia la

29 M U rb = M Ub / M U0 ; / / r e l a t i v e p e r m e ab i l i t y o f m a te r ia lb

30 M U rc = M Uc / M U0 ; / / r e l a t i v e p e r m e ab i l i t y o f m a te r ia lc

31

32 R a = Ha * La / ph i ; // r e l u c t a n c e o f m a t e r i a l a33 R b = Hb * Lb / ph i ; // r e l u c t a n c e o f m a t e r i a l b34 R c = Hc * Lc / ph i ; // r e l u c t a n c e o f m a t e r i a l c35

36 disp ( c )37 disp ( M U r a ,

r e l a t i v e p e r m e ab i l i t y o f m a te r i a l a = )

38 disp ( M U r b , r e l a t i v e p e r m e ab i l i t y o f m a te r i a l b = )39 disp ( M U r c , r e l a t i v e p e r m e ab i l i t y o f m a te r i a l c = )40 disp ( R a , r e l u c t a n c e o f m a t e r ia l a = )41 disp ( R b , r e l u c t a n c e o f m a t e r ia l b = )42 disp ( R c , r e l u c t a n c e o f m a t e r ia l c = )

Scilab code Exa 15.3 Find the mmf produced by the coil

1 m u0 = 4 * %p i * 10 ^ - 7;

2 A = 0 .0 02 5; // c r o s s s e c t i o n a l a re a o f t h e c o i l3 // d i me n si o ns o f t he c o i l ( i n m et er s )4 L g = 0 .0 02 ; // a i r ga p l e n g t h ( i n m e te r s )

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5 L bd = 0 .0 25 ;

6 L de = 0 .1 ;7 L ef = 0 .0 25 ;

8 L fk = 0 .2 ;

9 L bc = 0 .1 75 ;

10 L ca b = 0 .5 ;

11

12 Lbghc = 2*( Lbd + Lde + Lef + ( Lfk /2) ) - Lg ; / / l e n g t ho f t he f e r r o m a gn e t ic m a t e r i a l i n v o l v e d h er e

13

14 p h ig = 4 *1 0^ - 4 ; / / a i r g ap f l u x ( i n Wb)15 B g = phig /A ; / / a i r gap f l u x d e n s i t y ( i n t e s l a )

16 H g = Bg /mu0 ; // f e i l d i n t e n s i t y o f t he a i r gap17 mm fg = Hg * Lg ; //mmf p ro d uc ed i n t h e a i r g ap ( i n At )18

19 B bc = 1.38 ; // f l u x d e n si t y c o rr e sp o nd i n g t o c a s ts t e e l

20

21 H bg hc = 1 25 ; // f i e l d i n t e n s i t y c o r r es p o n d i ng t o f l u xd e n s i t y o f 0 . 1 6T i n t h e s t e e l

22 m m f bg h c = H bg hc * L b gh c ; // mmf c o r r e s p o n d i n g t o b gh c23

24 m mf bc = mm fg + m mf bg hc ; //mmf a c r o s s p at h b c25 H b c = m mf bc / L bc ;

26 p hi bc = B bc * A ; / / f l u x p ro du ce d i n bc27

28 p hi ca b = p hi g + p hi bc ; // t o t a l f i ux e x i s t i n g i n l e gcab

29 B c a b = p h ic a b / 0 . 0 0 37 5 ; // f l u x d e n s i t y30 H ca b = 6 90 ;

31 m mf c ab = H ca b * L ca b ; //mmf i n l e g c ab32

33 m mf = m mf bc + m mf ca b ; / /mmf p ro du ce d by t h e c o i l

3435 disp ( m m f , mmf p ro du ce d by t h e c o i l ( i n At ) = )

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Scilab code Exa 15.5 B Find the magnetic force exerted on the plunger

1 // b2 m u0 = 4* % pi * 10 ^ -7 ;

3 / / p l u n g e r m ag net d i m e n s i o n s ( i n m e t er s )4 x = 0 .0 25 ;

5 h = 0. 05 ;

6 a = 0 .0 25 ;

7 g = 0 .0 01 25 ;

8

9 m m f = 1 41 4; / / ( i n At )10

11 F = % pi * a * mu 0 *( m m f ^2 ) *( h ^ 2) * ( 1 /( x + h ) ^ 2) / g ; //m ag ni tu de o f t he f o r c e

12 disp (F , m ag ni tu de o f t he f o r c e ( i n Newtons ) = )

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Chapter 16

Transformers

Scilab code Exa 16.1 Find A equivalent resistance and reactance referredto both the sides B voltage drops across these in Volts and in per cent ofthe rated winding voltage C Repeat B for the low voltage side D equivalentleakage impedances referred to both the sides

1 // a2 V 1 = 1 10 0; / / h i g h e r v o l t a g e3 V 2 = 22 0; / / l o w er v o l t a g e

4 a = V1 / V2 ; // t u r ns r a t i o5 r 1 = 0. 1; / / h i gh v o l t a g e w in d in g r e s i s t a n c e ( i n ohms )6 x 1 = 0. 3; / / h i gh v o l t a g e l e a k a g e r e a c t a n c e ( i n ohms )7 r 2 = 0 .0 04 ; // l ow v o l t a g e w in di ng r e s i s t a n c e ( i n ohms

)8 x 2 = 0 .0 12 ; / / l ow v o l t a g e l e a k a g e r e a c t a n c e ( i n ohms )9

10 Re1 = r1 + ( a^2) *r2 ; // e q u i v a l e n t w in d in gr e s i s t a n c e r e f e r r e d t o t h e p r i m a r y s i d e

11 Xe1 = x1 + ( a^2) *x2 ; / / e q u i v a l e n t l e a k a ge r e a c t a n ce

r e f e r r e d t o t h e p r i m a r y s i d e12 Re2 = ( r1 /a ^2) + r2 ; // e q u i v a l e n t w in d in gr e s i s t a n c e r e f e r r e d t o t h e s e c o n d a r y s i d e

13 Xe2 = ( x1 /a ^2) + x2 ; / / e q u i v a l e n t l e a k a ge r e a c t a n cer e f e r r e d t o t h e s e c o n d a r y s i d e

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14

15 disp ( a )16 disp ( R e 1 , e q ui v a l e n t w i n di ng r e s i s t a n c e r e f e r r e d t ot h e p ri ma ry s i d e )

17 disp ( X e 1 , e q u i v a l e n t l e ak a ge r e ac t a nc e r e f e r r e d t ot h e p ri ma ry s i d e )

18 disp ( R e 2 , e q ui v a l e n t w i n di ng r e s i s t a n c e r e f e r r e d t ot he s e co n da r y s i d e )

19 disp ( X e 2 , e q u i v a l e n t l e ak a ge r e ac t a nc e r e f e r r e d t ot he s e co n da r y s i d e )

20

21 // b

22 P = 100; //pow e r ( i n kVA )23 I 2 1 = P * 1 00 0 / V1 ; // p r im ar y w in di ng c u r r e nt r a t i n g24 V r e1 = I 21 * R e1 ; // e q u i v a l e n t r e s i s t a n c e drop ( i n

v o l t s )25 V p e rR 1 = V re 1 * 1 00 / V 1 ; // % e q u i v a l e n t r e s i s t a n c e

dr op26

27 V x e1 = I 21 * X e1 ; // e q u i v a l e n t r e a c t a nc e d ro p ( i nv o l t s )

28 V pe r X1 = V xe 1 * 1 00 / V 1 ; // % e q u i v a l e nt r e a ct a n ce drop29

30 disp ( b )31 disp ( V r e 1 , e q u i v a l e n t r e s i s t a n c e drop e x pr e ss e d i n

t er ms o f p ri ma ry q u a n t i t i e s ( i n v o l t s ) = )32 disp ( V p e r R 1 , % e q u i v a l e n t r e s i s t a n c e dr op e x pr e ss e d

i n t e r m s o f p r i m a r y q u a n t i t i e s = )33 disp ( V x e 1 , e q u i v a l e n t r e ac t a nc e dr op e x pr e ss e d i n

t er ms o f p ri ma ry q u a n t i t i e s ( i n v o l t s ) = )34 disp ( V p e r X 1 , % e q u i v a l e n t r e a c t a n ce d ro p e x p r es s e d

i n t e r m s o f p r i m a r y q u a n t i t i e s = )35

36 // c37 I 2 = a * I21 ; // s e co n da r y w in di ng c u r r e n t r a t i n g38 V re 2 = I 2* Re 2 ; // e q u i v al e n t r e s i s t a n c e dr op ( i n

v o l t s )

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39 V p e rR 2 = V re 2 * 1 00 / V 2 ; // % e q u i v a l e n t r e s i s t a n c e

dr op4041 V xe 2 = I 2* Xe 2 ; / / e q u i v a l e n t r e a c t a n ce d ro p ( i n v o l t s

)42 V pe r X2 = V xe 2 * 1 00 / V 2 ; // % e q u i v a l e nt r e a ct a n ce drop43

44 disp ( c )45 disp ( V r e 2 , e q u i v a l e n t r e s i s t a n c e drop e x pr e ss e d i n

t er ms o f s ec on da ry q u a n t i t i e s ( i n v o l t s ) = )46 disp ( V p e r R 2 , % e q u i v a l e n t r e s i s t a n c e dr op e x pr e ss e d

i n t e r m s o f s ec o nd ar y q u a n t i t i e s = )

47 disp ( V x e 2 , e q u i v a l e n t r e ac t a nc e dr op e x pr e ss e d i nt er ms o f s e co n da r y q u a n t i t i e s ( i n v o l t s ) = )

48 disp ( V p e r X 2 , % e q u i v a l e n t r e a c t a n ce d ro p e x p r es s e di n t e r m s o f s ec o nd ar y q u a n t i t i e s = )

49

50 // d51 Z e1 = c om p le x ( Re 1 , X e1 ) ; // e q u i v a l e n t l e a k a g e

i mpeda nce r e f e r r e d t o t he p ri ma ry52 Ze2 = Ze1 /a ; / / e q u i v a l e n t l e a k a g e i mp ed an ce

r e f e r r e d t o t he s ec o nd ar y53

54 disp ( d )55 disp ( Z e 1 , e q u i v a l e n t l e a k a ge i mp ed an ce r e f e r r e d t o

t h e p ri m a ry = )56 disp ( Z e 2 , e q u i v a l e n t l e a k a ge i mp ed an ce r e f e r r e d t o

t h e s ec on da ry = )

Scilab code Exa 16.2 Compute the 6 parameters of the equivalent circuit

referred to the high and low sides

1 P l = 39 6; // w at tm et er r e ad i ng on open c i r c u i t t e s t2 V l = 12 0; // v o lt m et e r r e ad i ng on open c i r c u i t t e s t3 I l = 9 .6 5; // ammeter r e a d i n g o open c i r c u i t t e s t

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4 a = 2 40 0/ 12 0; // t u r ns r a t i o

56 t h e at a = acos ( P l / ( V l * I l ) ) ; // p ha se d i f f e r e n c e

b et we en v o l t a g e and c u r r e n t7 I rl = Il * cos ( t h e a t a ) ; // r e s i s t i v e pa r t of Im8 I xl = Il * sin ( t h e a t a ) ; // r e a c t i v e p ar t o f Im9

10 r l = V l/ I rl ; // l ow v o l t a g e w in di ng r e s i s t a n c e11 r h = ( a ^2 ) *rl ; / / r l on t h e h ig h s i d e12 x l = V l/ I xl ; / / m ag ne ti zi ng r e ac t a nc e r e f e r r e d t o t he

l ow er s i d e13 x h = ( a ^2 ) *xl ; // c o r r es p o n d i ng h ig h s i d e v a lu e

1415 P h = 81 0; // w at tm et er r e ad i ng on s h o rt c i r c u i t t e s t16 V h = 92; // v ol t m e t er r ea d i n g on s h o r t c i r c u i t t e s t17 I h = 2 0. 8; // ammeter r e a d i n g on s h o rt c i r c u i t t e s t18

19 Z e h = Vh / Ih ; // e q u i v a l e n t i mp ea da nc e r e f e r r e d t o t heh i g h e r s i d e

20 Z e l = Z eh / ( a ^2 ) ; // e q u i v a l e n t i mp ed an ce r e f e r r e d t ot h e l ow er s i d e

21 R e h = P h /( I h ^ 2) ; // e q u i v a l e nt r e s i s t a n c e r e f e r r e d t o

t h e h i g h er s i d e22 R e l = R eh / ( a ^2 ) ; // e q u i v a l e nt r e s i s t a n c e r e f e r r e d t ot h e l ow er s i d e

23 X eh = sqrt ( ( Ze h ^ 2) - ( R eh ^ 2 ) ) ; / / e q u i v a l e n tr ea ct an ce r e f e r r e d to th e h ig he r s i d e

24 X e l = X eh / ( a ^2 ) ; // e q u i v a l e n t r e ac t a nc e r e f e r r e d t ot h e l ow er s i d e

25

26 disp ( Z e h , e q u i v a l e n t i mp ea da nce r e f e r r e d t o t heh i g he r s i d e = )

27 disp ( Z e l , e q u i v a l e n t i mp ed an ce r e f e r r e d t o t he l o we r

s i d e = )28 disp ( R e h , e q ui v a l e n t r e s i s t a n c e r e f e r r e d t o t h e

h i g h er s i d e = )29 disp ( R e l , e q ui v a l e n t r e s i s t a n c e r e f e r r e d t o t h e

l ow er s i d e = )

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30 disp ( X e h , e q ui v a l e n t r e a c ta n ce r e f e r r e d t o t h e

h i g h er s i d e = )31 disp ( X e l , e q u i v a l e n t r e a ct a n ce r e f e r r e d t o t he l ow ers i d e = )

Scilab code Exa 16.3 For the transformer compute A efficiency B voltageregulation

1 // a

2 P = 50; / / p o we r r a t i n g ( i n kVA )3 P h = 81 0; // w at tm et er r e ad i ng on s h o rt c i r c u i t t e s t4 P l = 39 6; // w at tm et er r e ad i ng on open c i r c u i t t e s t5 I h = 2 0. 8; // ammeter r e a d i n g on s h o rt c i r c u i t t e s t6 p f = 0. 8; // power f a c t o r = 0 . 8 l a g g i n g7

8 l os se s = ( Ph + P l) / 10 00 ; // l o s s e s i n kW9 o ut pu tP = P * pf ; / / o u t p u t p ow er

10 i np ut P = o ut pu tP + l os se s ; / / i n p u t p o we r11

12 e f fi c ie n cy = o ut p ut P / i n pu tP ;

13 disp ( a )14 disp (efficiency , e f f i c i e n c y = )15

16 // b17 X eh = 4; // e q u i v a l e n t r e ac t a nc e r e f e r r e d t o t he

h i g h er s i d e18 R e h = 1 .8 7; // e q u i v a l e nt r e s i s t a n c e r e f e r r e d t o t h e

h i g h er s i d e19 Z e h = c om p le x ( Re h , X eh ) ; / / e q u i v a l e n t i mp ed an ce

r e f e r r e d t o t h e h i g he r s i d e

20 i h = c o mp l ex ( I h * pf , - Ih * sqrt (1 - ( pf ^ 2) ) );21 V 1 = 2400 + Zeh * ih ; // p r i ma ry v o l t a g e22

23 v o lt a ge R eg u la t io n = ( real ( V 1 ) - 2 4 0 0 ) * 1 0 0 / 2 4 0 0 ; //p e rc e n t v o l t a g e r e g u l a t i o n

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24 disp ( b )

25 disp (voltageRegulation , p e rc e nt v o l t a g e r e g ul a t on = )

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Chapter 18

The three phase Induction

motor

Scilab code Exa 18.1 Find A input line current and power factor B devel-oped electromagnetic torque C horse power output D efficiency

1 // a2 V 1 = 44 0/ sqrt ( 3 ) ;

3 s = 0 .0 25 ; / / s l i p4 r 1 = 0. 1;

5 r 2 = 0 .1 2;

6 x 1 = 0 .3 5;

7 x 2 = 0. 4;

8

9 z = c om pl ex ( r1 + r2 /s , x1 + x2 );

10 i 2 = V1 /z ; / / i np ut l i n e c u r r e nt11 I2 = sqrt ( real ( i2 ) ^2 + imag ( i 2 ) ^ 2 ) ; / / m a gn it ud e o f

i np ut l i n e c u r r e n t12 disp ( a )

13 disp ( i 2 , i np ut l i n e c ur r e nt = )1415 i 1 = c o mp l ex ( 1 8* cos ( - 1 . 4 84 ) , 1 8* sin ( - 1 . 4 8 4 ) ) ; //

m a g n et i z i ng c u r r e n t

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16 I1 = sqrt ( real ( i1 ) ^2 + imag ( i 1 ) ^ 2 ) ; / / m a gn it ud e o f

m a g n et i z i ng c u r r e n t17 i = i1 + i2 ; // t o t a l c u r r e n t drawn fro m t he v o l t a g es o u r c e

18 I = sqrt ( real ( i )^2 + imag ( i ) ^ 2 ) ; / / m a gn i tu d e o f t o t a l c u r r e n t

19 t h et a = atan ( imag ( i ) / real ( i ) ) ; // p ha se d i f f e r e n c eb et we en c u r r e n t and v o l t a g e

20 pf = cos ( t h e t a ) ; // p ower f a c t o r21 disp ( p f , power f a c t o r = )22 if t he ta >= 0 then

23 disp ( l e a d i n g )

24 e l s e d i sp ( l a g g i n g )25 end

26

27 // b28 f = 60; / / h e r t z29 n s = 1 80 0;

30 w s = 2 * %p i * ns / f ; // s t a t o r a n gu l ar v e l o c i t y31 P g = 3 * I2 ^ 2* r 2 / s; //pow e r32 T = Pg / ws ; // d e v el o p e d e l e c t r o m a g n e t i c t o r q ue33 disp ( b )34 disp (T ,

d e v e l o pe d e l e c t r o m a g n e i c t o r q ue ( i n Newton

m e t e r ) = )35

36 // c37 P ro t = 9 50 ; // r o t a t i o n a l l o s s e s ( i n w at ts )38 P o = Pg *(1 - s ) - Prot ; / / o u t p u t p ow er39 H Po = Po / 74 6; / / o u tp ut h o r s e p ower40 disp ( c )41 disp ( H P o , o u tp u t h o r s e p ower = )42

43 // d

44 P c = 1 20 0; // c o r e l o s s e s ( i n W)45 S CL = 3* I ^2 * r1 ; / / s t a t o r c op pe r l o s s46 R C L = 3 * I2 ^ 2 * r2 ; // r o t a r c op pe r l o s s47 loss = Pc + SCL + RCL + Prot ; / / t o t a l l o s s e s48 Pi = real ( 3 * V 1 * i ) ; / / i n p u t p o we r

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49 e ff ic ie nc y = 1 - ( l os s / Pi ) ;

50 disp (efficiency , e f f i c i e n c y = )

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Chapter 19

Computations of Synchronous

Motor Performance

Scilab code Exa 19.1 Find A induced excitation voltage per phase B linecurrent C power factor

1 // a2 e f fi c ie n cy = 0 .9 ;

3 P i = 2 00 * 74 6 / e f fi c ie n cy ; / / i n p u t p o we r4 x = 11; / / r e a c t a n c e o f t h e m oto r5 V 1 = 2 30 0/ sqrt ( 3 ) ; // v o l t a g e r a t i n g6 d el ta = 1 5* % p i / 18 0; / / po we r a n g l e7 E f = P i *x / ( 3* V 1 * sin ( d e l t a ) ) ; / / t h e i n d u c e d

e x c i t a t i o n v o l t a ge p er p h a s e8 disp ( a )9 disp ( E f , t he i nd uc ed e x c i t a t i o n v o l t a g e p er p ha se =

)10

11 // b

12 z = c om p le x ( 0 , x) ; // i m p ed an ce o f t h e m ot or13 e f = c o mp l ex ( E f * cos ( - d e l t a ) , E f * sin ( - d e l t a ) ) ;14

15 Ia = ( V1 - ef )/ z ; / / a rm at ur e c u r r e n t16 disp ( b )

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17 disp ( I a , a rm at ur c u r r e nt = )

1819 // c20 t h e at a = atan ( imag ( I a ) / real ( I a ) ) ; // p ha se d i f f e r e n c e

b et we en I a and V121 pf = cos ( t h e a t a ) ; // p ower f a c t o r22

23 disp ( c )24 disp ( p f , power f a c t o r = )25

26 i f s in ( t h ea t a ) > 0 then

27 disp ( l e a d i n g )

28 else29 disp ( l a g g i n g )30 end

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Chapter 20

DC machines

Scilab code Exa 20.2 Caculate A electromagnetic torque B flux per poleC rotational losses D efficiency E shaft laod

1 // a2 V t = 23 0; // ( i n v o l t s )3 I a = 73; / / a r m at u re c u r r e n t ( i n amps )4 I f = 1. 6; / / f e i l d c u r r e n t ( i n amps )5 R a = 0 .1 88 ; / / a rm at ur e c i r c u i t r e s i s t a n c e ( i n ohms )

6 n = 11 50 ; / / r a t e d s p ee d o f t h e r o t o r ( i n rpm )7 P o = 2 0* 74 6; / / o u t p ut p ow er ( i n w a t ts )8

9 Ea = Vt - ( Ia *Ra ); // a r m at ur e v o l t a g e10 w m = 2 * %p i * n /6 0; / / r a te d s pe ed o f t he r o t o r ( i n r ad /

se c )11 T = Ea *Ia /wm ; / / e l e c t r o m a g n e t i c t o r q ue12

13 disp ( a )14 disp (T , e l e c t r o m a g n e t i c t o rq u e = )

1516 // b17 a = 4 ; // no . o f p a r a l l e l a rm at ur e p at hs18 p = 4 ; / / no . o f p o l e s19 z = 882; / / no . o f a rm at ur e c o n d u c to r s

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20 f lu x = E a * 60 * a /( p * z *n ) ; // f l u x p e r p o l e ( i n Wb)

2122 disp ( b )23 disp ( f l u x , f l u x p er p o l e = )24

25 // c26 Pr ot = ( Ea * Ia ) - Po ; // r o t a t i o n a l l o s s ( i n wat t )27 disp ( c )28 disp ( P r o t , r o t a t i o n a l l o s s e s = )29

30 // d31 los ses = Prot + ( Ia ^2 * Ra ) + ( Vt * If ) ;

32 P i = ( E a * I a ) + ( I a ^ 2 * R a ) + ( V t * I f ) ; // i n putpower

33 e ff ic ie nc y = 1 - ( l os se s / Pi ) ;

34

35 disp ( d )36 disp (efficiency , e f f i c i e n c y = )

Scilab code Exa 20.3 Determine the new operating speed

1 / / f i n a l f l u x = 0 . 8 i n i t i a l f l u x2 I a1 = 73; / / i n i t i a l a r m a tu r e c u r r e n t ( i n amps )3 V t = 23 0; // ( i n v o l t s )4 R a = 0 .1 88 ; // a rm a t u r e c i r c u i t r e s i s t a n c e5 n 1 = 1 15 0; / / i n i t i a l r o t o r s p e e d ( i n rpm )6 E a1 = 2 16 .3 ; / / i n i t i a l a r m a t u r e v o l t a g e7

8 I a2 = ( 1/ 0. 8) * I a1 ; / / f i n a l a rm at ur e c u r r e nt9 Ea2 = Vt - ( Ia2 * Ra ) ; / / f i n a l a rm at ur e v o l t a g e

1011 n 2 = ( E a2 / E a1 ) * ( 1/ 0 .8 ) * n1 ; // f i n a l r o t o r s pe ed12

13 disp ( n 2 , f i n a l r o t o r s pe ed ( i n rpm ) = )

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Scilab code Exa 20.4 find A motor speed B required pulse frequency Crepeat part A and B for the given ON time to cycle time ratio

1 // a2 r op = 0 .4 ; / / r a t i o o f ON t im e T0 t o c y c l e t im e Tp3 V b = 55 0; / / r a te d t e r m i na l v o l t a g e o f t he dc motor4 I a = 30; / / c u r r e n t drawn by t h e m ot or ( i n amps )5 R a = 1; // a rm at ur e c i r c u i t r e s i s t a n c e ( i n ohms )6 t s = 5 .9 4; / / t o r q ue and s p ee d p a ra m et er o f t h e m oto r

( i n Nm/A)7

8 V m = r op * Vb ; // a v e r ag e v a l u e o f t h e a rm at ur et e r m i na l v o l t a g e

9 Ea = Vm - ( Ia *Ra ); / / i n d u ce d a r ma t ur e v o l t a g e10

11 w m = Ea / ts ; / / m ot or s p e e d ( i n r a d / s )12 disp ( a )13 disp ( w m , m ot or s p ee d ( i n r ad / s ) = )14

15 // b16 d e lt aI = 5; // c ha ng e o f a rm at ur e c u r r e n t d u ri n g t he

ON p e r i o d17 L a = 0. 1; / / a r ma tu re w in d in g i n d u c t a n ce ( i n H)18 T o = L a* d el ta I /( Vb - Ea ) ; //ON tim e19 T p = T o/ r op ; / / c y c l e t im e20

21 f = 1/ Tp ; // r e q u i r e d p u l s e s p er s ec on d22 disp ( b )23 disp (f , r e q ui r e d p u l s e s p er s ec on d = )24

25 // c26 r op = 0 .7 ; // new r a t i o o f ON ti me T0 t o c y c l e t im e

Tp

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27 V m = r op * Vb ; // a v e r ag e v a l u e o f t h e a rm at ur e

t e r m i na l v o l t a g e28 Ea = Vm - ( Ia *Ra ); / / i n d u ce d a r ma t ur e v o l t a g e29

30 w m = Ea / ts ; / / m ot or s p e e d ( i n r a d / s )31 disp ( c )32 disp ( w m , m ot or s p ee d w it h To/Tp e q u a l t o 0 . 7 ( i n r ad

/ s ) = )33

34 T o = La * d el ta I /( V b - Ea ) ; //ON tim e35 T p = To / ro p; / / c y c l e t im e36

37 f = 1/ Tp ; // r e q u i r e d p u l s e s p er s ec on d38 disp (f , r e q u i r e d p u l s e s p e r s e co n d w it h To/Tp e q u a l

t o 0 .7 = )

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Chapter 23

Principles of Automatic

Control

Scilab code Exa 23.1 Determine the new transfer gain and feedback factor

1 d el ta Gi = 42 0 - 3 80 ; / / v a r i a t i o n i n t he w it ho utf e ed b ac k g a in

2 G i = 40 0; / / w i th o ut f e e db a c k g a i n3 T = 400; // t r a n s f e r f u nc t i o n o f t he c l o se d l oo p

s y s t e m4 / / ( v a r i a t i o n i n T) /T = ( c ha ng e i n G) /G (1 / 1+HG)

= 0 . 0 25 / / 1 + HG = R6 R = ( d e lt a Gi / G i ) / 0. 02 ;

7

8 G = T * R ; // new d i r e c t t r a n sm i s s i on g ai n w it hf e e d b a c k

9 H = ( G / T - 1) /G ; / / f e e d ba c k f a c t o r10

11 disp (G , new d i r e c t t r a n sm i s s i on g ai n w it h f e e db ac k = )12 disp (H , f e e db ac k f a c t o r s = )

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Chapter 24

Dynamic behaviour of Control

systems

Scilab code Exa 24.2 find A dynamic response of the system B positionlag error C change in amplifier gain D damping ratio and maximum percentovershoot E output gain factor for maximum overshoot equal to 25percent

1 // a2 / / p a r am e te r v a l u e s3 K p = 0. 5; //V/rad4 K a = 10 0; //V/V5 K m = 2 *1 0^ - 4 ; / / l b f t /V6 F = 1 .5 *1 0^ - 4 ; / / l b f t / r a d / s7 J = 10^ -5 // sl ug f t 28

9 K = Kp *Ka *Km ; / / l o op p r o p o t i o n a l g a in10 d r = F /(2 * sqrt ( K * J ) ) ; // d amp in g r a t i o11 wn = sqrt ( K / J ) ;

12 t s = 5 /( d r* wn ) ;

13 w d = wn *sqrt (1 - dr ^2) ; / / f r e q u e n c y a t w hi c h dampedo s c i l l a t i o n s o c c u r14 disp ( a )15 disp ( w d , damped o s c i l l a t i o n s o c c u r a t a f r e q u e n c y =

)

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16 disp ( d r , damping r a t i o = )

1718 // b19 Tl = 10 ^ -3; // l o a d d i s t u r b a n c e ( l b f t )20 e = Tl /K ; / / p o s i t i o n l a g e r r o r21 disp ( b )22 disp (e , p o s i t i o n l a g e r r o r ( i n r a d ) = )23

24 // c25 K aN ew = ( e / 0 .0 2 5) * K a ; // new l o o p g a i n26 disp ( c )27 disp ( K a N e w , new l oo p g ai n f o r whi ch t h e p o s i t i o n l a g

e r r o r i s e qu al t o 0 .0 25 r a d = )28

29 // d30 d rN ew = F / (2 * sqrt ( K p * K a N e w * K m * J ) ) ; //new damping

r a t i o31 disp ( d )32 disp ( d r N e w , new da mp in g r a t i o = )33

34 // e35 / / f o r a maximum o v e r s h o o t o f 2 5% , ( F + Qo ) / 2s q r t (K

J ) = 0 . 436 Q o = ( 0. 4* 2* sqrt ( Kp * K aN ew * Km * J) ) - F ;37 K o = Qo / ( Ka Ne w *K ) ; // o ut pu t g ai n f a c t o r38 disp ( e )39 disp ( K o , o ut p u t g ai n f a c t o r = )

Electrical Engineering Fundamentals_V. Del Toro

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