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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 = )
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 3.12 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) = )
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 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
14 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 7.11 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
14 disp ( I b r , c u r r e nt f l o w i n g t hr ou gh Z3 = )
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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 = )
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