manipal institute of technology manipal 576...

MANIPAL INSTITUTE OF TECHNOLOGY

Manipal – 576 104

DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGG.

CERTIFICATE

This is to certify that Ms./Mr. …………………...……………………………………

Reg. No. …..…………………… Section: ……………… Roll No: ………………... has

satisfactorily completed the lab exercises prescribed for Electrical Circuits Lab

[E&E 2111] of Second Year B. Tech. Degree at MIT, Manipal, in the academic year

2015-2016.

Date: ……...................................

Signature Signature

Faculty in Charge Head of the Department

CONTENTS

LAB

NO. TITLE

PAGE

NO. REMARKS

COURSE OBJECTIVES AND OUTCOMES i

EVALUATION PLAN & INSTRUCTIONS TO THE STUDENTS i

COURSE PLAN ii

SIMULATION EXPERIMENTS

MODULE 1-MATLAB 1-14

1 MATLAB – TUTORIAL 1 15-18

2 MATLAB – TUTORIAL 2 19-25

3 MODELING WITH SIMULINK 26-28

MODULE 2-PSPICE 29-32

4 PSPICE-NETLIST 33-38

5 PSPICE -SCHEMATICS 39-41

HARDWARE EXPERIMENTS

6 SUPERPOSITION AND RECIPROCITY THEOREMS 42-46

7 THEVENIN‘S AND NORTON‘S THEOREMS 47-52

8 MAXIMUM POWER TRANSFER THEOREM 53-56

9

A. SELF AND MUTUAL INDUCTANCE OF A COIL

B. POWER, POWER FACTOR AND POWER FACTOR

IMPROVEMENT.

57-60

61-63

10 THREE PHASE POWER MEASUREMENT 64-68

Electrical Circuits Laboratory

i

Course Objectives

Use simulation tools like MATLAB/SIMULINK and PSPICE to solve

engineering problem

Apply network theorems for the analysis of given electrical systems

Measure power consumed by a three phase star/delta connected load

Improve power factor of the load using capacitor

Course Outcomes

At the end of this course, students will be able to

Implement simple electric circuits in MATLAB/SIMULINK platform to perform

steady state analysis and transient analysis

Design GUI using MATLAB

Implement electric circuits in PSPICE to perform steady state analysis and

transient analysis.

Verify network theorems like Thevenin, Superposition etc.

Measure self and mutual inductance of a coil and improve power factor of the coil

Measure 3 phase power for balanced and unbalanced load

Evaluation plan

Internal Assessment Marks : 60%

Continuous evaluation component (for each experiment):10 marks

The assessment will depend on punctuality, preparation, conduction, maintaining

the observation note and answering the questions in viva voce

Total marks of the10 experiments will be reduced to marks out of 60

End semester assessment of 3 hour duration: 40 %

INSTRUCTIONS TO THE STUDENTS

1. Students should carry the Lab Manual Book and Observation Book to every lab

session.

2. Be in time and follow the institution dress code.

3. Show the results to the instructors on completion of experiments and copy the

results in the Lab record.

4. For Hardware experiments, Get the circuit verified by the instructor before

switching on the supply.

5. The students should not go out of the lab without permission


Dept of E&E, MIT Manipal ii

Course Plan

ELE 2111: ELECTRICAL CIRCUITS LABORATORY [0 0 3 1]

Module I Circuit Simulation using MATLAB /

SIMULINK Introduction to MATLAB: Interactive computation,

script files, function files. Steady-state analysis of

circuits: Solution of algebraic equation.

Transient analysis of circuits: Solution of system

equations using ODE solvers. Introduction to GUIDE.

Introduction to SIMULINK and SIMSCAPE.

Week 1-3

Module II Circuit Simulation using PSPICE

Introduction to PSPICE, Steady state analysis of DC

circuits, single & three-phase AC circuits, and coupled

circuits.

Frequency response of circuits – series & parallel

resonance.

Week 4-5

Module III Circuits Hardware Experiments

Verification of Theorems: Superposition, Reciprocity,

Thevenin‘s, Norton‘s and Maximum power transfer

theorems.

Measurement of self and mutual inductances

Power, power factor and pf improvement, Three phase

power measurement

Week 6-10

Repetition Week 11

Test Week 12


Dept of E&E, MIT Manipal 1

Module I

CIRCUIT SIMULATION USING MATLAB/ SIMULINK

Introduction to MATLAB

MATLAB is a high-level language and interactive environment for numerical computation,

visualization, and programming. MATLAB helps to analyze data, develop algorithms, and

create models and applications. The language, tools, and built-in math functions enable to

explore multiple approaches and reach a solution faster than with spreadsheets or traditional

programming languages. MATLAB can be used for a range of applications, including signal

processing and communications, image and video processing, control systems, test and

measurement, computational finance, and computational biology. More than a million

engineers and scientists in industry and academia use MATLAB, the language of technical

computing.

Simulink is a block diagram environment for multi domain simulation and Model-Based

Design. It supports system-level design, simulation, automatic code generation, and

continuous test and verification of embedded systems. Simulink provides a graphical editor,

customizable block libraries, and solvers for modeling and simulating dynamic systems. It is

integrated with MATLAB, enable to incorporate MATLAB algorithms into models and

export simulation results to MATLAB for further analysis.

Getting started

Using Windows Explorer, create a folder user_name in the directory

c:\eclab\batch_index

For uniformity, let the batch_index be A1, A2, A3, B1, B2, or B3 and user_name be the

roll number

Invoke MATLAB

Running MATLAB opens the Matlab Desktop on your monitor. Of these the

Command window is the primary place where you interact with MATLAB. The

prompt >> is displayed in the Command window and when the Command window is

active, a blinking cursor should appear to the right of the prompt.

>> cd c:\eclab\batch_index\roll number % sets working directory

The working directory can also be selected using the Browse for folder icon on the

desktop

The desktop includes these panels: Current Folder — Access your files.

Command Window — Enter commands at the command line, indicated by the prompt (>>).

Workspace — Explore data that you create or import from files.

Command History — View or rerun commands that you entered at the command line.



MATLAB is available on a number of computing environments: Microsoft Windows, UNIX

operating systems such as Linux and Macintosh.

Number Display Formats

When MATLAB displays numerical results, it follows several rules. By default an integer

will be displayed as an integer and real number as real number with four digits to the right of

the decimal point. We can override this default behavior by specifying different formats as

given in the table below.

Table1: examples of number display formats

MATLAB Command pi comments

format short 3.1416 4 digits after decimal point

format long 3.141592653589793 15 digits after decimal point

format shortE 3.1416e+000 4 digits after decimal point plus

exponent

format longE 3.141592653589793e+000 15 digits after decimal point

plus exponent

format shortEng 3.1416 Best of format short or shortE

format longEng 3.141592653589793 Best of format long or longE

format hex 400921fb54442d18 Hexadecimal

Format blank 3.14 2 decimal digits

MATLAB Help

>> help inv

The online help system is accessible using the help command. Help is available for

functions eg. help inv. Demos and help can be invoked using the Help Menu at the top

of the window and from the Start icon at the bottom left of the Matlab Desktop.

Table 2: Common MATLAB functions Availability

Elementary math functions doc elfun

Data Analysis and Fourier transforms doc datafun

Elementary matrices and matrix manipulation doc elmat

Specialized math functions doc specfun

Help topics helpwin

Various MATLAB functions along with syntax and description can be obtained by typing

commands given in Table 2 to the command window.



Elementary Matrix Operations

To enter a matrix with real elements

A = [ 5 3 7; 8 9 2; 1 4.2 6e-2] A is a 3x3 real matrix

To enter a matrix with complex elements

B = [5+3j 7+8j; 9+2j 1+4j] B is a 2x2 complex matrix

Transpose of a matrix A_trans = A'

Determinant of a matrix A_det = det(A)

Inverse of a matrix A_inv = inv(A)

Matrix multiplication C = A * A_trans

Operators

Arithmetic operators

+ : Addition - : Subtraction

* : Multiplication / : Division

\ : Left Division ^ : Power

Relational operators

< : Less than <= : Less than or equal to

> : Greater than >= : Greater than or equal to

== : Equal ~= : Not equal

Logical operators

& : AND | : OR ~ : NOT

Array operations

Matrix operations preceded by a . (dot) indicates array operation.

Simple math functions

sin, cos, tan, asin, acos, atan, sinh, cosh ….

log, log10, exp, sqrt …

Special constants

pi, inf, i, j, eps …..

Control Flow Statements

for loop : for k = 1:m

.....



end

while loop : while condition

.....

end

if statement : if condition1

.....

else if condition2

.....

else

.....

end

Exercise: Use MATLAB command window as a calculator to compute the following

expressions

Compute the following

i) ii) iii) iv)

v) log10(

[Answers: 1.0323, 1.0323, 0.1180, 20.0855, 1.3029]

i) sin ( ) ii) +

[Answers: 0.5, 1, ]

Some useful commands

who whos clc clf clear load save pause help exist ver version

Function Calling Syntax

SCRIPT FILE (M-File)

File Menu – new – Script file

Output=function name (input)



Will open a new M – file where one can type the program save as ex_eclab1.m (end with an

extension .m)

Note: It is important to note that an existing function name should not be used as a file

name.

Example: Find the power and energy 0.1H when the current and voltages are given by

Plot current, voltage, power and energy

% to plot current, voltage, power and energy

clear; clc;

t=0:0.01:2;

for i=1:length(t)

if t(i)<=0

I(i) = 0;

v(i)=0;

elseif t(i)>0 &t(i)<=1

I(i)=20*t(i);

v(i)=2;

else t(i)>=1

I(i)=20;

v(i)=0;

end

end

L=0.1;

I;v;P=I.*v; W=0.5*L*I.^2; plot(t,I,t,v,t,P,t,W)

The script file can also be executed from the Save and Run command in Debug Menu of the

MATLAB Editor.

Exercise: plot the power and energy stored in a 0.1H inductor when

for

The ‘DOT’ preceding the standard *, /, ^… gives element by element operation

The colon is one of the most useful operators in MATLAB. It can create vectors,

subscript arrays, and specify for iterations.



Array operations

EXAMPLE:

Example: enter a 5x5 matrix using the function ‗magic‘ and do the following

A(:,j) jth

column of A

A(i,:) ith

row of A

A(:,:) same as A

A(j:k) A(j), A(j+1),...,A(k)

A(:,j:k) A(:,j), A(:,j+1),...,A(:,k).

A(:,:,k) kth

page of three-dimensional array A.

A(i,j,k,:)

vector in four-dimensional array A. The vector

includes A(i,j,k,1), A(i,j,k,2), A(i,j,k,3), and so on.

A(:) all the elements of A, regarded as a single column.

A(:,2)=[] A (1,:)=[] Deletion of rows and columns

B = A(5, 1:5); Read columns 1-5 of row 5

B = A(2:2:4, 1:5);

Read columns 1-5 of rows 2 & 4.

B = A(:, 1:5);

Read columns 1-5 of all rows

%array-array mathematics

Q1=A+A;Q2=A-A;Q3=A.*A;Q4=A./A;

%array Element by element

%exponentiation

P=A.^2

%reciprocal

R=1./A

%raise 2 to the power of each element

%of array

T=2.^A

size(A)

ones(size(A))

zeros(size(A))

zeros(2,5)

%identity matrix

eye(3)

%uniformly distributed random array

%b/w 0 &1

rand(3)

%optimum array addressing

d=pi;

O=repmat(d,3,4)

%Array Manipulation

A=[1 2 3;5 6 7;8 9 2]

%set element in 2nd

row 3rd

column to 0

A(2,3)=0

clc; clear;

%array operations

%enter row vector

A=[0 2 4 6 8 10 12 14 16 18 20]

%or

A=0:2:20

%column vector

A1=A'

A2=[0;2;4;6;8;10;12;14;16;18;20]

%array elements can be accessed using subscripts

B=A(5)

% access a block of elements at one time

C=A(1:5)

D=A(6:end)

%reverse order

E=A(3:-1:1)

%to extract the elements in any order

F=A([4 8 1 10])

A3=linspace(0,20,11)

A4=logspace(-1,2,10)

%complex column vector

G=A2+A2*j

%array of 2 rows and 4 columns

H=[2 3 4 5;7 8 9 6]

%scalar - array operations

%apply the operation to all elements of the array

S1=2*A; S2=2-A; S3=2+A;S4=A/2;



Relational Operators

The relational operators are <, >, <=, >=, ==, and ~=. Relational operators perform element-

by-element comparisons between two arrays. They return a logical array of the same size,

with elements set to logical 1 (true) where the relation is true, and elements set to logical 0

(false) where it is not. If one of the operands is a scalar and the other a matrix, the scalar

expands to the size of the matrix.

Examples :

Logical Operators: AND ( &), OR( |), NOT( ~)

These operators are commonly used in conditional statements. The expression operands for

AND, OR, and NOT are often arrays of non-singleton dimensions. The MATLAB software

performs the logical operation on each element of the arrays. The output is an array that is the

same size as the input array or arrays. Table 3 shows the output of AND, OR, and NOT

statements that use scalar and/or array inputs. In the table, S is a scalar array, A is a non-

scalar array, and R is the resulting array. If two or more operations have the same precedence,

the expression is executed in order from left to right. In order to avoid compatibility problem

between different versions of MATLAB, it is better to practice parentheses according to the

precedence. Table 4 shows the operation and order of precedence.

%relational operations

A = [0.53 0.67 0.01 0.38 0.07 0.42 0.69];

X=0.02; X>=A;

all(A>=X)

any(A>=X)

find(A)%non-zero elements

find(A>X)

find(0 < A & A < 0.1*pi)

strcmp('Yes', 'No')%returns 0

strcmp('Yes', 'Yes') %returns 1

A = {'MATLAB','SIMULINK'; ...

'Toolboxes', 'MathWorks'};

B = {'Handle Graphics', 'Real Time Workshop'; ...

'Toolboxes', 'MathWorks'};

C = {'handle graphics', 'Signal Processing'; ...

' Toolboxes', 'MATHWORKS'};

strcmp(A,C)%compare with sensitivity to case

strcmpi(B,C)%compare without sensitivity to case



Table 3: Description and output of logical operators

Operation Result Description

S1 & S2 R = S1 & S2 AND

If both are true ,the result is

true(1), otherwise the result

is false(0)

S & A R(1) = S & A(1); ...

R(2) = S & A(2); ...

A1 & A2 R(1) = A1(1) & A2(1);

R(2) = A1(2) & A2(2); ...

S1 | S2 R = S1 | S2 OR

If either one or both are true

the result is true(1),

otherwise the result is

false(0)

S | A R(1) = S | A(1);

R(2) = S | A(2); ...

A1 | A2 R(1) = A1(1) | A2(1);

R(2) = A1(2) | A2(2); ...

~S R = ~S NOT

Give 1 if the operand is

false(0) and 0 if the operand

is true(1)

~A R(1) = ~A(1);

R(2) = ~A(2), ...

xor(A1,A2) R(1)=xor(A1(1),A2(1)),

R(2)=xor(A1(2),A2(2)),…

Give 1 if one operand only

is true , otherwise 0

A&&B

Short circuit operators B is evaluated only of A is true

Returns logical 1 (true) if

both inputs evaluate to true,

and logical 0 (false) if they

do not.

A||B Returns logical 1 (true) if

either input, or both, evaluate

to true, and logical 0 (false)

if they do not.

Table 4: Order of Precedence of Operation

Operation Order of

precedence Parentheses ( for nested parentheses , inner

have precedence)

1 (highest)

Transpose (.'), power (.^), complex

conjugate transpose ('), matrix power (^)

2

Exponentiation 3

Logical NOT (~) 4

Multiplication , Division 5

Addition , subtraction 6

Relational operators (<, >, >=, <=, ==,~=) 7

Logical AND (&) 8

Logical OR (|) 9

Short circuit AND 10

Short Circuit OR 11



Examples on logical operators

Graphics Using MATLAB MATLAB is capable of producing two dimensional x-y plots and three-dimensional plots,

displaying images and even creating and playing movies. MATLAB plotting functions and

tools direct their output to a figure window. Each figure is a separate window that can dock in

the desktop, and collect together with other plots in a Figure Group. The most common for

plotting 2 dimensional data is the plot function which sets of data arrays on appropriate axes

and connects the points with straight line. The plot function uses a default line style and color

to distinguish the data sets plotted in the graph. Change the appearance of these graphic

components or add annotations to the graph helps to present the data in a particular way. The

following code illustrates the basic components of a graph.

Examples:

% examples of logical operators

x=-3; y=6;

-5<x<-1 %returns 0

-5<x&x<-1 % returns 1

~(y<7) % returns 0

~y<7 %returns 1

~((y>=8)|(x<-1))% returns 0

~(y>=8)|(x<-1)% returns 1

A = [0 1 1 0 1];

B = [1 1 0 0 1];

A&B

A|B

~A

xor(A,B)

P=28;Q=21

C=bitand(A,B)% bitwise and operation

b=1;a=20

x = (b ~= 0) && (a/b > 18.5)

b=0;a=20

x = (b ~= 0) && (a/b > 18.5)

% To plot y = 2x for -10 x 10

clear; clc;

x=-10:0.01:10;

y=x.^2;

plot(x,y,'r')

xlabel('x')

ylabel('y')

title('plot of y=x^2')



Exercise: Create line plot using specific line width, marker color, and marker size by

programming also using interactive plottools

Functions to generate data points x=linspace(1,100,10), generates 10 equally spaced points between 1 &100

x=logspace(-1,2,100), generates 100 logarithmically equally spaced points between to

Try the following 2D plotting functions Line Graphs: plot, plotyy, loglog, semilogx,semilogy,stairs, contour,ezplot,ezcontour

Bar Graphs: bar, barh, hist, pareto, errorbar,stem

Area Graphs: area, pie, fill, contour, image, pcolor, ezcontourf

Direction Graphs: feather, quiver, comet

Radial Graphs: polar, rose, compass, ezpolar

Scatter Graphs: scatter, spy, plotmatrix

Example of a filled circle

Exercise: obtain the following as subplots using suitable functions

-4 -2 0 2 40

0.2

0.4

0.6

0.8

1

0 1 2 3 4-0.2

0

0.2

0.4

0.6

0.25

0.5

30

210

60

240

90

270

120

300

150

330

180 0

11%

33%

6%

28%

22%

clc;clear;

t=0:pi/20:2*pi;

fill(cos(t),sin(t),'r',0.8*cos(t),0.8*sin(t),'y')

axis square



Table 5. Examples of 3D plots

Plot Type Plot Program surf plot

)( 22 yxxez

-2

-1

0

1

2

-2

-1

0

1

2-0.5

0

0.5

[X,Y] = meshgrid(-2:0.2:2, -2:0.2:2);

Z = X .* exp(-X.^2 - Y.^2);

surf(X,Y,Z)

‘waterfall’

plot for the

peaks function

using meshz

-4

-2

0

2

4

-4

-2

0

2

4-10

-5

0

5

10

[X,Y] = meshgrid(-3:.125:3);

Z = peaks(X,Y);

meshz(X,Y,Z)

contour plot

of

)( 22 yxxez

-0.4

-0.2 -0.2

-0.2

00

0

0.2

0.2

0.2

0.4

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

[X,Y] = meshgrid(-2:.2:2,-2:.2:3);

Z = X.*exp(-X.^2-Y.^2);

[C,h] = contour(X,Y,Z);

set(h,'ShowText','on','TextStep',get(h,'LevelSte

p')*2)

colormap cool

Sphere plot

-10

12

34

-3

-2

-1

0

1-1

-0.5

0

0.5

1

[x,y,z] = sphere;

surf(x,y,z) % sphere centered at origin

hold on

surf(x+3,y-2,z) % sphere centered at (3,-2,0)

Visualizing

vector fields

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

clc;clear;

%visualizing vector fields

[x,y]=meshgrid(-2:0.2:2,-2:0.2:2);

V=x.^2+y;

dx=2*x;

dy=dx; % to fix dy same size as dx

dy(:,:)=1;% dy is same size as dx but all 1's

contour(x,y,V), hold on

quiver(x,y,dx,dy), hold off



Polynomials, Curve Fitting & interpolation Example: Consider the polynomial

Curve fitting

Note: Also use Basic Fitting tool & Curve fitting App

Examples on Interpolation

Function Type Example

Interp1

Y1=Interp1(X,Y,XI)

1D interpolation x = 0:10; y = sin(x); xi = 0:.25:10;

yi = interp1(x,y,xi);

plot(x,y,'o',xi,yi)

Interp2

ZI = interp2(X,Y,Z,XI,YI)

2D interpolation [x,y,z] = peaks(10); [xi,yi] = meshgrid(-

3:.1:3,-3:.1:3);

zi = interp2(x,y,z,xi,yi);

mesh(xi,yi,zi)

%Polynomials with MATLAB

P=[1 -10 50 -20 -70 40]; %enter the polynomial coefficients

polyval(P,5); %evaluate the polynomial for x=5

x=-2:0.01:4;

y=polyval(P,x); % generate y for x = -2 to 4

plot(x,y); %plot the polynomial

R=roots(P); %roots of the polynomial

poly(R); % to find polynomial coefficients

P1=[0 0 2 0 -4 -8];

P2=P+P1; %addition

P3=conv(P,P1); %multiplication

[q,r]=deconv(P3,P); %division

k=polyder(P); %Derivative

k1=polyder(P,P1); %derivative of P*P1

[N,D]=polyder(P3,P); %Derivative of P3/P

I=polyint(P); %integrate the polynomial

clc; clear;

%example of curve fitting

x=[-5 -4 -2.2 -1 0 1 2.2 4 5 6 7];

y=[0.1 0.2 0.8 2.6 3.9 5.4 3.6 2.2 3.3 6.7 8.9];

p=polyfit(x,y,6);

f=polyval(p,x);

plot(x,y,'b',x,f,'r')

table=[x; y; f;y-f]'



Example 2:

%To plot a function its integral

and derivative

syms x;

f=sin(x)/x;

figure(1)

ezplot(f,[-15, 15])

F=int(f,x);

figure(2)

ezplot(F,[-15, 15])

G=diff(F)

figure(3)

ezplot(G,[-15, 15])

Symbolic Computation Symbolic Math Toolbox provides functions for solving and manipulating symbolic math

expressions and performing variable-precision arithmetic. It allows analytically perform

differentiation, integration, simplification, transforms, and equation solving. Also generate

code for MATLAB, Simulink, and Simscape from symbolic math expressions.

To declare variables x and y as symbolic objects use the syms command:

syms x y

diff; int ;solve; ezplot; ezplot3;ezsurf; dsolve;

Example 1

clc; clear;

%to declare the variables

syms x y w s z n S

(x-y)*(x+y)

expand(ans)

factor(ans)

w=(x-y)*(x+y)

subs(w,{x,y},{[1,2]})

%inverse laplace transform

ilaplace(1/(s-1))

%inverse z transform

iztrans(z/(z-2))

[x,y] = solve('x^2*y^2 - 2*x - 1 = 0','x^2 - y^2 - 1 = 0')

x=double(x); y=double(y);



Example: Current signal is given; find the voltage across the capacitor of 2F

Hint:

%generate a test signal

[u,t]=gensig('square',5,30,0.1);

H=tf(1,[2 0]);

lsim(H,u,t)

REFERENCES

1. William Palm III, Introduction to MATLAB 7.4 for Engineers, MGH 2007.

2. D. Hanselman and B. Littlefield, Mastering MATLAB , Prentice Hall, 2011

3. Brian R. Hunt, Ronald L. Lipsman, Jonathan M. Rosengurg, A Guide to MATLAB,

Cambridge University Press, 2011.

4. Amos Gilat, MATLAB - An Introduction with Applications, Wiley India Edition, 2010.

5. Rudra Pratap, Getting Started with MATLAB – A Quick Introduction for Scientists and

Engineers, Oxford University Press, 2010.

6. Shampine I.F, Solving ODEs with MATLAB, Cambridge University Press, 2003.

7. www.mathworks.com

TAH & Installation Manipal Institute of Technology is the first Institute in the country that has implemented a

campus-wide license for the MATLAB and Simulink product families. The campus-wide

license ensures access to world-class research infrastructure for our students, research

scholars and faculty members.

http://www.mathworks.com/



TUTORIAL 1

Objective: Solution of first and second order system equations using MATLAB(numerical)

and symbolic math toolbox(symbolic)

1. Find the loop currents of the circuit given in Fig. 1.1

VVo 10

+-

1I 2I

101R 103R

52R 204R

Fig. 1.1 Using Mesh Analysis

0

10

355

515

2

1

I

I

Sample Solution (using command line editor)

Z = [15 -5;-5 35]; v = [10; 0]; i = inv(Z) * v; Results: I1= 07, I2= 0.1

Using Symbolic math toolbox

syms I1 I2;

e1='15*I1-5*I2=10'; e2='-5*I1+35*I2=0'; s=solve(e1,e2,'I1','I2'); I=[s.I1,s.I2];



i(t)V=1

R =1

t=0

10 mF

2. Find the loop currents & node voltages of the circuit given in Fig. 1.2

(Write script files for solution) using command ‘inv’ and also using ‘solve’

Fig. 1.2

3. Find the transient response of the circuit given in Fig. 1.3

Given:

Sample Solution : (illustration of for loop)

clear; clc; disp(' RC transient analysis') v = input(' Enter source voltage : '); r = input(' Enter value of resistance :'); c = input(' Enter value of capacitor: '); T = r*c;

fprintf('\n The results are : \n\n') disp('t (sec) i (A) v_c (V)') for n = 1:10

t(n) = (n-1)*T/2; Fig.1.3

i(n) = (v/r)* exp(-t(n)/T); v_c(n) = v* (1 - exp(-t(n)/T)); fprintf('%6.4f\t%6.4f\t%6.4f\n', t(n),i(n),v_c(n))

end;

Colon operator: The colon operator is useful for creating index arrays, creating vectors of

evenly spaced values, and accessing sub-matrices. A regularly spaced vector of numbers is

obtained by means of

n = initial value : increment : final value

RCwhereeVtv

eR

Vti

t

c

t

1)(

)(



Sample Solution : (generating vectors)

clear; clc;

disp(' RC transient analysis')

v = input(' Enter source voltage : '); r = input(' Enter value of resistance : '); c = input(' Enter value of capacitor : '); T = r*c; fprintf('\n The results are : \n\n') disp('t (sec) i (A) v_c (V)') t = 0 : 0.5*T : 5*T; % start value : increment : final value i = (v/r) * exp(-t/T); v_c = v * (1 - exp(-t/T)); % To output the results in tabular form A = [t; i; v_c]; % concatenates the vectors fprintf('%6.4f\t%6.4f\t%6.4f\n', A);

4. The switch shown in the diagram in Fig.1.4 has been in position ‘a’ for long time.

At t = 0, the switch is moved to position ‘b’ where it remains for 2 s and then moves

back to position ‘a’, where it remain indefinitely.

i. Obtain analytical expression for vc(t) and ic(t) for all t

ii. Use MATLAB to obtain plots vc(t) versus time and current ic(t) versus time over

the range 0 s ≤ t ≤ 10 s

Fig 1.4

5. Plot the inductor current and the capacitor voltage for time, 0 t 10s, of an RLC

series circuit with R = 1, L = 1H, & C = 10mF connected to a dc source of 10V

through a switch. The switch is closed at t = 0 & the circuit elements are initially

relaxed.

Discussion:

The RLC circuit can be described by the following KVL equations in terms of the state

variables vc and i, where vc is the voltage across the capacitor and i is the current through

the inductor.

+-

71R

12R33R

FC 5.0

VVo 100

a

b )(tic



'

'

c

c

Cvi

VvLiRi

%solution

clc;clear;

syms I(t) vc(t)

V=10;R=1;C=.01;L=1;

[I(t),vc(t)]=dsolve(diff(I)==(V/L)-(R/L)*I-(vc/L),diff(vc)==(1/C)*I,...

I(0)==0,vc(0)==0);

%[I(t),vc(t)]=dsolve(diff(I)==1-I-vc,diff(vc)==I,I(0)==0,vc(0)==1)

I=simplify(I)

vc=simplify(vc)

figure(1)

ezplot(I)

figure(2)

ezplot(vc)



TUTORIAL 2

Objective: Solution of first and second order system equations using Ordinary Differential

Equation (ODE) solvers and introduction to GUIDE.

Function File:

A function file is also an m-file, just like a script file except it has a function definition line at

the top that defines the input and output explicitly. function <output_list> = fname <input_list>

Save the function file as fname.m The filename will become the name of the new command

for MATLAB. Variables inside a function are local. Use global declaration to share variables.

1. Plot the dc transient response of a series RL circuit with R = 1 , L = 1 H, and

V=10 V. Switch is closed at t = 0.

Discussion:

The RL circuit can be described by the following KVL equations in terms of the state

variable i, where i is the current through the inductor.

VLiRi '

In order to use the MATLAB function ode23, the first order differential equations must be

placed in a .m file that returns the derivative of the state variable. The equation is:

ixwhereLRxVx /)('

Sample Solution

% Transient analysis RL series circuit % Solution using Ordinary Differential Equation (ODE) solver "ode23" % This program uses the function "rl_sys" global V R L; % define circuit parameters V=10; R=1; L=1; % define initial conditions IL0 x0=0; % define solution parameters t0=0; tf=10; tspan=[t0,tf]; % Numerical integration using MATLAB function "ode23" % User defined function "rl_sys" describes the system [t,x]=ode23('rl_sys',tspan,x0); % o/p results plot(t,x); %************************************************************************************



% open a new function file by name rl_sys.m % function "rl_sys" describes the system function xdot = rl_sys(t,x); global V R L; xdot=(V-R*x)/L; %************************************************************************************

2. Plot the dc transient response of a series RC circuit with R = 1 , C = 1mF, and

V = 1V. Switch is closed at t = 0. Measure Vc at t = 1s.

global V R C V=1; R=1; C=1e-3; options = odeset('RelTol',1e-4,'AbsTol',1e-4); [t,VC] = ode23(@RC,[0 10],[0],options); % o/p results plot(t,VC); %************************************************************************************ %function file %rc.m function vcdot=RC(t,VC) global V R C; vcdot=(V-VC)/(R*C);

%************************************************************************************

3. Obtain the capacitor voltage vo(t) and the current i(t) for 0 t 2 second. Circuit is

initially relaxed. At t=0, the switch is closed to position 1 and at t=0.5s the switch is

moved to position 2.

Fig.2.1

%Example: RC with switch clc; clear; global V R C; V=10;R=1;C=100e-3; options=odeset('RelTol',1e-6,'AbsTol',1e-6); %switch in position 1 VCi1=0; [t1,VC1]=ode45(@RC,[0 0.5],VCi1,options); I1=(V-VC1)./R; %switch in position 2 V=-20;

1

2

10 V 20 V

1 ohm

100 mFi(t)

Fig 9_2



for k=1:length(t1) if (t1(k)==0.5) VCi2=VC1(k); end end [t2,VC2]=ode45(@RC,[0.5 2],VCi2,options); I2=(V-VC2)./R; VC=[VC1;VC2];t=[t1;t2];I=[I1;I2]; plot(t,I,t,VC) grid;

Exercise: Use Publish to get the report in html/ pdf format

4. Plot the dc transient response of a series RC circuit with R = 1 , C = 1 F, and

V = 10 V. Switch is closed at t = 0 and Vc(0) = -5 V. Measure Vc at t = 2s.

ixvxwhereLRxxVx

Cxx

c

2121

2

, ,/)(

/

'

2

'

1

global V R L C; % define circuit parameters V=10; R=1; L=1; C=10e-3; % define initial conditions VC0=0; IL0=0; x0=[VC0;IL0]; % define solution parameters t0=0; tf=10; tspan=[t0,tf]; % Numerical integration using MATLAB function "ode23" % User defined function "rlc_sys" describes the system [t,x]=ode23('rlc_sys',tspan,x0); % o/p results plot(t,x); grid; %********************************************************************** % rlc_sys.m % function "rlc_sys" describes the system function xdot = rlc_sys(t,x); global V R L C; xdot=[x(2)/C;(-x(1)-R*x(2)+V)/L]; %************************************************************************************

Graphical User Interface (GUI) Design a GUI that will add together two input numbers and displays answer in a designated

text field.

GUIDE - Graphical User Interface Development Environment.

To invoke GUIDE from MATLAB command window

>>guide

Create a blank GUI and drag & drop the required objects on to the layout editor.



GUI Design Steps:

1. Type guide in the command window and this will open the GUIDE QUICK START

window. Choose the first option Blank GUI (Default) and click OK to open the layout

editor.

2. Drag and drop Two Edit Text components (for two input numbers), Three Static

Text components (for +, = and result field) and a Pushbutton component (for

addition operation) to the GUI by clicking on the respective component icon.

Rearrange the components accordingly.

3. Edit the properties of each of the components by double clicking on it. This will open

the Property Inspector Window. In the Property Inspector Window, edit the string

parameter appropriately (say the string parameter of two Edit Text components must

be set to 0, the static Text components to +, = and 0 and Pushbutton to Add). Also

observe the tag parameter of each of these components as these will be used in GUI

callbacks code.

4. Save the designed GUI and this will open the automatically generated .m file.

5. Within the .m file, go to function pushbutton1_Callback and type the

following code: Finally, execute the GUI program by pressing the icon on the GUIDE editor

and test the GUI by inputting some numbers.

Congratulations on creating your first GUI!!!!



Num1= get(handles.edit1,'String');

Num2 = get(handles.edit2,'String');

total = str2num(Num1) + str2num(Num2);

c = num2str(total);

set(handles.text3,'String',c);

6. Modify the above GUI and make a scientific calculator arithmetic, trigonometric

and logarithmic functions.

7. Create a GUI to show the addition of two sinusoidal components with different

frequencies.



% Get user input from GUI

f1 = str2double(get(handles.f1_input,'String'));

f2 = str2double(get(handles.f2_input,'String'));

t = eval(get(handles.t_input,'String'));

% Calculate data

x = sin(2*pi*f1*t) + sin(2*pi*f2*t);

y = fft(x,512);

m = y.*conj(y)/512;

f = 1000*(0:256)/512;

% Create frequency plot in proper axes

plot(handles.frequency_axes,f,m(1:257))

set(handles.frequency_axes,'XMinorTick','on')

grid on

% Create time plot in proper axes

plot(handles.time_axes,t,x)

set(handles.time_axes,'XMinorTick','on')

grid on

8. Create a GUI as given below to plot different 3D graphs with options to change

shading, colormap and axis



Assignment:

1. Plot the inductor current and the capacitor voltage for time, 0 t 10s, of an RLC

series circuit with R = 1, L = 1H, & C = 10mF connected to a dc source of 10V

through a switch. The switch is closed at t = 0 & the circuit elements are initially

relaxed.( using ode solver)

2. Steady- state condition exist in the network shown in figure below at t = 0-, when

the V2 =10 V is connected to the RL circuit. At t = 0+, the switch is moved

downward and the source V1 is then connected to the RL circuit, where it remains

for t ≥ 0.

i. Write and solve theoretically the differential loop equation for current i(t), and

the voltage vR(t) and vL(t) for t ≥ 0.

ii. Create the script file RL_IC that returns the solution for part 1 by using

MATLAB symbolic solver dsolve.

iii. Repeat part 2 by using the MATLAB numerical solver ode45.

iv. Also obtain the plots of i(t), vR(t) and vL(t) versus t.

+-

2R

HL 1

VV 201

b

)(ti+-VV 102

0t

Fig 2.2

3. Create GUI for transient analysis with one pushbutton for RL, one for RC and one

for RLC circuit. Enter the filename which is used to solve RL circuit (Q.No 1) in the

pushbutton callback meant for RL. Use global declaration to share variables.



TUTORIAL 3

Objectives: Transient Analysis of simple circuits using SIMULINK and SIMSCAPE.

SIMULINK - Graphical modeling, dynamic system simulation

To invoke SIMULINK from MATLAB Command Window

>> simulink or

In File menu, Select New Model

Draw the block schematic in the worksheet by copying the components from

Simulink library.

Set simulation parameters and then start simulation. After the simulation click on

scope to view the result.

1. An RLC series circuit with R = 1, L = 1H, & C = 10mF is connected to a dc

source of 10V through a switch. Plot the inductor current and the capacitor voltage

for time, 0 t 10s, if the switch is closed at t = 1s & the circuit elements are

initially relaxed. Repeat the question taking R=20 and 100.

Fig. 3.1

2. Using SIMULINK, obtain the capacitor voltage vo (t) and the current i(t) for

0 t 2s. Circuit is initially relaxed. At t=0, the switch is closed to position 1 and at

t=0.5s the switch is moved to position 2.

Fig.3.2

output

To WorkspaceStep

Scope

Mux

s

1

Integrator1

s

1

Integrator

1/C

Gain2R

Gain1

1/L

Gain

1

2

10 V 20 V

1 ohm

100 mFi(t)

Fig 9_2



Fig.3.2.1

3. Find the inductor current and capacitor voltage if the switch is moved to position 1

at time t=0s and switched to position 2 at time t=2s. Initial capacitor voltage

Vc(0)= -20 V.

+-VVo 100

I

FC 001.0

200cR

HL 5513.0

1LR

a

b

Fig.3.3

Fig.3.3.1



4. Plot the transient response (0 t 10s) of a series RL circuit with R = 1 ,

L = 100mH and square wave excitation with V= 10V, period 1s and 50% duty cycle.

5. Plot the dc transient response of a series RC circuit with R = 1, C = 1 F and

V= 10 V. Switch is closed at t = 0 and Vc (0) = -5 V. Measure Vc at t = 2s. Create

subsystem and mask it.

Introduction to SIMSCAPE

SIMSCAPE- Extension of Simulink - represents physical components or relationships

directly.

Open SimulinkNewModel. Get all the components from foundation library and

utilities of Simscape blockset.

PS-S block: used to interface simscape with simulink blocks.

S-PS block: used to interface simulink with simscape blocks

1. Plot the dc transient response of a series RLC circuit with R = 1 , L = 1 H, C=10mF,

and V=10 V. Switch is closed at t = 0.

FIFf

Fig 3.4

Repeat exercises 4 and 5 using SIMSCAPE.

+V -

f(x)=0

Scope

R

+ -

PSS

PSS

L

+ - +I

-

C

+-



Module II

CIRCUIT SIMULATION USING PSPICE What is PSPICE?

SPICE is an acronym for Simulation Program with Integrated Circuit Emphasis and

PSPICE is a PC version of SPICE. The program SPICE was developed at University of

California; Berkeley in early 1970‘s and has become a standard tool in the area of circuit

simulation. Over the years many mainframe and PC versions of SPICE have evolved.

PSPICE contains circuit models for common circuit elements, active as well as passive,

analog as well as digital, and is capable of simulating most of the electrical and electronic

circuits. The software forms a set of analysis equations from the circuit description and is

solved using numerical methods.

OrCAD PSPICE, a Windows based package, comes as part of Cadence PCB System

Division's OrCAD series products consisting of tools for analog and digital circuit simulation,

waveform analysis, and PCB design. In this laboratory module we will be using OrCAD

PSPICE Student version 9.2 for circuit simulation. This public domain software has most of

the capabilities of its full version, except for a limitation on circuit size. In OrCAD PSPICE,

the circuit can be described either as a netlist or as a schematic. However, we will be using

the netlist approach for this laboratory module. The circuit is then analyzed and the waveform

are displayed interactively using the waveform viewer, Probe. If we visualize PSPICE as a

software breadboard, then the Probe can be viewed as a software oscilloscope.

References:

1. Rashid M.H, Spice for Circuits and Electronics using PSPICE, PHI, 2004

2. Conant Roger., Engineering Circuit Analysis with Pspice and Probe, MGH, 1993

3. Al Hashimi Bashir, Art of Simulation using pspice analog and digital, CRC, 1995.

4. Tuinenga P.W.SPICE a Guide to circuit simulation and analysis using PSPICE, PHI, 1990



A Quick Reference to OrCAD PSPICE Format of Input Data file

The input data file is to be created with extension .cir. It consists of 5 modules

1. Title and comment statements

Title statement - first line of the file, serves as identification

Comments - * as first character

2. Data statements - netlist

3. Solution control statements

4. Output specification statement

5. END statement

Format of Data Statements

Passive elements

Resistor : Rname node1 node2 value

Capacitor : Cname node1 node2 value IC = value

Inductor : Lname node1 node2 value IC = value

Coupled ckt : Kname L1 L2 value

- Initial condition, IC = value, is optional

- Suffixes for specifying value:

P - pico, N - nano, U - micro, M - milli, K - kilo, MEG - mega

- Units optional

- Coupled circuits, value – coefficient of coupling, dot end should be the first node of L1

and L2.

Independent sources

Voltage source : Vname +node -node type value

Current source : Iname +node -node type value

Type - options

- DC value

- AC magnitude phase



- PWL (t1 v1 t2 v2 ....)

v(t)

t t1,v1

t2,v2

t3,v3

t4,v4

- PULSE(v1 v2 td tr tf tw per)

v(t)

t

v1

v2

td tr tw tf

per

- SIN(Vo Va freq td alpha theta)

Vo

td

Vo+Va

t

V0(t)=Vo + Va Sin(2. (freq.(t-td)-theta/360)). exp(- alpha.(t-td))

- EXP(V1 V2 td1 tau1 td2 tau2)



Dépendent sources

VCVS Ename +node -node +nc -nc value VCCS Gname +node -node +nc -nc value CCVS Hname +node -node vn value CCCS Fname +node -node vn value

Electronic devices

Diode D(name) n1 n2 DNAME

.model DNAME D model parameters

BJT Q(name) NC NB NE QNAME

.model QNAME NPN model parameters

FET J(name) ND NG NS JNAME

.model JNAME NJF model parameters

Format of Solution Control Statement

DC operating point . OP

DC Analysis .DC Sname ivalue fvalue inc

AC Analysis .AC options npoints fstart fstop

options : LIN, DEC and OCT

Transient Analysis .TRAN tstep tstop SKIPBP

Transfer function .TF Vout Vin

Fourier Analysis .FOUR freq N V1 V2 …..

Format of Output Specification Statement

. PRINT type list

type - type of analysis

list - variable list

. PROBE invokes software oscilloscope for interactive display

Note: All the exercises can be simulated either by describing the netlist or using the

schematic capture mode. Students are advised to use the netlist mode for Tutorial 4 and

Schematic capture mode for Tutorial 5.



TUTORIAL 4

Getting started with PSPICE AD:

Invoke PSpice AD Lite Edition OrCAD PSpice A/D Demo window appears.

Open text editor File Menu New Text File. Edit the PSpice input data file. Save the file in

your directory with file extension .cir

Now the circuit described as a netlist is ready for simulation.

Simulating the circuit

File Open Simulation and open

Simulation Run

Simulation complete message appears if the netlist was free from errors.

Viewing output file

View Output file

Electrical Circuit Simulation using PSPICE

Objective: This tutorial aims at introducing PSPICE as an electrical circuit simulation tool.

Exercises: DC Circuit Steady State Analysis

1. Find the load voltage and load current of the circuit given in Fig. 4.1(a).

+

-

2

1

2

4A 1

RL=2

2A

12V

Fig.4.1(a)

+-

2

1

2

4A 1

2A

Fig. 4.1(b)

1

RL=2

2

3 4

0

Name all circuit elements.

Mark all the nodes of the network.

Choose a reference node (0) see Fig. 4.1(b) .

Create input file dc_ckt1.cir as listed.



Sample solution (using dc_ckt1.cir)

* DC steady state analysis * CIRCUIT DESCRIPTION V1 1 0 DC 12 I1 2 4 DC 2 I2 0 3 DC 4 R1 2 1 2 R2 2 3 1 R3 3 4 2 R4 4 0 1 RL 2 0 2 * SOLUTION CONTROL STATEMENT .DC V1 12 12 1 * OUTPUT SPECIFICATION STATEMENT .PRINT DC I(RL) V(2) * END STATEMENT .END

Simulate dc_ckt1.cir and view the output file dc_ckt1.out.

Results: V1 I(RL) V(2) 1.200E+01 3.000E+00 6.000E+00

Repeat the problem using. .OP statement instead of .PRINT and compare the outputs.

Results: NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE (1) 12.0000 (2) 6.0000 (3) 8.0000 (4) 4.0000 VOLTAGE SOURCE CURRENTS NAME CURRENT V1 -3.000E+00 TOTAL POWER DISSIPATION 3.60E+01 WATTS

In PSpice, source current is defined as current flowing into the source.

2. Find the galvanometer current in the circuit given in Fig. 4.2. (Given Rg = 100 )

Fig. 4.2



Exercises:AC Circuit Steady State Analysis

3. Find the output voltage Vo of the circuit given in Fig. 4.3

C1

=3

30 u

F

V1=11.3 sin (800t+45)

L1

= 4

mH

R1

= 1

k

Vo

Fig.4.3

I1=

4 s

in (

80

0t

)

Sample Solution (using ac_ckt1.cir)

ac_ckt1.cir * AC circuit analysis using PSPICE V1 1 2 AC 11.3 45 I1 0 1 AC 4 0 R1 2 0 1K C1 2 0 330U L1 1 0 4M .AC LIN 1 400 400 .PRINT AC VM(2) VP(2) .END

4. Find the potential difference between the nodes A and B of the circuit given in

Fig. 4.4 (ω=1rad/sec).

3 3

V1

100

V2

1030

j4 10-j4

A B

+

+

Fig. 4.4



Exercises:Three-phase AC Steady State Analysis

5. Find the neutral shift Von of the circuit given in Fig. 4.5 (ω=1rad/sec).

Fig. 6.5

220240

+

++

2200

220120

R1 = 10

R2 = 5 R3 = 6

XC = -j12 XL = j8

1

0

32

4

5 6

Fig. 4.5

Sample Solution (using ac_3ph1.cir) ac_3ph1.cir * Three phase AC analysis V1 1 0 AC 220 0 V2 2 0 AC 220 240 V3 3 0 AC 220 120 R1 1 4 10 R2 4 5 5 R3 4 6 6 L1 6 2 8 C1 5 3 0.08333 .AC LIN 1 0.159115 0.159115 .PRINT AC VM(4) VP(4) .END

6. A three phase, three wire, star connected 415V symmetrical ABC system supplies a

delta connected load where Zab = 5 , Zbc = 5 + j12 , Zca = 6 - j8 . Find the line

currents.

Exercises: Analysis of Coupled Circuits

7. Find the voltage across the capacitor in the circuit given in Fig. 4.6 (ω=1rad/sec)

Fig. 6.6

1001090

V1V2

++R1

5 5

R2

XC

-j10

XL1 = j5 XL2 = j5

XM = j2

Fig. 4.6



Sample Solution (using cup_ckt1.cir) cup_ckt1.cir * Analysis of coupled circuits using PSPICE V1 1 0 AC 10 0 V2 5 0 AC 10 90 R1 1 2 5 R2 4 5 5 L1 2 3 5 C1 3 0 0.1 L2 4 3 5 K1 L1 L2 0.4 .AC LIN 1 0.159115 0.159115 .PRINT AC VM(3) VP(3) .END

8. Find the current delivered by the source in the circuit given in Fig. 4.7 (ω=1rad/sec)

5

500

+

j8

Fig. 6.7

j3

2 4

M = j4

Fig. 4.7

Exercises:Analysis of Circuits with Controlled Sources

9. Find the output voltage Vo of the circuit given in Fig. 4.8 (ω=1rad/sec)

V1

120

V2

00

2Ix

C1

1F

C2

1F

Ix

R1 = 1 R2 = 1 L1 = 1H

Vo

+

+

R3

1

Fig. 4.8

Sample Solution (using ctl_ckt1.cir) ctl_ckt1.cir * Analysis of controlled circuits using PSPICE V1 1 0 AC 12 0 V2 5 0 AC 0 0 F1 0 3 V2 2 R1 1 2 1



R2 2 3 1 R3 4 0 1 C1 2 0 1 C2 4 5 1 L1 3 4 1 .AC LIN 1 0.159115 0.159115 .PRINT AC VM(4) VP(4) .END

10. Find the current Io of the circuit given in Fig. 4.9 (ω=1rad/sec)

-j1

0

0

1 1

1

j1

Io

Ia

+

+

2Ia

Fig. 4.9



TUTORIAL 5 Familiarization of PSPICE schematic

Invoke Pspice Capture Lite Edition OrCAD PSpice Capture Demo window appears.

A. Drawing the schematic

File New Project; Create a new project: Analog or Mixed A/D; Specify name of

project and location (user directory) – create blank project-Worksheet appears

i. Get all components and place them on the worksheet.

Place Part Part - R (analog.olb library) To change the resistance value double click on the value on the screen, type the value

required in value dialog box.

Get the source from source.olb library. The schematic should have GND as a reference node.

Place Ground (capsym). Change its name to 0

ii. Connect all components as in circuit diagram Place Wire

iii. Save the schematic in your directory.

B. Analysis

i. Pspice Create netlist

ii. PSpice New Simulation Profile-specify the name-click the type of analysis you want to

perform. Ex: click transient analysis for performing transient analysis, click AC sweep analysis for getting frequency response etc.

iii.Pspice Run

C. Results

To see the waveforms, place the voltage and current markers on the circuit. Trace Add Trace

Exercises: DC Transient Analysis

1. Observe the DC transients of the RLC series circuit of Fig. 5.1 for 10s in steps of 0.01

when the switch is closed at t=0 s.

+-10 V

R = 1 L = 1H

C =

10

mF

1

0

2 3

Fig. 7.1

Fig. 5.1

2. Repeat the above problem with resistance equal to (i) 20 and (ii) 100 for 10s in steps of

0.01s. Observe the differences.



3. Observe the various voltage and current transients of the circuit given in Fig. 5.2 for

2s in steps of 1ms. The switch is closed to position 1 at t=0 and it is changed to position

2 at t=0.5s.

10 V

R = 200

C =

1m

F

Fig. 7.2

a

b

1

0

2

10 V

Fig. 5.2

4. Observe the various voltages and current in the circuit given in Fig. 5.3.

R = 200C

= 1

mF

Fig.7.3

1

0

2

v(t)

-10

10per = 100 m

tw = 50 m

tr = tf = 1n

t

v(t)

IC =

10V

+

Fig. 5.3

Exercises: AC Transient Analysis

5. Observe two cycles of the AC transient response of the RL series circuit given in

Fig. 5.4. The switch is closed at = 60o.

+

Fig. 7.4

100

0.55

133

H

10si

n(1

00t

+60

0)

Fig. 5.4

Also observe the active, reactive, and apparent power associated with the above circuit.

Exercises: Frequency Response of Circuits

6. Observe the frequency response of the RLC series circuit of Fig. 5.5 as the frequency

is varied from 100Hz to 100kHz.Plot the frequency response of the current in the loop

and the voltage across the capacitor. From the current plot, find the resonant

frequency (fr), half power frequencies (fL, fU), and bandwidth (BW). Compare with the

theoretical values.



10 V

R = 2 L = 50H

C

10F

Fig. 10.5

f

+

Fig. 5.5

Observations

Trace the following waveform - I vs f ; Vc vs f, XL vs f ; XC vs f ; R vs f ; Z vs f ; impedance

angle vs f.

Mathematical Expressions for theoretical verification

For current response

i. Resonant frequency, fLC

r 1

2

ii. Q factor, QR

L

C

f L

R f RC

r

r

1 2 1

2

iii. Lower half-power frequency, f fQ QL r

1

2

1

21

2

iv. Upper half-power frequency, f fQ QU r

1

2

1

21

2

v. Bandwidth, BW f ff

QU L

r

2

7. Observe the frequency response of the circuit given in Fig.5.6 as the frequency is varied

from 300 to 3kHz. Plot the frequency response of the voltage across the capacitor and

find the resonant frequency, half power frequencies and bandwidth. Repeat for R=1

and R=6 and comment on the responses obtained.

1A

1mH

22F

Fig. 7.6

f

2

Fig 5.6

Observations

Trace the following waveform - V vs f ; BL vs f ; BC vs f ; G vs f ; Y vs f ; impedance angle

vs f.



Module III

Experiment No.1

SUPER POSITION & RECIPROCITY THEOREMS

OBJECTIVE:

To verify (a) super position theorem and (b) reciprocity theorems as applied to

electric circuits.

APPARATUS:

Sl. No. Apparatus Type Range Quantity

1 Standard resistors of different ranges - - 4

2 Milliammeters D.C. Suitable 2

3 Voltmeter D.C. 0 – 10V 1

4 T.P.S. Units D.C - 2

CIRCUIT DIAGRAM:

a) Superposition theorem:

(i) With both energy sources acting:

Fig. (i)

Current Source

0- 10v

R 1 = 100

R 2 = 100 R 3 = 50

R L = 25

TPS - 1

-

+

-

+

-

+

A

V

A

TPS - 2

Voltage Source



ii) With only current source acting:

Fig. i(a)

iii) With only the Voltage source acting:

Fig. i(b)

b) Reciprocity theorem:

Voltage excitation & Current response:

Fig. ii(a)

Voltage

source

Current

Source

RL

R3R2

R1

TPS - 1

-

+

-

+-

+

A

V

A

TPS - 2

R1

Voltage

Source

R3R2

RL=

-

+-

+

A

V

R2= _____ R3= _____

I1

-

+

A

R1= ___

R4= _____

TPS -

+

V



After exchanging the excitation and response

Fig. ii(b)

STATEMENTS:

a) Superposition theorem: It states that the response in any element of a linear, bilateral,

active network containing two or more energy sources is the sum of the responses

obtainable when each source is acting separately and all other energy sources set to

zero with their internal impedances, if any, remaining in the circuit.

b) Reciprocity theorem: It states that in a linear, bilateral, single source network, the

ratio of excitation to the response is constant, even when the positions of excitation

and response are interchanged.

PROCEDURE:


1. Connect the circuit as shown in Fig. (i), after selecting suitable standard resistors

and milliammeter ranges.

2. Switch on both the voltage and current sources (TPS-1&2) and set them to

convenient values.

3. Note down the current through the load resistance RL and voltage across it.

4. Repeat the experiment for different voltage & current settings on the two energy

sources.

5. Rewire the circuit as shown in Fig. i(a).[ Here only the current source is acting in

the circuit and the voltage source is suppressed].

6. Switch on the current source, set it to previous fixed values and note down the

current through & voltage across the load resistance RL) each time.

7. Rewire the circuit as shown in Fig. i(b). [Here only the voltage source is acting in

the circuit and the current source is suppressed].

8. Switch on the voltage source, set it to previous fixed values and each time, note

down the current through and voltage across the load resistance RL.

0-10V

+

V

I2

R1

R2 R3

R4 TPS+

-

A



(b) Reciprocity theorem:

1. Connect the circuit as shown in Fig. ii(a)

2. Switch on the d.c. Supply from TPS and adjust it to some convenient value.

3. Note down the milliammeter reading as ―I1‘.

4. Repeat the expt. for different fixed values of voltage from TPS (excitation voltage).

5. Interchange the positions of excitation and response i.e. the TPS and milliammeter,

as shown in Fig. ii(b).

6. Note down the milliammeter readings (I2) for different fixed values of voltage from

TPS.

TABULAR COLUMN:


Trial

No.

Both sources

acting

Current source

acting

Voltage source

acting

V1+V2

(Volts)

I1+I2

(mA)

V

(Volts)

I

(mA)

V1

(Volts)

I1

(mA)

V2

(Volts)

I2

(mA)

1

2

3

4

Verification: Superposition Theorem is verified when V = V1 + V2

& I = I1 + I2

(b) Reciprocity Theorem:

Trial

No.

Excitation

V 1(Volts)

Response

I1 (mA)

Excitation

‗V2‘ Volts

Response

I2 (mA)

Verification

1

1I

V 2

2I

V

1

2

3

4

Verification: Reciprocity theorem is verified if 2

2

1

1

I

V

I

V



CALCULATIONS:

Verify the practical results obtained, analytically by theoretical calculations.

SUPPLEMENTARY STUDY:

1. Can Superposition theorem be applied to determine the power consumed by an element?

Discuss.

2. Can Superposition and reciprocity theorems be applied to networks with dependent

sources? Discuss.

3. What are the applications of Superposition and Reciprocity theorems?

4. Verify the practical results obtained using PSPICE analysis.

-----------------------------------------------******-----------------------------------------------



R4=50 R3=25

R2=100 R1=50

TPS R5=10

B

A

RL=25

Experiment No.2

THEVENIN’S & NORTON’S THEOREMS

OBJECTIVE:

To verify (a) Thevenin‘s theorem and (b) Norton‘s theorem, as applied to electric

circuits.

APPARATUS:


1 Standard resistors - Suitable

ranges

5

2 Decade resistance box - 1

3 TPS Unit D.C - 1

4 Voltmeter D.C. 0 - 10V 2

5 Milliammeters D.C. Suitable

Ranges 2

CIRCUIT DIAGRAM:

Given Circuit:

Fig. (i)



Verification of Thevenin’s theorem:

i. To find Load current in the Original circuit:

Fig. i(a)

ii. To find Open circuit Voltage ‘VTh:

Fig. i(b)

0-10V

R3 R4

RL

IL

B

TPS R5

+ A

R2 R1

A

+ V

R3 R4

TPS R5

B

R2 R1

VTh

-

+

V

0-10V

+ V

A



iii. To determine Thevenin’s equivalent resistance, ‘RTh’:

Fig. i(c)

iv. To find load current from Thevenin’s equivalent circuit:

Fig. i(d)

v. To find load current from Norton’s equivalent circuit:

RL

Rth

TPS - 1

-

+

-

+-

+

A

V

A

Isc

Fig. i(e)

+

-

V1

I1

R5

R3 R4

R2 R1

TPS

A1

B1

- +

0-10V V

A

RTh

IL1

RL

-

+

A

B

A A1 B

1

VTh

-

+

V

TPS



STATEMENTS:

a) Thevenin’s theorem:

In a linear bilateral active network containing many independent and controlled

sources, the current through the load can be obtained by replacing the entire network

by a single voltage source in series with an impedance, the voltage being equal to the

open circuit voltage across the load terminals, with the load removed and the

impedance is equal to the equivalent impedance of the network with respect to the

load terminals, with the load removed and with all the internal independent sources

set equal to zero.

b) Norton’s Theorem:

In a linear bilateral active network containing many independent and controlled

sources, the current through the load can be obtained by replacing the entire network

by a single current source in parallel with an impedance, the current being equal to

the short circuit current through the load terminals, with the load removed and the

impedance is equal to the equivalent impedance of the network with respect to the

load terminals, with the load removed and with all the internal independent sources

set equal to zero.

PROCEDURE:

a) Thevenin’s theorem:

1. Rig up the circuit as shown in Fig. i(a)., after selecting proper ranges for the meters.

2. Apply a convenient voltage from TPS and note down the load current ‗IL‘.

3. Repeat the above step for different fixed source voltage magnitudes.

4. Rig up the circuit as shown in Fig. i(b) & note down the value of open circuit voltage

VTH, across the terminals A&B for each fixed value of source voltage.

5. Rig up the circuit, as shown in Fig. i(c). Apply a reduced voltage from TPS and note

down the voltmeter and milliammeter readings. Calculate the value of Thevenin‘s

equivalent resistance.

6. Rewire the Thevenin‘s equivalent circuit, as shown in Fig. i(d). Note down the value

of load current IL1 for each value of VTh obtained from circuit in Fig. i(b).

b) Norton’s Theorem:

1. Rig up the circuit as shown in Fig. i(a)., after selecting proper ranges for the meters.

2. Apply a convenient voltage from TPS and note down the load current ‗IL‘.

3. Repeat the above step for different fixed source voltage magnitudes.

4. Rig up the circuit as shown in Fig. i(b) and replace the voltmeter connected across

terminals A & B by an ammeter.



5. Note down the value of short circuit current ISC, through the terminals A&B for each

fixed value of source voltage.

6. Rig up the circuit, as shown in Fig.i(c). Apply a reduced voltage from TPS and note

down the voltmeter and milliammeter readings. Calculate the value of Thevenin‘s

equivalent resistance.

7. Rewire the Norton‘s equivalent circuit, as shown in Fig. i(e). Note down the value of

load current IL1 for each value of ISC obtained from circuit in Fig. i(b).

TABULAR COLUMN:

(a) Thevenin’s Theorem:

Table (i) Table (ii)

Mean RTh =

Table (iii) Table (iv)

Trial

No.

Source voltage

‗V‘ in volts

‗IL‘ in mA

1

2

3

Trial

No.

‗V‘ in

volts

‗I‘ in

mA

RTh =

V/I in

Ohms

1

2

3

Trial

No.

Source voltage

‗V‘ in volts

VTh in volts

1

2

3

Trial

No.

‗VTh‘ in volts ‗IL1‘ in mA

1

2

3



Verification: Thevenin‘s theorem is verified if IL = IL1

(b) Norton’s Theorem:

Table (v) Table (vi)

Verification: Norton‘s theorem is verified if IL = IL1

Trial

No.

Source voltage

‗V‘ in volts

ISC in mA

1

2

3

Trial

No.

‗ISC‘ in volts ‗IL1‘ in mA

1

2

3



R4=50 R3=25

R2=100 R1=50

TPS R5=10

B

A

RL=25

Experiment No.3

MAXIMUM POWER TRANSFER THEOREM

OBJECTIVE:

To verify Maximum power transfer theorem, as applied to electric circuits.

APPARATUS:


1 Standard resistors - Suitable

ranges

5

2 Decade resistance box - 1

3 TPS Unit D.C - 1


5 Milliammeter D.C. Suitable

Ranges 1

CIRCUIT DIAGRAM:

Given Circuit:

Fig. (i)



Maximum power transfer theorem:

Fig. i(a)

STATEMENT:

It states that in any linear, active, bilateral network, maximum power is transferred

from the source to the load, when the load impedance becomes equal to the complex

conjugate of the Thevenin‘s impedance. As applied to D.C. circuits, it can be stated

that max. Power transfer takes place when the load resistance equals the Thevenin‘s

resistance (RTh) of the network.

PROCEDURE:

1. Connect the circuit, as shown in Fig. i(a).

2. Switch on the D.C. supply from TPS and keep it fixed at some convenient value.

3. Vary load resistance RL (decade resistance box) in steps, from minimum value and

note down the corresponding milliammeter and voltmeter readings.

4. Continue the variation of load resistance so that the readings are taken both before and

after the maximum power transfer.

5. Draw the graph of Power transferred verses load resistance.

R3 R4

RL

IL

B

TPS R5

+ A

R2 R1

-

A

+

V V



TABULAR COLUMN:

Trial

No.

V

(Volts)

IL in

(mA) L

LI

VR

()

LL RIP 2( )

(mW)



Verification:

From the graph, find out the value of load resistance RL corresponding to maximum power as

Rcr. At max. power transfer, this particular value of RL must be equal to equivalent internal

resistance of the network as viewed from the output terminals A&B, which is nothing but the

already calculated Thevenin‘s (or Norton‘s) equivalent resistance ‗RTH‘. Verify the same.

Also verify the practically determined max. power value with the theoretical value,

Pmax=Th

Th

R

V

4

2

= ___________ mW.

Load resistance, RL in

Specimen curve for Maximum power transfer theorem


1. Verify the practical results obtained using PSPICE analysis.

2. What are the applications of Maximum Power Transfer theorem?

Rcr = RTh

Pmax

P in mW



Experiment No.4A

MEASUREMENT OF SELF & MUTUAL INDUCTANCE

OBJECTIVE:

To measure the self and mutual inductances of the given two inductive coils

APPARATUS:



2 Ammeter D.C. 0 – 500mA 1

3 Regulated power supply unit D.C - 1

4 Inductive coils - - 2

5 Autotransformer Single phase

- 1

6 Voltmeter A.C. 0 – 100V 1

7 Ammeter A.C. 0 – 1A 1

CIRCUIT DIAGRAM:

(i) To find d.c. resistances of the two coils :

Fig. (i)

+

0-5V

Coil

TPS

+

V

0-500mA

A



(ii) To find self inductances of the two coils :

Fig. (ii)

(iii) To find mutual inductance between the two coils :

a) Series Addition:

Fig. iii(a)

b) Series Opposition:

Fig. iii(b)

E

0 - 1A

N

P

C

B

A

0 - 100V

Coil

1Ph.

50Hz

230V

Supply

A

V

A

B Coil - 2

0–100V

Coil - 1

E

C

N

0-1A

4 3 2 1

1 Ph.

50Hz

230V

Supply

P

V

A

0 – 100V

0 - 1A

4 3 2 1

A

E

C

B

N

P

1 Ph.

50Hz

230V

Supply

V

Coil - 2

Coil - 1

A



PROCEDURE:

1) Connect the circuit as shown in Fig.(i).

2) Apply a low voltage d.c. and note down the d.c. meter readings to determine d.c.

resistance R1 of the first coil.

3) Repeat the steps (1) & (2) with the second coil to determine its d.c. resistance ‗R2‘

4) Rig up the circuit as shown in Fig. (ii) using coil-1.

5) Keeping the autotransformer in zero output position, switch on the a.c. supply.

6) Adjusting the autotransformer, apply a reduced voltage and note down the a.c. meter

readings to calculate the self-inductance L1 of the first coil.

7) Repeat the steps (5) & (6) with coil-2, to determine its self-inductance ‗L2‘.

8) Connect both the coils in series addition, as shown in Fig.iii(a). and repeat the steps (5)

& (6) to determine the reactance ‗m1‘. Care must be taken to see that the two coils are

co-axial by keeping them exactly one above the other.

9) Connect both the coils in series opposition, as shown in Fig. iii(b) and repeat the steps

(5) & (6) to determine the reactance ‗m2‘.

TABULATION & CALCULATIONS:

(i) To find D.C. resistances of the two coils

COIL - 1 COIL - 2

V1 Volts I1 Amp

1

11

I

VR

V2 Volts I2 Amp

2

22

I

VR



(ii) To find Self inductances of the two coils :

COIL – 1 COIL - 2

V1 (

Volt

s)

I 1 (

Am

p)

1

11

I

VZ

()

2

1

2

11 RZX L

()

f

XL L

2

11

(H)

V2 (

Volt

s)

I 2 (

Am

p)

2

22

I

VZ

()

2

2

2

22 RZX L

() f

XL L

2

22

(H)

(iii) To find mutual inductance between the two coils:

Series Addition Series Opposition

'

1V Volts '

1I Amps

'

1

'

1'

1I

VZ

'

2V Volts '

2I Amps

'

2

'

2'

2I

VZ

Here 2

21

2

21

'

1 )2()( MXXRRZ LL

2

21

2'

1 )()(221

RRZMXX LL = m1 = _________ ----------------(1)

Similarly, 22

21

1

2 )2()(21

MXXRRZ LL

2

21

2'

2 )()(221

RRZMXX LL = m2 = _________ ----------------(2)

Equations (1) – (2) gives, 4M = m1 m2

Mutual inductance ‗M‘ = 4

21 mm =

f

mm

8

21 = __________ H.

RESULTS:

1. Self inductance of coil-1 = ___________ H

2. Self inductance of coil-2 = ___________ H

3. Mutual inductance between the two coils = ___________ H



Experiment No.4B

POWER, POWER FACTOR & P.F. IMPROVEMENT

OBJECTIVE:

To measure the power consumed by fluorescent lamp, determine the operating power

factor, striking voltage, holding voltage and study the power factor improvement.

APPARATUS:


1 Fluorescent tube with choke and starter - - 1



4 Wattmeter L.P.F. 1A/300V 1

5 Auto transformer Single phase 5A, 230V 1

6 Capacitor box - - 1

7 Switch S.P.S.T. - 1

8 Connecting wires - - -

CIRCUIT DIAGRAM:

Fig. (i)

Capacitor

C

CH2 CH1

B

C

F3 F4

S2

S1

Glow

type

starter

F2 F1

E

W

1Ph., Auto

Transformer

Fuse

Choke

V

N

P A

1Ph.

230V

Supply

0 – 1A

0 – 300 V

1A / 300V, LPF

S1

M L

V



PROCEDURE:

1) Rig up the circuit as shown in the circuit diagram

2) Keeping the S.P.S.T. switch ‗S1‘ in open position and the autotransformer in zero output

position, switch on the supply.

3) Adjust the Autotransformer till the fluorescent lamp just strikes and note down the

striking voltage.

4) Increase the voltage up to rated value of 230V and note down all the meter readings.

5) Decrease the autotransformer voltage gradually till the fluorescent lamp goes off. Note

down the corresponding holding voltage.

6) Close the S.P.S.T. switch and take down all the meter readings at rated voltage.

7) Repeat the experiment for different values of capacitances.

8) Bring the autotransformer setting to zero output position and switch off the supply.

TABULAR COLUMN:

readingDeflectionScaleFull

FactorPowernCalibratioCCofRangeCoilessureofRange

Wattmeterthe

offactortionMultiplicai

..Pr)(

=

(ii) Striking Voltage = Volts.

(iii) Holding Voltage = Volts

Trail

No.

V

Volts

I

amps

W

Watts

Power factor

cos VI

W

Remarks

With out capacitor

With capacitor:

C1 = F



ANALYSIS OF POWER FACTOR IMPROVEMENT:

Referring to the power triangle shown below, let ‗P‘ be the active power consumed by the

lamp circuit at rated voltage V. Let cos1 be the original power factor and cos2 be the

improved p.f. Without the external capacitor, the lagging reactive power drawn by the circuit,

Q1= P tan1.

With the capacitor, the reactive power is reduced to

Q2= P tan 2

The leading reactive power of the capacitor is

given by

Q1 Q2= cX

V 2

where ‗Xc‘ is the capacitive reactance

Xc = 21

2

QQ

V

or

Capacitance ‗C‘ = cfX2

1Farads.

Thus, given the value to which the power factor is to be improved, the required capacitance

can be determined and compared with the practical values.

RESULTS:

(i) The power consumed by the fluorescent lamp at rated voltage = __________Watts

(ii) Corresponding power factor, ‗cos1‘ = _________

(iii) Improved power factor ―cos2‖ = _______ with C = _________ F.

(iv) Corresponding theoretical value of ‗C‘ = _________ F


1. Discuss the other methods used for determination of self-inductance of a coil?

2. Discuss the causes of low power factor, disadvantages of low power factor and

methods of improving power factor.

-----------------------------------------------******-----------------------------------------------

2

S2

1 Q2

S1

P

(Q1 – Q2 )

Q1



Experiment No.5

THREE PHASE POWER MEASUREMENT

OBJECTVE:

To measure the active power in three phase star and delta connected circuits by Two

Wattmeter method.

APPARATUS:




3 Voltmeter A.C 0 - 100V 1


5 Wattmeter U.P.F. 10A/600V 2

6 Resistive load 3 phase - -

CIRCUIT DIAGRAM:

(i) Star Connection :

Fig. (i)

O

0–600V

W1

0–10A

IB

IA

0–300V

VA V

B1

0–10A

10A/600V, UPF

IC

V C

L M

W2

V C

L M

10A/600V, UPF

3Ph., Star Connected

Load

C1

B2 C2

A2

A1 0–10A

VL

F

F

F

C

A

B

3Ph.

400V

50Hz

Supply

TPST

Mains

A

A

A

V

N



(ii) Delta Connection:

Fig. (ii)

PROCEDURE:

(i) Star connection:

1) Rig up the circuit as shown in Fig (i).

2) Keep the load switches off and switch on the three-phase supply.

3) Switch on the load in steps such that same current flows through all the ammeters

(balanced load) included in the three phases. Note down all the meter readings.

4) Now, switch on the load in such a way that unequal currents flow through the

ammeters connected in the three phases (unbalanced load). Note down all the meter

readings. Also note down the neutral shift voltage between the load star point ‗O‘

and supply neutral ‗N‘ (in case of 3 Ph., 3 wire, star connection only).

5) Switch off the load and then switch off the three phase mains.

(ii) Delta connection:

6) Rig up the Circuit as shown in Fig (ii).

7) Repeat the steps marked (2) to (5).

3Ph., Delta

Connected Load

0–10A

IC A

B2 B1 0–10A

IB

A

C2

C1

A2

A1

IA

0–10A

A

IL

0–10A

A

0–600V

V

F

F

F

C

A

B

3Ph.

400V

50Hz

Supply

TPST

Mains

V C

W1

L M

10A/600V, UPF

V C

L M

W2

10A/600V, UPF



TABULAR COLUMN:

Multiplication factor of wattmeter, wattmeterofreadingDSF

ncalibratiooffpCCofrangeCurrentCPofrangeVoltageW

...

......1

= _____________

Multiplication factor of wattmeter, W2 = ___________

Trial

No.

Star Connected Load (Assuming power factor of resistive load to be unity)

Line

Voltage

VL Volts

Phase voltages PhV.3

volts

Phase or line

currents

W1

watts

W2

watts

Total Power

W=(W1+W2)

watts

Total

power

consumed

W=(VAIA+

VBIB+

VCIC)

watts

Power factor, cos =

21

211 3tancos

WW

WW

VA

Volts

VB

Volts

VC

Volts

IA

Amp

IB

Amp

IC

Amp

Balanced Load

1

2

Unbalanced Load

1

2

Neutral

shift

voltage

(volts)



Trial

No.

Delta Connected Load (Assuming power factor of resistive load to be unity)

Line

Voltage

VL = phase

voltage, VPh =

VA=VB=VC

Volts

Line

Current

IL amp

Phase currents

PhI.3

amps

W1

watts

W2

watts

Total Power

W=(W1+W2)

watts

Total power

consumed

W=(VAIA+

VBIB+ VCIC)

watts

Power factor, cos

21

211 3tancos

WW

WW

IA

amps

IB

amps

IC

amps

Balanced load:

1

2

Unbalanced load:

1

2

Circuits Lab


RESULTS:

To be verified:

(i) For Star Connected balanced load, VL = PhV3

(ii) For Delta Connected balanced load, IL = PhI3

(iii) Star & Delta Connections – balanced & unbalanced loads,

(W1+W2) = (VAIA+VBIB+VCIC)


1. Draw and explain the phasor diagrams for

a) Star connected, balanced, lagging p.f. load being supplied with

3 Ph., balanced, ABC sequence of supply.

b) Delta connected, balanced, U.P.F. load being supplied with 3 Ph.,

balanced, ABC sequence of supply.

2. While measuring three phase power by two wattmeter method, following

observations were made. Indicate the type of load and load power factor in

each case. Justify your answer.

i) One wattmeter reads negative and the other positive.

ii) Both meters read positive.

iii) Both the meters show same reading.

iv) One of the wattmeter shows zero reading.

v) One meter shows double the other.

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manipal institute of technology manipal 576...

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