ELECTRICAL PRINCIPLE

CE00436-1Electrical Principles Individual Assignment Page1 of 46

Level 1 Asia Pacific Institute of Information Technology 2014

TABLE OF CONTENTS

S.No Topic Page No

1. EXPRIMENT NO-1

2-8

2. EXPRIMENT NO-2 8-15




b. REFERENCE 45-447

s



Experiment no 1.

Aim:-

To Find V0 in the circuit of Fig.1 using Thevenin’s theorem and simulate it using Multisim

Simulator. Compute the error in two results.

Theory: -

Thevenin’s Theorem states that it is possible to simplify any linear circuit, no matter how

complex, to an equivalent circuit with just a single voltage source and series resistance

connected to a load. The qualification of “linear” is identical to that found in the

Superposition Theorem, where all the underlying equations must be linear. If we're dealing

with passive components, this is true. However, there are some components which are

nonlinear: that is, their opposition to current changes with voltage and/or current.

In electrical circuit theory, Thevenin’s theorem for linear electrical networks states that any

combination of voltage sources, current sources and resistors with two terminals is

electrically equivalent to a single voltage source V and a single series resistor R. For single

frequency AC systems, the theorem can also be applied to general impedances, not just

resistors. Any complex network can be reduced to a Thevenin's equivalent circuit consist of a

single voltage source and series resistance connected to a load.

Circuit:-

Figure (1)



Design of simulation Circuit using Multisim:-

Since, we have to find voltage across k2 resistor according to thevenin theorem. So, we

remove the load resistance k2 .

Figure (2)

Now find the Vo.c across AB terminal:

Figure (3)



Now short voltage source and open the current source and find Rth.

Figure (4)

Now find Vo.c across ab terminal.

Figure (5)



Now connect thevenin resistance with Vo.c:

Figure (6)

Again load resistance also connect with Vo.c:

Figure (7)

Now find the Vo

Figure (8)



Calculation:-

Figure (9)

321 || RRRRth

kkk 4)3||6(

k6

Figure (10)



Applying KVL to mesh 2:

0103)(10612 2

3

12

3 III ------------------------------------------------ (1)

In mesh 1:

mAI 41

Now put the value of 1I in equation (1);

0103)104(10612 2

33

2

3 II

After solving we get,

mAI 42

Now applying KVL to the path a-b or open path:

0103104 2

3

.1

3 IVI co

After solving we get,

VV co 28.

Using voltage divider rule,

Lth

Lcoo

RR

RVV

.

vkk

k7

26

228

Result

After simulating this circuit in multisim, we get

Vo.c=28V

Rth=6Kohm

Vo = 7V

After solving this circuit we get



Vo.c=28V

Rth=6Kohm

Vo = 7V

And error =0%

Conclusion:-

Thevenin theorem is the method to solve the electric circuit and calculate voltage or current

across any any component very easily. In this circuit, valve of all the Vo by calculation and

by simulation is same so, error is zero.

Experiment no 2

Aim: - Find V0 in the circuit using Nortan’s Theorem and also simulate it using Multisim

Simulator. Compute the error in two results.

Theory:

Norton's Theorem states that it is possible to simplify any linear circuit, no matter how

complex, to an equivalent circuit with just a single current source and parallel resistance

connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical

to that found in the Superposition Theorem: all underlying equations must be linear. In some

ways Norton’s Theorem can be thought of as the opposite to “Thevenins Theorem”, in that

Thevenin reduces his circuit down to a single resistance in series with a single voltage.

Norton on the other hand reduces his circuit down to a single resistance in parallel with a

constant current source.

Nortons Theorem states that “Any linear circuit containing several energy sources and

resistances can be replaced by a single Constant Current generator in parallel with a Single

Resistor“.

As far as the load resistance, RL is concerned this single resistance, RS is the value of the

resistance looking back into the network with all the current sources open circuited and IS is

the short circuit current at the output terminals.



Circuit:

Figure (11)


For finding the Vo first remove load resistor.

Figure (12)

To find Isc short the terminal ab:



Figure (13)

According to source transformation, we change 6V into 2mA.

Figure (14)

Now change k6 or k3 into k2 .

Figure (15)



According to source transformation convert 2mV into 4V.

Figure (16)

Now find the Is.c.

Figure (17)

Now find RN



Figure (18)

Connect RN parallel with Is.c

Figure (19)

Now connect load resistor all in parallel an find finally V0



Figure (20)

Calculation:-

Figure (21)

21 || RRR

kkk 26||3



In mess 1

mAII 421 --------------------------------------------------------- (1)

KVL across super mess

01041024 2

3

1

3 II

KVL around mess 3,

0)(102)(108 13

3

23

3 IIII

Since csII .3 the above equation becomes:

0)(102)(108 1.

3

2.

3 IIII cscs

After solving the equation we get,

AmAI cs 333.133133.0.

)28(||)46||3( kkkKkRin

k75.3

)||(.0 LNcs RRIV

mv

kk

258

4||75.31333.0

Result:-

After simulation

V0 = 258.058 mV

Is.c =0.1333mA

After calculation

V0 = 258.058 mV

Is.c =0.1333mA



Error =0.038%

Conclusion:- Nortons Theorem states that “Any linear circuit containing several energy

sources and resistances can be replaced by a single Constant Current generator in parallel

with a Single Resistor“. In this circuit value of all the component becomes approximate

equal.

Experiment no 3

Aim:-

Analyze the natural and step response of series RL and RC circuit mathematically as well as

with Multisim Simulator and compare the two results.

Theory:-

Transient analysis is the most general method for computing dynamic response. The purpose

of a transient analysis is to compute the behaviour of a structure subjected to time- varying

excitation is explicitly defined in the time domain. Transient analysis of behaviours of

electric circuit reveals that as soon as a circuit is switched from one condition to another

either by change of source or by alteration of circuit element branch currents and voltage

drops charges from their initial values to new values. These charges take a short spell of time

settle to steady state values till further switching or circuit alteration is attempted. This brief

spell of timse is called transient time and the value of the current and voltage drop during this

period is called transient value

The transient analysis of the circuit responds to energies stored in storage elements, such as

capacitors. If a capacitor has energy stored within it, then that energy can be dissipated by a

resistor. And this circuit resistor and capacitor are connected in series and series combination

is connected across a voltage source. When t=0 time capacitor behaves as a short circuit. The

initial current in the circuit is I0 =R

V form Ohm’s law. After some time the capacitor starts

charging as

Natural response:-

It is defined as the response which occurs solely from initial conditions with no other inputs.

The natural response is also known as the unforced response or characteristic response. The

model differential equation for such a system is homogeneous, in that there is no forcing



term. It is composed of weighted sums of functions est

,where s is possibly complex (or most

generally such functions multiplied by polynomials in the time variable t). This is a statement

about the solution of differential equations. However, it is a remarkable empirical result that

such differential equations well-describe many physical systems. Said another way, the types

of natural responses discussed below can be easily observed in an experimental context, and

in observations of many physical phenomena. The natural response ties things together.


Figure (22)

Formula used to find current inside the circuit.

i (t) =t

L

R

eR

V

R

V

i (t) =

t

L

R

eR

V1

Given the fig 24,

R= 10Ω, L= 20mH, V= 12V



i (t) =

t

e31020

10

110

12

i (t) = tke 5.012.1

Verification,

At, t = 0

i (t) = tke 5.012.1

i (0) = 05.112.1 ke

= 0

i (0) = 0 (show in figure 26)

At, t = 1mS

i (t) = tke 5.012.1

i (1m) = mke 15.012.1 = 15.012.1 e

=0.47216

i (1m) = 472.16mA (show in figure 26)



Now, Voltage across Inductor,

VL =Ldt

di

Put the value i=

t

L

R

eR

V1

VL = L

L

Re

R

V tL

R

0 = L

tL

R

eL

V

VL = Vt

L

R

e

= 12 tke 5.0

Verification,

At, t = 0

VL = tke 5.012

= 05.012 ke = 012 e = 112 = 12V

VL = 12V

At, t = 1.mS

VL = tke 5.012



= mke 15.012 = 5.012 e =7.3V

VL = 7.3V (show in figure 28)

Now, Voltage across Resistor,

VR =iR

VR =

t

L

R

eR

V1 R =

t

L

R

eV 1

=

t

e31020

10

112 = tke 5.0112

Verification,

At, t = 0

VR = tke 5.0112

= 05.0112 ke = 0112 e

= 0

VR = 0

At, t = 1mS



VR = tke 5.0112

= mke 1.15.0112 = 5.0112 e

=7.72V

VR = 4.72V

Now,Transient Analysis of RC series Circuits for step input signal using

Multisim.


Figure (23)

Formula used to find current inside the circuit.

i (t) = K RC

t

e

i (t) = RC

t

eR

V

(Final equation of current)

But give this circuit is,



R=15Ω, C=100µF, V=12V

i (t) =61010015

15

12

t

e = 66.6668.0 te

i (t) = 66.6668.0 te

Verification,

At, t = 0

i (t) = 66.6668.0 te

i (0) = 66.66608.0 e

= 0.8

i (0) = 800mA

Now, Voltage across Resistor,

VR = iR

Put the, i= RC

t

eR

V



VR = ReR

VRC

t

= RC

t

Ve

VR = 6101001512

t

e

VR = 66.66612 te V

Verification,

At, t = 0

VR =66.66612 te

= 66.666012 e = 012 e

= 12 1 = 12

VR = 12V

At, t = 1mS

VR = 66.666101 3

12

e



= 6.2V

VR = 6.2V

Now, Voltage across capacitor,

VC = idtC

1

Put the, i= RC

t

eR

V

VC =

0eeV RC

t

=

1RC

t

eV =

RC

t

eV 1

VC =

61010015112

t

e = 66.666112 te

Verification,

At, t = 0

VC = 66.6660112 e

= 0112 e = 1112 = 012 = 0V



VC = 0V

At, t = 1.1mS

VC = 66.666101 3

112

e

= 5.9 V

VC = 5.9V

Result:-

In RL circuit

At t =0, I =0

At t=1, I= 472 mA

% error =0.639%

In RC circuit

At t=0, I =800 mA

% error =1.23%

Conclusion:- Transient analysis of behaviours of electric circuit reveals that as soon as a

circuit is switched from one condition to another either by change of source or by alteration

of circuit element branch currents and voltage drops charges from their initial values to new

values.

Experiment no 4

Aim: -

Analyze the under-damped, over-damped and critically damped response of a series RLC

circuit mathematically as well as with Multisim Simulator and compare the two results.

Theory:-

In Transient Analysis also called time-domain transient analysis. And Multisim computes the

circuit’s response as a function of time. This analysis divides the time into segments and



calculates the voltage and current in each given element. Then finally the results, current

versus time and voltage versus time, are presented in Graphic View. And the Capacitors and

inductors are represented by energy storage elements. Then numerical integration is used to

calculate the current and voltage in the circuit. There are three possible cases first are under

damped case, second is over damped case and third is critically damped case. And in this

circuit I am using resistor, inductor, and capacitor are connected with 12V DC Source in

series. And also calculate the voltage and current in each element in help of Kirchhoff’s law.

Under damping

In this experiment I am explain that under damped case that means

LCL

R 1

2

2

. Where,

L

R

2 the damping factor and

LC

1 the resonant frequency

Let dt

d= m

Put the value m in equation (1),

= iLC

imL

Rim

12

0 =

LCm

L

Rmi

12

LCm

L

Rm

12 = 0

Now apply the formula a

acbbx

2

42

Give, a = 1, b =L

R, c=

LC

1,

12

114

2

LCL

R

L

R

m



2

42

LCL

R

L

R

m

LCL

R

L

R 1

22

2

(Dived 2 in both upper and lower)

Spouse, LCL

R

L

Rm

1

22

2

1

and

LCL

R

L

Rm

1

22

2

2

Let L

R

2 and

LCL

R 1

2

2

So, m1 = α+ jβ and m2 = α- jβ

(a. )Design of simulation Circuit using Multisim:-

Figure (24)

In this circuit is given, VVFCHLR 12,1,1,100

i (t) =

t

LCL

R

L

Rt

LCL

R

L

R

ee

LCL

RL

V1

22

1

22

2

22

1

22



i (t) =

tt

ee11

10

12

100

12

100

11

10

12

100

12

100

6

2

6262

1011

1

12

10021

12

i (t) =

tt

ee6262

105050105050

6210502

12

Verification,

At, t = 0

i (0) =

01050500105050

62

6262

10502

12ee

i (0) =

00

6210502

12ee

=

11

10502

12

62

=

0

10502

12

62

= 0

i (0) = 0A

Calculate the Voltage across Resistor,



VR = iR

=

ReemmL

V tmtm

12

12

1m 610250050

2m 610250050

VR =

ReemmL

V tmtm

12

12

VR =

1001025002

12 66 1025005010250050

6

tt ee

Verification:-

At, t=0S

Put the value of t in equation VR,

VR =

1001025002

12 66 1025005010250050

6

tt ee

VR =

1001025002

12 010250050010250050

6

66

ee



= 0V

VR = 0V (show in fig 33)

Calculate the Voltage across capacitor,

VC = idtC

1

Put the value of i (t) =

tmtmee

mmL

V12

12

VC =

1212

121

m

e

m

e

mmL

V

C

tmtm

=

1212

11112

me

me

mmL

V

C

tmtm=

1212

11112

me

me

mmL

V

C

tmtm

VC =

1212

121

m

e

m

e

mmL

V

C

tmtm

Calculate the Voltage across inductor,

VL= dt

diL



Put the value of i (t) =

tmtmee

mmL

V12

12

,

VL =

tmtmemem

mm

V12

12

12

Now, put the value of

6

1 10250050 m , 6

2 10250050 m and V= 12V

VL =

tmtmemem

mm

V12

12

12

=

tt emem

66 10250050

1

10250050

266 1025005010250050

12

Verification:-

At, t=0S

Put the value of t in equation VL,

VL =

01025005060102500506

6

66

10250050102500501025002

12

ee



=

66

61025005010250050

1025002

12

=

66

61025005010250050

1025002

12

= 12V

VL = 12V

Over damping

In this circuit I am using resistor, inductor, and capacitor are connected with 12V DC Source

in series. In this experiment I am enplane that over damped case that means

LCL

R 1

2

2

.Where,L

R


LC

1 the resonant frequency.

Again

Spouse, LCL

R

L

Rm

1

22

2

1

and

LCL

R

L

Rm

1

22

2

2

Let L

R

2 and

LCL

R 1

2

2

(a. )Design of simulation Circuit using Multisim:-



Figure (25)

i (t) =

tmtmee

mmL

V12

12

Put the value of LCL

R

L

Rm

1

22

2

1

and

LCL

R

L

Rm

1

22

2

2

VVFCmHLR 12,100,1,10

63

2

33110100101

1

1012

12

1012

12

m

7233

1 10106106 m 02.5099106 3

98.9001 m

63

2

33210100101

1

1012

12

1012

12

m

02.110992 m



Verification:-

At, t=0S

i (t) = tt ee 98.90002.1109918.1

i (t) = 098.900002.1109918.1 ee

= 1118.1 = 018.1 = 0A

Now, Voltage across resistor,

VR = Ri

= Ree tt 98.90002.1109918.1

= 1018.1 98.90002.11099 tt ee

VR = tt ee 98.90002.110998.11



Verification:-

At, t=0S

VR= tt ee 98.90002.110998.11

= 098.900002.110998.11 ee

= 118.11

= 08.11 = 0V

Now, Voltage across inductor

VL= dt

diL

Put the value of i (t) = tt ee 98.90002.110998.11 ,

VL = tt ee 98.90002.110993 98.90002.110998.11101

VL = tt ee 98.90002.11099 45.11997.130



Verification:-

At, t=0S

VL = tt ee 98.90002.11099 45.11997.130

VL = 098.900002.11099 45.11997.130 ee

= 145.119197.130 = 45.11997.130

VL = 45.11997.130 =11.52V


VC = idtC

1

Put the value of i (t) = tt ee 98.90002.110998.11

VC = dtee

C

tt 98.90002.110998.111

=

dtee tt 98.90002.11099

68.11

10100

1

= dtedte tt 98.90002.110994108.11 =

98.90002.11099108.11

98.90002.110994

tt ee



VC = dtee

C

tt 98.90002.110998.111

=

dtee tt 98.90002.11099

68.11

10100

1

At t = 0,

= dtedte tt 98.90002.110994108.11 =

98.90002.11099108.11

98.90002.110994

tt ee

= 0

Critical damping

In this circuit I am using resistor, inductor, and capacitor are connected with 12V DC Source

in series. In this experiment I am enplane that critical damped case that means

LCL

R 1

2

2

.Where,L

R


LC

1 the resonant frequency.

Now, critical damped case,

LCL

R 1

2

2

Now apply the formula a

acbbx

2

42

Give, a = 1, b =L

R, c=

LC

1,



12

114

2

LCL

R

L

R

m

2

42

LCL

R

L

R

m

LCL

R

L

R 1

22

2

(Dived 2 in both upper and lower)

Spouse, LCL

R

L

Rm

1

22

2

1

and

LCL

R

L

Rm

1

22

2

2

Let L

R

2 and

LCL

R 1

2

2

So, m1 = α+ jβ and m2 = α- jβ

In this case β is zero. Hence roots m1 and m2 are real but equal. System will become critical

damped.

m1= m2= α


Figure (26)



R= 20Ω, L= 1mH, C= 10µF and L

R

2

31012

20

= 31010 = 410

Now, put the value of in the equation i (t) = tte

L

V

i (t) = tte

L

V 410

Verification:-

At t =0S

i (t) = tte

L

V 410

= 0104

0 eL

V

i (t) = 0A

Calculate the voltage across resistor:-

VR = Ri

Put the value, i (t) = tte

L

V 410



VR = RteL

V t 410

Now put the value of R=20Ω, L=1mH, and V =12V

VR = 20101

12 410

3

tte = tte 410310240

Now put the value of R=20Ω, L=1mH, and V =12V

VR = 20101

12 410

3

tte = tte 410310240

Verification:-

At t =0S

VR = 0103 4

010240 e

VR = 0V

Now, Voltage across inductor

VL= dt

diL

Put the value of i (t) = tte

L

V 410



VL= tt eteV 44 1010410

Now put the value of V = 12V

= tt ete 44 101041012

VL = tt ete 44 10104 121012

Now put the value of V = 12V

= tt ete 44 101041012

VL = tt ete 44 10104 121012

Verification:-

At t =0S

VL = tt ete 44 10104 121012

= 010104 44

1201012 ee t

= 112

VR = 12V




VC = idtC

1

Put the value of i (t) = tte

L

V 410

VC =

dtedt

dtet

L

V

C

tt

4

4

10

4

10

10

1 =

4

10

4

10

1010

44 tt eet

LC

V

dtedt

dtet

L

V

C

tt

4

4

10

4

10

C10

1V =

4

10

4

10

1010

44 tt eet

LC

V

Result:-

Under damping

At t=0, VL=12V, I= 0A

Error =0%

Over damping

At t = 0, I=0, VL=11.5

Error =4.166%

Critical damping



At t=0, I=0, VL=12V

Error =0%

Conclusion:- This analysis divides the time into segments and calculates the voltage and

current in each given element. Then finally the results, current versus time and voltage versus

time, are presented in Graphic View. And the Capacitors and inductors are represented by

energy storage elements. Then numerical integration is used to calculate the current and

voltage in the circuit.

Experiment no 5

Aim:-

Analyze the frequency response of parallel RLC circuit to obtain its resonant frequency

mathematically as well as with Multisim Simulator and compare the two results.

Theory:-

RLC circuits are classical example of second order systems. Together with their mass-spring

dashpot mechanical analog, they are used use to illustrate fundamental system theory

concepts and techniques, such as Laplace transformation techniques and resonance.

Resonance in electrical consisting of passive and active elements represents a particular state

of the circuit when the current or voltage in the circuit is maximum and minimum with

respect to the magnitude of excitation at a particular frequency, the circuit impedance being

either minimum or maximum at the power factor unity. The phenomenon of resonance is

observed in both series or parallel a.c. circuits comprising of R,L and C and excited by an a.c.

source.



Figure (27)

Properties of resonance of parallel LRC circuit

Power factor is unit.

Current at resonance is (V/L/CR) and is in phase with applied voltage. The value of

current at resonance is mininum.

Net impedance at resonance of the parallel circuit is minimum and equal to (L/CR).

The resonance frequency of the this circuit is given by

2

2

0

1

2

1

L

R

LCf

Resonance frequency by graph = 90HZ

Calculation:

Resonance frequency 2

2

0

1

2

1

L

R

LCf



2

2

5

2

4005

1

14.32

1

m

k

m

= 92.80 HZ

Result:-

Resonance frequency = 92.80 HZ

% error = 3.0172%

Conclusion:- in this simulation the variation of current with frequency for RLC circuit

having the same L and C but different value of R. These curves are referred to as a frequency

response curve.



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ELECTRICAL PRINCIPLE