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2-1ECE ET LABMANUAL

Dept of E.E.E Page 1

SERIES AND PAREALLEL RESONANCE

AIM: To determine the performance of the series and parallel circuit at resonance.

SERIES RESONANCE

GIVEN CIRCUIT: MODEL GRAPH:

PARALLEL RESONANCE

Exp - 1

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2-1ECE ET LABMANUAL


CIRCUIT DIAGRAMS:

FOR SERIES RESONANCE :

FOR PARALLEL RESONANCE:



2-1ECE ET LABMANUAL


APPARATUS:-

S. No Name of the apparatus Range Type Quantity

1 Signal generator (0 – 1M)Hz,

(0-20) VPP digital 01No

2 Decade inductance Box (0-40mH) - 01No

3 Voltmeters (0-30V) MI 03No

4 Capacitor 0.1uf - 01No

5 Resistors 30 Ω - 01No

6 Ammeter (0-200m) A MI 01No

7 Experimental board - - 1No

8 C.R.O 30 MHz - 1No

9 Connecting wires - - Required

Number

THEORY:



2-1ECE ET LABMANUAL


Resonance is a particular type of phenomenon inherently found normally in every

kind of system, electrical, mechanical, optical, Acoustical and even atomic. There are several

definitions of resonance. But, the most frequently used definition of resonance in electrical

system is studied state operation of a circuit or system at that frequency for which the resultant

response is in time phase with the forcing function.

Series resonance:

A circuit is said to be under resonance, when the applied voltage ‘V’ and current are in

phase. Thus a series RLC circuit, under resonance behaves like a pure resistance network and the

reactance of the circuit should be zero. Since V & I are in phase, the power factor is unity at

resonance.

The frequency at which the resonance will occur is known as resonant frequency.

Resonant frequency, fr =

√

Thus at resonance the impedance Z is minimum. Since I = V/Z. The current is maximum . So

that current amplification takes place.

Parallel Resonance:

The parallel circuit consisting branches with single pure elements R,L & C is

an ideal circuit. How ever the performance of such a circuit is of interest in the general subject

of resonance. This ideal parallel circuit is of interest in the general subject of resonance.

Lower cut-off frequency is above the resonant frequency at which the current is reduced

to

√ times of it’s minimum value. Upper cut-off frequency is above.

Quality factor is the ratio of reactance power inductor (or) capacitor to its resistance. Selectivity

is the reciprocal of the quality factors.

PROCEDURE:

1. Connections are made as per the circuit diagram.



2-1ECE ET LABMANUAL


2. By varying the frequency note down the corresponding values of current in both cases and

note down f, VC, VL and VR.

3. At a particular value of frequency the current reaches its Maximum /minimum in

Series/Parallel resonance. That instant of frequency VC = VL and VR = VS in series Resonance

circuit.

THEORITICAL CALCULATIONS:

For Series Resonance circuit:

1. Resonant frequency fr =

√

=2516 Hz

2. Lower cut-off frequency fl = +

= 2457.85 Hz

3. Upper cut-off frequency f2 =

+

= 2576.14 Hz

4. Band width = f2-f1 = 119.66 Hz

5. Quality factor Q =

= .

= 21.07

6. Current at Resonance Io = VRo/R

TABULAR FORMS:

FOR SERIES RESONANCE :

S.NO FREQUNCY V(L) V(C) V(R) I(mA)

1

2

...



2-1ECE ET LABMANUAL


TABULAR FORMS:

FOR PARALLEL RESONANCE:

S.NO FREQUNCY V(L) V(C) V(R) I(mA)

1

2



2-1ECE ET LABMANUAL


For parallel Resonance circuit:

1. Resonant frequency fr =

= 2514 Hz

2. Lower cut-off frequency fl =

= 2456. 77 Hz



2-1ECE ET LABMANUAL


3. Upper cut-off frequency f2 =

=2577.6 Hz

4. Band width =

= 120.07 Hz 5. Quality factor Q =

= 0.0474 Hz

6. Current at resonance Io = VRo/R

PRECAUTIONS:

• Meter reading should be taken with out parallax error.

• Connection should be made tight.

RESULT:-

Review Questions:- 1. Definition of resonance?

2. Define the series resonance?

3. Define the parallel resonance?

4. Applications of resonance?

5. What is the condition of voltage &current at the resonance condition?

CONCLUSIONS: 1. Since the current at resonance is maximum, the series resonant circuit is

called as acceptor circuit.

2. As the resistance of the circuit decreases, the Q-factor increases and

selectivity of the circuit will be better.

3. Since the current at resonance is minimum, the parallel resonant circuit is

called as rejector circuit.

4. The variation of the resistance does not affect the resonant frequency.

TWO PORT NETWORK PARAMETERS

AIM:

S.No Parameter

Series Resonant circuit Parallel Resonant circuit

Theoretical

Values

Practical

Values

Theoretical

Values

Practical

Values

1 Resonant

Frequency, fr 2516 Hz 2514 Hz

2 Band width 119.66 Hz 120.07 Hz

3 Quality factor 21.07 0.0474 Hz

Exp - 2



2-1ECE ET LABMANUAL


To determine open circuit impedance parameters (Z) and short circuit admittance

parameters (Y) of the given two port network .

.

GIVEN CIRCUIT:

APPARATUS:


1 Regulated power supply (0 – 30) V/2A digital 01

2 Voltmeters (0-30) V MC 01

3 Ammeters (0-200m)A MC 01

4 Resistors

330 Ω

470Ω

630Ω

-

01

01

01

5 Experimental board - - 01

6 Connecting wires - - Required

Number

CIRCUIT DIAGRAMS:

BASIC CIRCUIT



2-1ECE ET LABMANUAL


WHEN V1=0

WHEN I1=0

WHEN V2=0

WHEN I2=0



2-1ECE ET LABMANUAL


THEORY:

A port is normally referred to a pair of terminals of a network though which we

can have access to network of calculating current in any part of network. Frequently the problem

is move restried in nature and may be that of calculating the response at a terminal pair

designated an input when excitation is applied at another terminal pair designated as input

terminals. It is a problem of terminal through which it is accessible, is called “Two Port

Network.“

If we relate the voltage of one port to the current of the same port, we get driving

point immitance. On the other hand, if we relate the voltage of one port to the current at another

port, we get transfer immittance. Immitance is a general term used to represent either the

impedance or the admittance of a network.

We will consider a general two-port network composed of linear, bilateral

elements and no independent sources. Dependent sources are permitted. It is represented as a

black box with two accessible terminals pairs as shown in. The voltage and current at port -1 are

V1 and I1 and at port =II are V2 and I2. The position of V1 and V2 and the directions of I1 and I2

are customarily selected. out of four variables, I1I1V2 and I2 only two are independent. The other

two are expressed in terms of the independent variable of network parameters.

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. Open the port – I i.e, I1=0 find the values of I1,I2, V1.

3. Short circuits the port V2 =0 find the values of V2,I1, I2.

4. Repeat steps 2,3 for port – II and find the values of V1,I1,I2 and V2,I1,I2 respectively.

5. Find all the parameters of two port networks I,e, Z,Y, ABCD, AI B

I C

I D

I, h, g

parameters from the above data.

PRECAUTIONS:

1. Initially keep the RPS output voltage knob in zero volt position.

2. Set the ammeter pointer to zero position.

3. Take the readings without parallax error.

4. Avoid loose connections.

5. Do not short-circuit the RPS output terminals.



2-1ECE ET LABMANUAL


TABULAR FORMS:

Theoritical Values

V1

(volts)

I1

(mA)

V2

(mA)

I2

(mA)

V1=0

0 61 20 92

I1=0

11.42 0 20 57

V2=0

20 107 0 61

I2=0

20 66 13.2 0

Practical Values

V1

(volts)

I1

(mA)

V2

(mA)

I2

(mA)

V1=0

I1=0

V2=0

I2=0



2-1ECE ET LABMANUAL


Theoretical calculations:

1. When I1 = 0 (i.e.,) When port is open circuited:

RL = 216.66 ohms

V2 = 20 V

I2= = 57 mA

V1 = I2.R = 11.42 V

2. When port -2 is open circuited (I2=0):

V1 = 20V

Rt = 185.74 ohms

I1 = = 66 mA

V2 = 13.2V

3. When port -1 is short circuited ( V1=0):

V2 = 20 V

Rt = 216.66 ohms

I2 = = 92 mA

I1 = I2 ! "# - 61 mA

4. When port –II is short – circuited (V2 = 0 ) :

V1 = 20 V

Rt = 185.74 ohms

I1 = V1/Rt = 107 mA

I2 = 61 mA

calculations for parameters:

Z-parameters:

Z11 =$ / I2=0 =

Z12 =$ / I1=0 =

Z21 =$ / I2=0 =

Z22 =$ / I1=0 =



2-1ECE ET LABMANUAL


Y – Parameters

Y11 = $ / V2=0 =

Y12 = $ / V1=0 =

Y21 = $ / V2= 0 =

Y22 = $ / V1= 0 =

ABCD parameters:

A = / I2= 0 =

B =

$ / V2= 0 =

C = $ / I2= 0 =

D = $$ / V1=0 =

H – Parameters:

h11 = $ / V2 =0 =

h12 = / I1=0 =

h21 = $$ / V2 =0 =

h22 = $ / I1 =0 =

g- parameters:

g11 = $ / I2 =0 =

g12 = $$ / V1=0 =

g21 = / I2=0 =

g22 = $ / V1=0 =

A1B

1C

1D

1 parameters:

A1 =

/ I1=0 =

B1 =

$ / V1=0 =

C1 =-

$ / I1=0 =

D1 =

$$ / V1=0 =



2-1ECE ET LABMANUAL


RESULT:

S. No Parameter Theoretical Values Practical Values

1 Z11 303. Ω

2 Z12 199.99 Ω

3 Z21 200 Ω

4 Z22 350 Ω

5 Y11 5.38 m mhos

6 Y12 3.07 m mhos

7 Y21 3.05 m mhos

8 Y22 4.6 m mhos

Reviw Questions:-

1. Write the 2-port network equations in terms of hybrid parameter?

2. Define image impedance?

3. What is Z- parameter?

4. Write the network equations of Y- parameter



2-1ECE ET LABMANUAL


CONCLUSIONS:

1. Since Z12 = Z21 and Y12 = Y21 the given circuit is reciprocal.

2. Since Z11 = Z22 and Y11 = Y22 the given circuit is symmetrical.

3. There is a small deviation between theoretical and practical values because internal

resistances of source and meters are not considered.



2-1ECE ET LABMANUAL


SUPERPOSITION AND RECIPROCITY THEOREM

Exp - 3



2-1ECE ET LABMANUAL


3(a).VERIFICATION OF SUPERPOSITION THEOREM

AIM: To verify the superposition theorem.

GIVEN CIRCUIT:

STATEMENT:

SuperPosition Theorem:

In any linear, bilateral, multi source network the response in any

element is equal to the algebraic sum of the responses obtained by each source acting

separately while all other sources are set equal to zero.

APPARATUS:


1 Dual channel regulated

power supply

(0 – 30)

V/2A digital 01No

2 Ammeter (0 – 200m) A MC 01No

3 Resistors

100Ω

150Ω

200Ω

-

01No

01No

01No



2-1ECE ET LABMANUAL



5 Connecting wires - - Required number

CIRCUIT DIAGRAM:

WhenV1&V2 source acting(To find I):-

WhenV1 source acting(To find I1):-

WhenV2 source acting(To find I2):-



2-1ECE ET LABMANUAL


THEORY:-

Superposition theorem:-

Superposition theorem states that in a linear bilateral network consisting N number of

sources each branch current is the algebraic sum of N currents through ( branch voltage), each of

which is determined by considering one source at a time and removing all other sources. In

removing the sources, voltage sources are short circuited or replaced by resistances equal to their

internal resistances for no ideal sources, while the ideal current sources are open circuited.

PROCEDURE:

1. Connect the circuit as per the fig (1).

2. Adjust the output voltage of sources X and Y to appropriate values (Say 30V and20V

respectively).

3. Note down the response (current, IL) through the branch of interest i.e. AB ammeter

reading.

4. Now set the source Y (20V) to 0V.

5. Note down the response (current, ILl) through the branch AB (ammeter reading).

6. Now set the source X (20V) to 0V and source Y to 20V.

7. Note down the response (current, ILll) through the branch AB (ammeter reading).

8. Reduce the output voltage of the sources X and Y to 0V and switch off the supply.



2-1ECE ET LABMANUAL


9. Disconnect the circuit.

TABULAR FORMS:

From Fig(1)

S. No

Applied

voltage

(V1) Volt

Applied

voltage

(V2) Volt

Current

IL

(mA)

From Fig(2)

S. No Applied voltage

(V1) Volt

Current

IL

(mA)

From Fig(3)


(V2) Volt

Current

IL

(mA)



2-1ECE ET LABMANUAL


THEORITICAL CALCULATIONS

From Fig(2)

I1=V1/(R1+(R2//R3)) = 161 mA

ILl =I*R2/(R2+R3) = 69 mA

From Fig(3)

I2=V2/(R2+(R1//R3)) = 92 mA

ILl1

=I*R1/(R1+R3) = 30 mA ; IL = ILl

+ ILl1

=69 mA+30 mA =99mA

PRECAUTIONS:


2. Set the ammeter pointer at zero position.



5.Avoid short circuit of RPS output terminals.

RESULT:

S.No Load current Theoretical Values Practical Values

1 When Both sources are acting, IL 99 mA

2 When only source X is acting, ILl 69 mA

3 When only source Y is acting, ILll 11

LI 30 mA



2-1ECE ET LABMANUAL


Review Questions:-

1) What do you man by Unilateral and Bilateral network? Give the limitations of

Superposition theorem.

2) What are the equivalent internal impedances for an ideal voltage source and for a

Current source?

3) Transform a physical voltage source into its equivalent current source.

4) If all the 3 star connected impedance are identical and equal to ZA, then what is the

Delta connected resistors?

CONCLUSION:

1. The given circuit is linear, since the response is algebraic sum of the individual

responses.

2. Superposition theorem is not valid for power responses.

3( b) RECIPROCITY THEOREM

AIM :

To verify reciprocity theorem for the given circuit .

GIVEN CIRCUIT:

STATEMENT:

Reciprocity theorem



2-1ECE ET LABMANUAL


In any linear, bilateral, single source network, the ratio of excitation to the response is same

even though the positions of excitation and response are interchanged.

APPARATUS:


1 Regulated power supply (0 – 30) V/2A digital 01No

2 Ammeter (0 – 200m) A MC 01No

3 Resistors

100Ω

150Ω

200Ω

-

01 No

01No

01No



CIRCUIT DIAGRAMS:-

CIRCUIT – 1:

CIRCUIT -2



2-1ECE ET LABMANUAL


THEORY:-

Reciprocity Theorems:-

This theorem permits in to transfer source from one position in the circuit to another and

may be stated as under.

In any linear bilateral network, if an e.m.f E acting in a branch causes a current ‘I’ in

branch ‘Y’ then the same e.m.f E located in branch ‘Y’ will cause a current I in branch.

However, currents in other branches will not change.



2-1ECE ET LABMANUAL


PROCEDURE:-

Reciprocity Theorems:-

1. Connect the circuit as per the fig (1).

2. Adjust the output voltage of the regulated power supply to an appropriate value (Say

30V).

3. Note down the response (current, IL) through 150Ω resistor (ammeter reading) .

4. Reduce the output voltage of the signal generator to 0V and switch-off the supply.

5. Disconnect the circuit and connect the circuit as per the fig (2).


30V).

7. Note down the response (current, IL1) through 100Ω resistor (ammeter reading) .

8. Reduce the output voltage of the signal generator to 0V and switch-off the supply.


TABULAR FORM:

RECIPROCITY THEOREM :

From Fig(1)


(V1) Volt

Current

IL

(mA)



2-1ECE ET LABMANUAL


From Fig(2)


(V2) Volt

Current

IL1

(mA)

THEORITICAL CALCULATIONS :

From Fig(1)

I1=V/(R1+(R2//R3)) = 161 mA

IL= I1*R3/(R2+R3) = 92 mA

From Fig(2)

I2=V/(R2+(R1//R3)) = 138 mA

IL1= I2*R3/(R1+R3) = 92 mA

PRECAUTIONS:


2. Set the ammeter pointer at zero position.



2-1ECE ET LABMANUAL




5.Avoid short circuit of RPS output terminals.

6.If voltmeter gives (-) ve reading then interchange the terminals connections of a voltmeter.

RESULT:

CONCLUSION:

1. The given circuit is bilateral, since the ratio of excitation to the response is same before

and after interchanging the positions of excitation and response.

S.No Parameter

Theoretical Value Practical Value

1 V/IL 92 mA

2 V/ILl 92 mA

Exp - 4



2-1ECE ET LABMANUAL


VERIFICATION OF MAXIMUM POWER TRANSFER THEOREM

AIM: To verify maximum power transfer theorem on d.c circuit .

STATEMENT:

It states that, maximum power will be transferred from source to load when the load

resistance is the complex conjugate of source resistance.

GIVEN CIRCUIT:

APPARATUS:


1 Regulated power supply (0 – 30)V/2A digital 01No

2 Voltmeter

(0-30) V MC 01No

3 Ammeter (0-1) A MC 01No

4 Rheostats 100 Ω/5A

50Ω/5A Wound Wire

01No

01No




2-1ECE ET LABMANUAL


CIRCUIT DIAGRAM:-

Fig(1)

Model Graph for Maximum Power Transfer Theorem:



2-1ECE ET LABMANUAL


THEORY:

The statement of maximum power transfer is “ In d.c circuits, maximum power is

transferred from a source to load when the load resistance is made equal to the internal resistance

of the source as viewed from the load terminal with load removed and all e.m.f sources replaced

by their internal resistance.

Consider a voltage source of V of internal resistance R delivering power to aload

RL. We shall prove that when RL = RS the power transferred is maximum.

Circuit current = %

&'"&(

Power delivered P = I2 RL

= ) %&'"&*+

RL

,&' &(- ,&' &( &'- =0

RL+Ri cannot be zero,

Ri – RL = 0

PROCEDURE:

1. Connect the circuit as per the circuit diagram fig(1).


30V).

3. Vary the load rheostat. in steps, and note down the response (current) through the load for

each step (ammeter reading) & load voltage.

4. Reduce the output voltage of the regulated power supply to 0V and switch-off the supply.


6. Calculate the power absorbed by the load, PL for each step using the formula PL=IL2

RL.

7. Plot the graph by taking ‘RL’ on X-axis and PL on Y-axis.

RS ==RL



2-1ECE ET LABMANUAL


8. Get the practical value of the load resistance for which it will gain the maximum power

from the source.

Tabulation for Maximum power transfer theorem:

SNO.

LOAD

RESISTENCE

In ohms

VOLTAGE

VL (in Volts)

CURRENT

IL (in amps)

POWER

P=VL*IL(watt)

1

2

…….



2-1ECE ET LABMANUAL


Theoretical calculations

Total current I = ./

&*"&'

P = I2 RL

= 0*&*"&'2. RL

Power is maximum dp/dRL =0

→ 1

1&' 2 ./&/"&' . &'3 = 0

→0/ 2 0/&/"&' &' 4&/ &'53 =0

→4&/ &'5 &/&' &' =0

&/ &' &' =

&/ &' =

&/ = &'

PRECAUTIONS:

→ Avoid loose connections

→ Ammeter should always connected in series with the circuit.

Rs=RL



2-1ECE ET LABMANUAL


RESULT:-

Review Questions:-

1. Derive the condition for maximum power transfer theorem.

2. Where and why maximum power transfer theorem is applied?

3. What is the efficiency of the circuit at the maximum power transfer

Condition & why?

4. Derive the condition for maximum power transfer theorem for a.c.

Circuits.

5.Define a dependent source.

VERIFICATION OF THEVENIN’S THEOREM AND NORTON’S THEOREM

AIM: To verify Thevenin’s & Norton’s theorems for the given circuit.

GIVEN CIRCUIT:

STATEMENTS:

Thevenin’s theorem

It states that any linear, active network with two open terminals can be

replaced by an equivalent circuit consisting of Thevenin’s equivalent voltage source Vth in series

with Thevenin’s equivalent resistance Rth. Where Vth is the open circuit voltage across the two

terminals and Rth is the resistance seen from the same two terminals.

Exp - 5



2-1ECE ET LABMANUAL


Norton’s theorem

It states that any linear, active network with two open terminals can be

replaced by an equivalent circuit consisting of Norton’s equivalent current source IN in parallel

with Norton’s equivalent resistance RN. where IN is the short circuit current through the two

terminals and RN is the resistance seen from the same two terminals

APPARATUS:

S. No Name of the

apparatus Range Type Quantity

1 Regulated power

supply

(0 –

30)V/2A Digital 01

2 Voltmeter

(0-30)V MC 01

3 Ammeter

(0-

2000m)A MC 01

4 Resistors

100Ω

150Ω

200Ω

Carbon

Composition

02

01

01

5 Experimental board --- --- 01

6 Connecting wires --- ---- Required

number

CIRCUIT DIAGRAMS:-

TO FIND IL:



2-1ECE ET LABMANUAL


FIG(1)

TO FIND VTH:

FIG(2)

TO FIND Rth:

TO FIND IN:



2-1ECE ET LABMANUAL


Fig(4)

THEORY:

Thevenin’s theorem:

The values of VTh and RTh are determined as mentioned in thevenin’s theorem.

Once the thevenin equivalent circuit is obtained, then current through any load resistance RL

connected across AB is given by, I = %6

&6"&'

Thevenin’s theorem is applied to d.c. circuits as stated below.

Any network having terminals A and B can be replaced by a single source of

e.m.f. VTh in series with a source resistance RRh.

(i) The e.m.f the voltage obtained across the terminals A and B with load, if any

removed i.e., it is open circuited voltage between terminals A and B.

(ii) The resistance RTh is the resistance of the network measured between the terminals A

and B with load removed and sources of e.m.f replaced by their internal resistances.

Ideal voltage sources are replaced with short circuits and ideal current sources are

replaced with open circuits.

To find VTh, the load resistor ‘RL’ is disconnected, then VTh = %

&"&7 Χ R3

To find RTh,

RTh = R2 + & &7&" &7

Thevenin’s theorem is also called as “Helmoltz theorem”

NORTON’S THEOREM:



2-1ECE ET LABMANUAL


Nortom’s theorem is applied to d.c circuits may be stated as below.

Any linear network having two terminals ‘A’ and ‘B’ can be replaced by a current source

of current output IN in parallel with a resistance RN.

(i) The output IN of the current source is equal to the current that would flow through

AB when A&B are short circuited.

(ii) The resistance RN is the resistance of network measured b/wn A and B with load

removed and the sources of e.m.f replaced by their internal resistances.

Ideal voltage source are replaced with short circuits and ideal current sources are replaced

with open circuits .

Norton’s theorem is converse of thevenin’s theorem in that Norton equivalent

circuit uses a current generator instead of voltage generator and the resistance RN is parallel

with generator instead of being series with it.

for source current,

II = %&8 =

%4& "&7 5&&"&&7"&&7

for short-circuit current,

IN = Χ &7

&"&7 = %&7

&&"&&7"&&7

PROCEDURE:

Thevenin’s Theorem

1. Connect the circuit as per fig (1)


30V).

3. Note down the response (current, IL) through the branch of interest i.e. AB (ammeter

reading).




2-1ECE ET LABMANUAL


5. Disconnect the circuit and connect as per the fig (2).

6. Adjust the output voltage of the regulated power supply to 30V.

7. Note down the voltage across the load terminals AB (Voltmeter reading) that gives Vth.



10. Adjust the output voltage of the regulated power supply to an appropriate value (Say V =

30V).

11. Note down the current (I) supplied by the source (ammeter reading).

12. The ratio of V and I gives the Rth.



15. Adjust the output voltage of the regulated power supply to 30V

16. Note down the response (current, IN) through the branch AB (ammeter reading).



THEORITICAL CALCULATIONS :

Thevinen’s theorem:

VTH=(V/(R1+R3) )R3 = 20 V

RTH =((R1*R3)/(R1+R3))+RL = 216.6 Ω



2-1ECE ET LABMANUAL


IL= VTH / (RTH +RL) = 63 mA

Norton’s theorem:

RN = R2 +(R1*R3)/(R1+R3) = 216.6 Ω

VTH = V /(R1+R3) ) R3 = 20 V

IN= VTH /RTH = 92 mA

IL = IN *RTH/(RTH+RL) = 63 mA

Tabulation for Thevinen’s theorem:

THEORITICAL VALUES PRACTICAL VALUES



2-1ECE ET LABMANUAL


RTH=RL= 216.6 Ω

IL=63 mA

Vth=20 V

RTH=RL=

IL=

Vth=

Tabulation for Norton’s theorem:

THEORITICAL VALUES PRACTICAL VALUES

RN=RL=216.6 Ω

IL=63 mA

IN=92 mA

RN=RL=

IL=

IN=

RESULT:-



2-1ECE ET LABMANUAL


Review Questions:-

1) The internal resistance of a source is 2 Ohms and is connected with an

External load of 10 Ohms resistance. What is Rth ?

2) In the above question if the voltage is 10 volts and the load is of 50Ω.

What is the load current and Vth? Verify IL?

3) If the internal resistance of a source is 5 Ω and is connected with an

External load of 25 Ohms resistance. What is Rth?

4) In the above question if the voltage is 20V and the load is of 50 Ohms,

What is the load current and IN ? Verify IL ?



2-1ECE ET LABMANUAL


OPEN CIRCUIT CHARACTERISTICS OF D.C SHUNT GENERATOR

AIM:

To obtain the no load characteristics of a DC shunt generator and to determine the critical

field resistance.

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor D.C Generator

1 Voltage 220V 220V

2 Current 19. A 13.6 A

3 Speed 1500 R.P.M 1500 R.P.M

4 Power 5 HP 3 KW

5 Field current 1 A 1 A

APPARATUS REQUIRED:

S.NO Description Type Range Quantity

1 Volt meter M.C 0-300v 1

2 Ammeter M.C 0-2A 1

3 Tachometer Digital 0-10,000 R.P.M 1

4 Rheostat Wire wound 300Ω/2A

2

Exp - 6



2-1ECE ET LABMANUAL


5 Connecting

wires ------ ------- As required

CIRCUIT DIAGRAM:

OPEN CIRCUIT CHARACTERISTICS OF D.C SHUNT GENERATOR



2-1ECE ET LABMANUAL


THEORY:

Magnetization curve is relation between the magnetizing forces and the

flux density B. this is also expressed as a relation between the field current and the induced

emf , in a D.C machine. Varying the field current and noting corresponding values of

induced emf can determine this.

For a self-excited machine the theoretical shape of the magnetization

curve is as shown in the figure. The induced emf corresponding to residual magnetism

exists when the field current is zero. Hence the curve starts, a little above the origin on y-

axis. The field resistance line Rsh is a straight-line passing through the origin.

PROCEDURE:

1) All the connections are as per the circuit diagram.

2) 220V, DC supply is given to the motor by closing DPST switch.

3) Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motor starts

running.

4) Adjust the speed of the motor to rated value by the adjusting the field rheostat of

motor.

5) By using field rheostat vary the field current of generator.

6) By varying the filed current in steps note down all the readings of generated voltages

at constant speed.

7) Now the field current & field rheostat of motor is removed slowly and the power is

switched OFF.

TO FIND CRITICAL FIELD RESISTANCE:

1) Draw the shunt field resistance line

2) Draw tangent to the OCC

3) The slope of this tangent gives the Rfc

Critical field resistance, Rc=Eg/ IF =



2-1ECE ET LABMANUAL


TABULAR COLUMN:

Residual Voltage = Speed=

GRAPH:

Draw the graph between generated voltage at no load and field current. By taking

generated voltage Eg in volts on Y axis and field current If in amps on X-axis.

S.NO Eg in Volts If in Amps



2-1ECE ET LABMANUAL


MODEL GRAPH:

TO FIND FIELD RESISTANCE:

S.NO Voltage Field current Field resistance



2-1ECE ET LABMANUAL


PRECAUTIONS:-

1) The rheostat is connected such that minimum resistance is included in field circuit of

motor

2) The rheostat is connected such that maximum resistance is included in field circuit of

generator.

3)Starter handle is moved slow

RESULT:

.

REVIEW QUESTIONS:

1.What is meant by critical field resistance?

2.What is meant by critical speed?

3.Residual magnetism is necessary for self excited generators or not.

4.Why this test is conducted at constant speed?



2-1ECE ET LABMANUAL


SWINBURNE’S TEST ON D.C SHUNT MACHINE

AIM:

To conduct the field test on two identical series machines and to find the efficiency at full

load.

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor

1 Voltage 220V

2 Current 19A

3 Speed 1500 R.P.M

4 Power 5 HP

5 Field Current 0.6 A

6 Insulation Class B

APPARATUS REQUIRED:


1 Voltmeter M.C 0-300V 01

Exp – 7



2-1ECE ET LABMANUAL


2 Ammeter M.C 0-2A

0-5 A

0-1A

01

01

01

3 Tachometer Digital 0-10000 R.P.M 1

4 Rheostat Wire wound 300Ω/2A

01

5 Connecting

Wires

---- ---- As required



2-1ECE ET LABMANUAL


CIRCUIT DIAGRAM: SWINBURNS’S TEST ON D.C SHUNT MACHINE

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2-1ECE ET LABMANUAL


THEORY:

It is a simple indirect method in which losses are determined separately and

from their knowledge, efficiency at any desired load can be predetermined. the only test needed

is no-load test. However, this test is applicable to those machines in which flux is practically

constant.

The constant losses in a dc shunt machine= Wc = stray losses(magnetic & mechanical

losses) +shunt field copper losses

Wc = No load input – No load armature copper losses

= VIL0 - Iao2 where Ra is the armature resistance

and Iao=IL0-Ish

PROCEDURE

1) Make all the connections are as per the circuit diagram.

2) Keep the field rheostat in minimum resistance position and armature rheostat in

maximum position.

3) Excite the motor with 220V, DC supply by closing the DPST switch and start the

motor by moving the handle of 3-point starter from OFF to ON position.

4) By adjusting the rheostats in motor armature and field bring the speed of the motor to

its rated value. Note down the readings of Ammeter and Voltmeter at no load

condition

5) The necessary calculations to find efficiency of machine as motor & generator at any

given value of armature current is done.

TO FINDARMATURE RESISTANCE(RA):

1) Connect the circuit per the circuit diagram

2) Keep the rheostat in maximum position.

3) Now excite the motor terminals by 30V supply by closing DPST switch.

By varying the rheostat & motor down the readings of Ammeter and voltmeter



2-1ECE ET LABMANUAL


MODEL CALCULATIONS:-

For motor:

IL= Ia+If

No load losses = Wo =VIo – Iao2Ra

Input = VI

Cu losses = Ia2 Ra

Total losses =No load losses + cu losses

Efficiency( η) = Output/Input

Output = input-total losses

For generator:-

I a = IL +If

No load losses = Wo = V Io – Iao2Ra

Input = VI

Cu losses = Ia2 Ra

Total losses =No load losses + cu losses

Efficiency ( η) = Output / Input

Output = input - total losses



2-1ECE ET LABMANUAL


CIRCUIT DIAGRAM TO FIND ARMATURE RESISTANCE:

TABULAR COLOUMN:

S.NO Voltmeter

reading

V Volts

Ammeter

Reading

I in Amps

Ammeter

reading

If in Amps

Speed in

RPM

ARMATURE RESISTANCE (Ra):

S.No Voltage Current



2-1ECE ET LABMANUAL


CALCULATION TABLE:

As a Motor:

S.NO Load IL in A Ia=(IL-Ish)

in A

Win =Ia2

Ra in

watts

Total

losses in

w

%Efficiency

As a Generator:

S.NO Load IL in A Ia=(IL+Ish)

in A

Win =Ia2

Ra in

watts

Total

losses in

w

%Efficiency

GRAPH:

The graph is drawn between

(a)Output in Watts Vs Efficiency(%η)

By taking output in Watts on X axis current, Efficiency on Y-axis

MODEL GRAPH:



2-1ECE ET LABMANUAL


THEORY:

Advantages:

1) It is convenient and economical method of testing of DC machines since power required

to test a large machine is very small.

2) The efficiency of the machine can be predetermined at any load. Since stray losses are

known.

Disadvantages:

1) This test cannot be performed with dc series motors.

2) This test is only applicable to those machines in which flux and speed remain

constant.

3) As the test is performed on no load it is impossible to know whether at full load

commutation would be satisfactory and the temperature raise would be with in

specified limits or not.

4) No account is taken for change in iron losses form no load to full load on account of

distribution of flux due to armature reaction. On full load the flux distribution is very

much affected due to armature reaction and is some case to an extent that iron losses

become 1.5 times of iron losses at no load .

PRECAUTIONS:

1.We should start the motor under no load

2.Take the reading without parallax error.

3.The connections must be tight.

4. If voltmeter gives –ve reading then interchange voltmeter terminal connecting of

voltmeter.



2-1ECE ET LABMANUAL


RESULT:

REVIEW QUESTIONS:

1.Why the magnetic losses calculated by this method are less than the actual value?

2.Is it applied to D.C series machines?

3.Comment on the efficiency determined by this method.

BRAKE TEST ON D.C SHUNT MOTOR

AIM:

To conduct the brake test on a given D.C shunt motor and to draw its performance

curves .

NAME PLATE DETAILS:

S.NO Characteristic D.C Motor

1 Voltage 220V

2 Current 19A

3 Speed 1500 R.P.M

4 Power 5 HP

5 Field current 0.6 A

APPARATUS REQUIRED:

Exp – 7



2-1ECE ET LABMANUAL



1 Volt meter M.C 0-300v 01

2 Ammeter M.C 0-20A 01

3 Rheostat Wire wound 300Ω/2A 01

4 Tachometer Digital 0-10000 R.P.M 01

5 Connecting

Wires

---- ---- As required



2-1ECE ET LABMANUAL


CIRCUIT DIAGRAM: BRAKE TEST ON D.C SHUNT MOTOR

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2-1ECE ET LABMANUAL


THEORY:

The speed of a D.C motor Nα V.Ia Ra

ϕ The speed of the D.C motor is inversely proportional to the flux produced by the

field. Decreasing the flux the speed of the machine can be increased. The flux of the field

winding can be changed by changing shunt field current with the help of shunt field rheostat.

Another method for speed control is to keep a variable resistance in series with the armature. By

increasing the resistance the voltage drop also increases and hence the voltage applied across the

armature decreases which result in the decrease in speed of the motor.

PROCEDURE:

1. All the connections are as per the circuit diagram.

2. 220V, DC supply is given to the motor by closing DPST switch.

3. Move the 3-point starter handle form ‘OFF’ to ‘ON’ position slowly and motor

starts running.

4. Vary the field rheostat and armature rheostat until the motor reaches its rated

speed and take voltmeter and ammeter readings.

5. Apply the land by break drum pulley and for each applications of load the

corresponding Voltmeter (V), Ammeter (I), Spring forces S1 & S2 and Speed (N)

readings are noted.

6. Calculate output & efficiency for each reading.

7. Note down all the readings in the tabular form carefully.

8. Remove the load slowly and keep the rheostat as starting position and switch

‘OFF’ the supply by using DPST switch



2-1ECE ET LABMANUAL


Tabular column

S.NO

Voltmet

er

Reading

V volts

Amme

ter

Readi

ng I

amps

Input =

VI

watts

Forces in

KG Net forces

F = S1~S2

in kg

Torque

=

f*r*9.81

(N-M)

Speed in

RPM

(N)

O/p=

2πNT/60

(Watts)

%ή=

(o/p)/(i/p)*100 S1 S2

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Tirumala Engineering College Electrical Technology lab


GRAPH:

The graph is drawn between

a) Output in Watts Vs Speed(N) in RPM

b) Output in Watts Vs Torque(T) in N-m

c) Output in Watts Vs Current(I) in A

d) Output in Watts Vs Efficiency(%η)

By taking output in Watts on X axis and speed, Torque, current, Efficiency on

Y- axis .

MODEL GRAPH:

FORMULAE:

Torque=:F*Re*9.81 N

Power output=[(2*Π*N*T) /60] W





PRECAUTIONS:

1. Initially 3-point starter should be kept at ‘OFF’ position and later it must be varied

slowly and uniformly from ‘OFF’ to ‘ON’ position.

2. The field regulator must be kept at its minimum output position.

3. The brake drum of the motor should filled with cold water.

4. The motor should be started without load.

APPLICATIONS:

• Essentially for constant speed applications requiring medium starting torque.

• May be used for adjustable speed not greater than 2:1 range.

• For lathes, centrifugal pumps, reciprocating pumps, fans, blowers, conveyors, wood working

machines, machine tools, printing presses, spinning and weaving machines etc.

RESULT:

REVIEW QUESTIONS:

1 .Why a 3-point starter is used for starting a D.C shunt motor?

2 . If a 3-point starter is not available ,how can a D.C motor be started?

3 . Explain the function of overload release coil in 3-point starter .





OC & SC TESTS ON 1 – PHASE TRANSFORMER

AIM:

To conduct Open circuit and Short circuit tests to pre-determine the performance of the

single phase transformer

NAME PLATE DETAILS:

Voltage 230 V (HV), 115 V (LV)

Current 26 A(HV) ,13.04 A (LV)

KVA RATING 3KVA

APPARATUS:

S.NO Name Type Range Quantity

1 Ammeter MI

MI

(0- 5) A,

(0- 20) A

1

1

2 Voltmeter MI

MI

(0- 300 )V,

(0- 50 )V

1

1

3 Wattmeter UPF 150 V/ 20 A 1

4 Wattmeter LPF 300 V/5 A 1

5 Auto

Transformer

1-Φ 230 V / (0- 270) V/ 15

A

1

Exp – 8 This PDF was made with the DEMO version of PDFtypewriter. The full version will notadd this text to your PDFs. You can purchase the full version athttp://ctdeveloping.com




CIRCUIT DIAGRAMS









OC Test Observations

Where

M. F. = Multiplication factor = FSD

VI φcos

FSD Full scale divisions

SC Test Observations

SVSC (V) ISC (A) WSC = W x M.F (w)

S.No. Vo (V) Io (A) Wo = W x M.F (w)





PROCEDURE (OC TEST):

1. Connections are made as per the circuit diagram

2. Initially variac should be kept in its minimum position

3. Close the DPST switch

4. By varying Auto transformer bring the voltage to rated voltage

5. When the voltage in the voltmeter is equal to the rated voltage of LV winding note down all

the readings of the meters

6. After taking all the readings bring the variac to its minimum position

7. Now switch off the supply by opening the DPST switch

PROCEDURE (SC TEST):

1. Connections are made as per the circuit diagram

2. Short the LV side and connect the meters on HV side

3. Before taking the single phase, 230 V, 50 Hz supply the variac should be in minimum

position

4. Now close the DPST switch so that the supply is given

5. By varying the variac when the ammeter shows the rated current then note down all the

readings

6. Bring the variac to minimum position after taking the readings and switch off the supply





MODEL CALCULATIONS:

(a)Calculation of Equivalent circuit parameters:

Let the transformer be the step-up transformer

Primary is L. V. side.(V1) , Secondary is H. V. side (V2)

(i) Parameters calculation from OC test

cos φφφφ0 = oo

o

IV

W =

Iw = I0 cos φφφφ0 = KII ww /1 = =

wI

VR 1

0 = = 2

0

1

0 KRR = =

Iµ = I0 sin φφφφ0 =

µI

VX 1

0 = = 2

0

1

0 KXX = =

K = 1

2

V

V =

(ii) Parameters calculation from SC test

202

sc

SC

I

WR = = 2

02

2

0202 RZX −= =

SC

SC

I

VZ =02 =

2

0201 / KXX = =

2

0201 / KRR = =

2

0201 / KZZ =





Tabulation:

(a) Efficiency at different loads and P.f s

cos φφφφ1 = ___________ cosφφφφ2 = ___________

S.N

o.

Load Cu.lo

ss

(W)

Outp

ut

(W)

Inp

ut

(W)

η

(%

)

X

x

S.N

o.

Loa

d

Cu.lo

ss

(W)

Outp

ut

(W)

Inp

ut

(W)

η

(%

)

1.

2.

3.

4.

¼

F.

L.

½

F.

L.

¾

F.

L.

F.

L.

1.

2.

3.

4.

¼

F.L

.

½

F.L

.

¾

F.L

.

F.L

.

(b) Regulation at full load

Lagging Pf Leading Pf

S.

No. P.F. % Reg.

S.

No. P. F. % Reg.

1. 0.3 1. 0.3

2. 0.4 2. 0.4

3. 0.5 3. 0.5

4. 0.6 4. 0.6





5. 0.7 5. 0.7

6. 0.8 6. 0.8

7. Unity 7. Unity

(b) Calculations to find efficiency:

For ½ full load

Cupper losses = Wsc x (1/2)2 watts =

where Wsc = full – load copper losses

Constant losses = W0 watts =

Output = ½ KVA x cos φφφφ = [cos φφφφ may be assumed]

Input = output + Cu. Loss + constant loss =

% 100xInput

Outputefficiency = =

(C)Calculation of Regulation at full load:

I2 = Load (KVA) X 103 / V2 =

100sincos

Re%2

022022 xV

XIRIgulation

φφ ±= =

‘+’ for lagging power factors

‘-‘ for leading power factors





PRECAUTIONS:

1. Loose connections should be avoided

2. Variac should be in minimum position initially

3. During the SC test make sure that constant current is supplied

MODEL GRAPHS:

1) Load Vs Efficiency

2) Pf Vs Regulation





RESULT:

REVIEW QUESTIONS:

1) The regulation calculated is exact or approximate?

2) Is it direct or indirect test?

3) What are the parameters to be calculated by using this test?

4) What are the conditions for maximum regulation and zero regulation?





BRAKE TEST ON 3 -PHASE SQUIRREL CAGE INDUCTION MOTOR

AIM:

To perform Brake test on 3- phase Slip ring induction motor to determine performance

characteristics.

NAME PLATE DETAILS:

Voltage 415 V

Current 7 A

Power 3 H.P

Speed 1500 Rpm

APPARATUS:

S.NO NAME TYPE RANGE QUANTITY

1 Voltmeter MI 0-600 V 01No

2 Ammeter MI 0-10 A 01No

3 Wattmeter Dynamometer 600 V/ 10 A

UPF(F/R)

02No

4 Tachometer Digital (0-10000)RPM 01NO

5 Connecting

Wires

---- ---- Required

number

Exp – 9





CIRCUIT DIAGRAM













PROCEDURE:

1. Connect the circuit as per the circuit diagram

2. Insert the proper rating fuses

3. Initially we place the rotor rheostat in maximum position

4. Initially there is no load on the motor

5. Close the TPST switch by giving 415 V, 50 Hz AC supply using 3-phase Auto Transformer

6. Measure no load voltage, current and power

7. Apply the load by tightening the brake drum measure the values of voltage, current ,

wattmeter, speed, weights reading i.e. S1 and S2.

8. After taking the values make brake drum in initial position(voltage 415 to 0 position) then

switch off the supply (Remove TPST switch)

9. Tabulate the results and draw the graphs.

MODEL CALCULATION:





MODEL GRAPHS:

1. Output Vs Efficiency

2. Output Vs Torque

3. Slip Vs Torque





PRECAUTIONS:

1. While starting the motor the load should not be there on the drum

2. The applied load on the brake drum does not exceed its rated value

3. The readings should be noted without parallelex errors

4. Always the drum should be kept cool

RESULT:

REVIEW QUESTIONS:

1) Is it possible to realize maximum torque at starting of induction motor?

2) What should be the frequency of injected emf in rotor circuit?





OC & SC TESTS ON 3 – PHASE ALTERNATOR

AIM:

To conduct Open circuit and Short circuit tests on a three- phase Alternator and to determine

the voltage regulation and synchronous impedance using EMF and MMF methods

NAME PLATE DETAILS:

DC SHUNT MOTOR ALTERNATOR

VOLTAGE (0-220 )V VOLTAGE: 415 V

CURRENT: 2.8 A CURRENT: 7 A

SPEED: 1500 RPM SPEED: 1500 RPM

POWER: 7.5 HP POWER: 7.5 HP

EXCITATION CURRENT: 2 A EXCITATION CURRENT: 1.55 A

APPARATUS:

S.NO NAME TYPE RANGE QUANTITY

1 Voltmeter MI 0-600 V 01No

2 Voltmeter MC 0-50 V 01No

3 Ammeter MI 0-10 A 01No

4 Ammeter MC 0-2 A 01No

5 Rheostat Wire

Wound

300 Ω/ 2 A

100Ω/ 5 A

01No

01No

6 Connecting

Wires

---- ----- Required

number

7 Tachometer

Digital

(0-

10,000)RPM

01No

Exp – 10





CIRCUIT DIAGRAM:-





Tabulation:

a) OC & SC Test:

O. C. Test S. C. Test

Speed = Speed =

S.No. Field

current (A)

Phase

voltage (V)

S.No. Field

current,

(If) (A)

Short circuit

current (ISC), (A)









b) Armature Resistance:

S.No. I (A) V (volts) Rdc = V/I ΩΩΩΩ





PROCEDURE:

(OC TEST):


2. Bring the D.C shunt motor to rated speed by varying rheostat (say field).

3. Alternator is brought to the synchronous speed by using shunt motor.

4. Initially the stator terminals are open circuited.

5. Now increase (Potential divider) the field current of alternator and note the values of ammeter,

voltmeter.

6. Slowly reduce the (Potential divider ) excitation and field rheostat of shunt motor to their

initial positions. switch off supply

(SC TEST):


2. The Stator terminals are short circuited using a switch

3. Now close the DPST switch and by varying the field rheostat bring DC shunt motor to its

rated speed

4. Alternator is synchronized using shunt motor

5. Now vary the DC excitation of alternator in steps to field current rated value and note A1 and

A2 values

6. Bring back the excitation and field rheostat to their initial positions





MODEL CALCULATIONS:

EMF METHOD:

From Graph SC

OC

SI

VZ = for the same If and speed: =

Ra = (1.6) RdC =

22

aSS RZX −= =

Assume p.f. (CosΦ) =

Assume armature current (Ia) =

Generated emf of alternator on no load is

( ) ( )22

0 sincos Saaa XIvRIvE ±++= φφ =

+ for lagging p.f.

- for leading p.f.

The percentage regulation of alternator for a given p.f. is

100Re% 0 xV

VEg

−= =

where

E0 – Generated emf of alternator per phase voltage

V – Full load, rated terminal voltage per phase.

PRECAUTIONS:

1. The excitation should not exceed the field current

2. The rheostat in shunt motor should be at minimum position

3. Take the readings without parallelox errors





MODEL GRAPHS:

1. Field current ( If ) Vs Generated EMF ( E )

2. Field current ( If ) Vs Short circuit current ( I sc )

3. Power factor Vs % Regulation













RESULT:

REVIEW QUESTIONS:

1) What is meant infinite bus bar?

2) How to minimize hunting effect?

3) At what condition frequency collapses?

4) What is synchroscope?



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