Electrical Technology Manual JWFILES

7/30/2019 Electrical Technology Manual JWFILES

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(Affiliated to JNTU Kakinada), Jonnada, Vizianagaram:535 005Website : www. lendi.org Ph no : +91 08922 241111, 241144 E mail id : lendi _2008 @yahoo.com Fax No: +91 08922 241112

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(Affiliated to JNTU Kakinada) JONNADA , VIZIANAGARAM:535005

Class : B.Tech Year: II/IV Branch: ECE

ELECTRICAL TECHNOLOGY LABORATORY

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Experiment no: 1

VERIFICATION OF SUPERPOSITION

& RECIPROCITY THEOREMS

AIM : To verify Superposition & Reciprocity theorems for the given network.

APPARATUS :

THEORY:-

I. Superposition Theorem Statement:

In a linear network with several independent sources which include

equivalent sources due to initial conditions and linear dependent sources, the overall

response in any part of the network is equal to the sum of the individual responses dueto each independent source, considered separately, with all other independent sources

reduced to zero.

Note: 1. The sources which are considered one at a time making all other sources zero,are the independent sources including sources due to initial conditions only. Thedependent sources are retained as they are in the network.

2. When one independent source is considered & all other independent sourcesare reduced to zero means that all the other independent voltage source are replaced withshort circuit and all the other independent current sources are replaced with open circuit.If the sources contain internal impedances, that sources are replaced by their internal

impedances.

II. Reciprocity Theorem Statement:

The Reciprocity theorem states that the ratio of response to excitation is

invariant to an interchange of the position of the excitation and response in a single

source network. However if the excitation is a voltage source, the response should be a

current and vice versa.

S. No Name of Apparatus Type Range Quantity

1 Voltmeter PMMC 0-300V 2

2 Ammeter PMMC 0-2.5A 1

3 RheostatWWWWWWWW

50/5A110/2A

300/1.7A300/2A

2112

4. Fuse TCC 5A 4

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PROCEDURE:-

I. SUPERPOSITION THEOREM:

1. Connect the circuit as per the Circuit diagram.2. Close Switch S1 on to the Supply mains & remain Switches S2 & S3 open and

Switch S4 closed.3. Note down the Voltmeter readings V1 ,V2 & Ammeter reading as I' in the S.No1

of Table14. Now close Switch S2 on to the Supply mains & remain Switches S1 & S4 open and

Switch S3 closed.5. Note down the Voltmeter readings V1 ,V2 & Ammeter reading as I" in the S.No2

of Table16. Now Close Switches S1 & S2 on to the Supply mains & remain Switches S3 &

S4 open.7. Note down the Voltmeter readings V1 ,V2 & Ammeter reading as I in the S.No3

of Table1

8. Finally disconnect the circuit from the Supply mains by open all the Switches.

II. RECIPROCITY THEOREM:

CASE : I

1. Connect the circuit as per the Circuit diagram.2. Close Switch S1 on to the Supply mains3. Note down the Voltmeter V1& Ammeter A1 readings in S. No. 1 of Table 24. Disconnect the circuit from the Supply mains by opening the Switch S1.

CASE : II

1. Connect the circuit as per the Circuit diagram.2. Close Switch S2 on to the Supply mains3. Note down the Voltmeter V2 Ammeter A2 readings in S. No. 2 of Table 24. Disconnect the circuit from the Supply mains by opening the Switch S2.

OBSERVATION TABLE:-

TABLE 1 SUPERPOSION THEOREM:

S.No.Voltmeter Reading

(Volts)

Voltmeter Reading

(Volts)

Ammeter Reading

(Amps)

1. V1 = V2 = I' =2. V1 = V2 = I" =3. V1 = V2 = I =

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TABLE 2 RECIPROCITY THEORE:

S.No. Voltmeter Reading (Volts) Ammeter Reading(Amps)

1. V1 = I1 =

2. V2 = I2 =

PRECAUTIONS:

1. Avoid Loose Connections .2. Readings must be taken without parallax error.3. Before switching on the supply for the circuit , ensure that all rheostats are at

maximum position and during the experiment these should not be disturbed.

RESULTS:

I. SUPERPOSITION THEOREM:

1. I' =

2. I" =

3. I =

II. RECIPROCITY THEOREM:

1. V1/ I1 =

2. V2/ I2 =

CONCLUSIONS:

VIVA QUESTIONS:

1) What are the Statements of the above theorems?2) What is a linear network?3) Where the above theorems are used practically?4) What are the practical applications of the above theorems?5) What is a bilateral network? Give examples.6) What are the limitations of above theorems?

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Experiment no: 2

VERIFICATION OF MAXIMUM POWER

TRANSFER THEOREM

AIM : To verify Maximum Power transfer theorem for the given circuit.

APPARATUS:


1 VoltmeterMCMC

0-300V0-150V

11

2 Ammeter MC 0-2A 1

3 RheostatWWWWWW

100/5A50/5A200/2A

211

4. Fuse TCC 5A 2

THEORY:

Statement: The Maximum Power transfer theorem states that A Resistance load RL,

being connected to a DC network, receives maximum power when it is

equal to the internal resistance of the source network as seen from the

load terminals i.e. Rth

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With reference to Fig(B)

VthIL =

Rth+RL

While the power delivered to the resistive load is

PL = IL2 RL = (Vth)

2

x RL(Rth+RL)

2

PL can be maximized by varying R and hence, maximum power can be delivered when(dPL/dRL) = 0

(Rth +RL)2 d (Vth

2RL)-Vth2 RL d (Rth+RL)

2

dRL dRL==> ____________________________________ = 0

(Rth+RL)4

(Rth +RL)2 (Vth

2) - Vth2 RL 2. (Rth+RL)

==> ____________________________________ = 0

(Rth+RL)4

==> (Rth +RL) -

RL 2. = 0

==> RL = Rth

Hence it has been proved that power transfer from a dc source network to aresistive network is maximum when the load resistance of the network is equal to theinternal resistance of the dc source

Again with RL=Rth, the system being perfectly matched for load and source,power transfer becomes maximum and this amount of power (Pmax) can be obtained as

Pmax = Vth2 RTh = Vth

2(RTh+RTh)

2 4RTh

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The total power supplied is thus

P in = 2 VTh2 = VTh

24RTh 2RTh

During maximum power transfer the efficiency of the circuit becomes,.

= Pmax x 100= 50%

Pin

PROCEDURE:-

I) TO FIND POWER VARIATIONS WITH RL

1. Connect the circuit as per the Circuit diagram 1.2. Apply 220 V DC Supply to the circuit by closing the DPST Switch.

3. Note down the readings of Ammeter & Voltmeter in Table 1 which are connectedacross the load after keeping the load rheostat, RL at its minimum value.

4. Increase the load resistance in steps and for each step, note down the correspondingAmmeter and Voltmeter readings in Table 1.

5. Disconnect the circuit from the supply by opening the DPST Switch.

II) TO FIND Rth

1. Connect the circuit as per the Circuit diagram 2.2. Apply 220 V DC Supply to the circuit by closing the DPST Switch.3. Note down the readings of Ammeter & Voltmeter in Table 2 .

4. Disconnect the circuit from the supply by opening the DPST Switch.

OBSERVATION TABLE:-

TABLE 1

S No VL (volts) IL (amps) RL = VL/ IL () PL = IL2RL ()

(W)

1.2.3.

.

.

.10.

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TABLE 2

S No VS (volts) IS (amps) Rth= VS/IS()

1.

MODEL GRAPH:-

PRECAUTIONS:-

1. Avoid loose connections.2. Avoid Parallax error.3. Take more number of readings for a better plot

RESULTS:-

1. Pmax = ---------- W

2. RL = ---------

3. Rth = ---------

4. = ---------

CONCLUSIONS:-

VIVA QUESTIONS:-

1) What is the Statement of Maximum Power Transfer theorem?2) What is a linear network?3) What is a bilateral network?4) What are the applications of the above theorem?5) What are the advantages & disadvantages of the above theorem?

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Experiment no: 3

VERIFICATION OF THEVENINS

& NORTONS THEOREMS

AIM : To verify Thevenins & Nortons theorems for the given circuit.

APPARATUS:


1 Voltmeters 0-300V MI 22 Ammeter 0-2A MI 2

3 Rheostats50 , 5A

110 , 2A200 , 2A

WWWWWW

211

4 1- Variac230V / (0-270)V,

15A---- 1

5. SPST 5A ---- 26. Fuse 5A TCC 2

THEORY:-

I) Thevenins Theorem Statement:

Any combination of linear bilateral circuit elements and active sources,

regardless of the connection or complexity, connected to a given load RL, may bereplaced by a simple two terminal network consisting of a single voltage source of Vth

volts and single resistance Rth in series with the voltage source, across the two

terminals of the load RL . The Vth is the open circuit voltage measured at the two

terminals of interest, with load resistance RL removed. This voltage is also called

Thevenins equivalent voltage. The Rth is the Thevenins equivalent resistance of the

given network as viewed through the open terminals with RL removed and all the

active sources are replaced by their internal resistances . If the internal resistances are

not known then independent voltage sources are to be replaced by the short circuit

while the independent current sources must be replaced by open circuit.

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II) Nortons Theorem Statement:

Any combination of linear bilateral circuit elements and active sources,

regardless of the connection or complexity, connected to a given load RL, can be

replaced by a simple two terminal network, consisting of a single current source of IN

amperes and a single resistance RN in parallel with it, across the two terminals of theRL. The INis the short circuit current flowing through the short circuited path, replaced

instead of RL. It is also called Nortons current. The RN is the equivalent resistance of

the given network as viewed through the load terminals, with RL removed and all the

active sources are replaced by their internal resistances. If the internal resistances are

unknown then the independent voltage sources must be replaced by short circuit while

the independent current sources must be replaced by open circuit.

PROCEDURE:-

I) FOR CIRCUIT 1:

1. Connect the circuit as per the circuit diagram.2. Apply 230 V AC Supply to the Variac (with its variable position at C ) by closing the

DPST Switch.3. Gradually vary the variable position of the Variac until the Voltmeter1 reads 200 V.4. Note down the corresponding readings of Ammeter & Voltmeter2 in Table 1 with the

conditionsi) SPST 1 Closed & SPST 2 Openii) SPST 1 Open & SPST 2 Openiii) SPST 1 Closed & SPST 2 Closed

5. Gradually vary the variable position of the Variac until the Voltmeter1 reads 0 volts6. Disconnect the Variac from the supply by opening the DPST Switch.

II) FOR CIRCUIT 2:


DPST Switch.3. Gradually vary the variable position of the Variac until the Voltmeter reads 150 V &

note down the corresponding reading of Ammeter in Table 2.4. Gradually vary the variable position of the Variac until the Voltmeter reads 0 volts5. Disconnect the Variac from the supply by opening the DPST Switch

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III) FOR CIRCUIT 3:


DPST Switch.3. Gradually vary the variable position of the Variac until the Voltmeter reads Vth , as

obtained in Table 14. Close the SPST Switch & vary the rheostat until the Ammeter reads current I for

which Vth / I gives Rth , the value as obtained in Table 2 .5. Once the Rheostat set to Rth , open the SPST Switch & note down the reading of the

Ammeter in Table 36. Gradually vary the variable position of the Variac until the Voltmeter reads 0 volts7. Disconnect the Variac from the supply by opening the DPST Switch

IV) FOR CIRCUIT 4:

1. Connect the circuit as per the circuit diagram.

2. Use the same Rheostat which set to Rth as in the Circuit 33. Apply 230 V AC Supply to the Variac (with its variable position at C ) by closing the

DPST Switch.4. Gradually vary the variable position of the Variac until the Ammeter1 reads current IN

as obtained in Table 1 & note down the corresponding reading of the Ammeter2 inTable 4.

5. Gradually vary the variable position of the Variac until the Voltmeter reads 0 volts6. Disconnect the Variac from the supply by opening the DPST Switch

OBSERVATION TABLE:-

TABLE 1 (For Circuit 1)

S.No Switch conditions Voltmeter V1(Volts)

Voltmeter V2(Volts)

Ammeter

(Amps)

1. SPST 1 ClosedSPST 2 Open

VS = VL = IL =

2. SPST 1 OpenSPST 2 Open

VS = Vth = IL = 0

3. SPST 1 ClosedSPST 2 Closed

VS = VL = 0 IN =


S.No Voltmeter (Volts) Ammeter (Amps) Rth = VS/IS()

1. VS = IS = Rth =

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S.No Voltmeter (Volts) Ammeter (Amps)

1.Vth = IL =


S.No Ammeter I1 (Amps) Ammeter I2 (Amps)

1. IN =IL =

PRECAUTIONS:-

1. Avoid loose connections.2. Avoid Parallax error.3. Before switching on the supply for each circuit ensure that all rheostats are at

maximum position and during the experiment these should not be disturbed.4. Variable position of the Variac(auto transformer) should be at minimum position

before switching on the power supply.

RESULTS:-

1. IL from the Main circuit =

2. IL from the Thevenins Equivalent Circuit =

3. IL from the Nortons Equivalent circuit =

CONCLUSIONS:-

VIVA QUESTIONS:-

1) What is the Statement of Thevenins theorem?2) What is a linear network?3) What is a bilateral network?4) What are Active & Passive elements?5) What are the applications of the above theorem?6) What are the limitations of application of this theorem ?

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SWINBURNES TEST ON

D.C. SHUNT MACHINE

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Experiment no: 4

SWINBURNES TEST ON A DC SHUNT MACHINE

AIM : To Pre-determine the efficiency and performance characteristics of aDC Shunt machine. (both as a generator & motor)

NAME PLATE DETAILS:

S.No Type DC Shunt Motor

01 Ratings 3.0 HP

02 Volts. 220 V DC03 Current 12 A

04 Exc. Volts. 220 V DC

05 Exc. Current 0.6A

06 Duty S1

07 Ins. Class B

08 Speed 1500 rpm

APPARATUS:

S.No Apparatus Required Rating Type Qty.

01 Voltmeter (0-300) V M.C 1

02 Ammeter (0-1) A M.C. 1


04 Rheostat 300, 2 AWIRE

WOUND1

05 Tachometer 0-10,000 RPM ANALOG 1

06 Fuse 6A T.C.C. 2

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THEORY:

SWINBURNES TEST:-

It is a simple method in which losses are measured separately and from theirknowledge, efficiency at any load can be pre-determined in advance. The onlyrunning test needed is a no load test.

Swinburnes test is applicable to those machines in which flux is practicallyconstant i.e. Shunt wound and Compound wound machines.

The machine is running as a motor on no-load at its rated voltage and its speed beadjusted to its rated value using Shunt regulator.

The no-load armature current Iao is measured using an ammeter, where as shuntfield current Ish is given by another ammeter. The no-load input current is given

by Io = Iao + Ish

Let the supply voltage be V voltsNo-load input = V Io watts

Power input to armature = V Iao watts

Power input to shunt = V Ish watts

No-load input supplies Copper losses (Armature & Field), Iron losses (Hysteresis& Eddy current) & Mechanical losses ( Friction losses & Windage).

Constant losses = No load input power Armature copper losses

Wc = V Io Iao Ra watts .

Predetermination of efficiency of a motor at any loadInput = V I watts.

Armature Cu losses = (I - Ish) Ra

Constant losses = Wc

Total losses = Wc + ( I - Ish) Ra

= (Input Total losses) / (Input)

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Predetermination of efficiency of a generator at any load

Output = V I watts.

Armature Cu losses = (I + Ish) Ra

Constant losses = Wc

Total losses = Wc + ( I + Ish) Ra

= (Output) / (Output + Total losses)

Maximum Efficiency : Variable losses (Ia Ra) = Constant losses ( Wc)

PROCEDURE:

1. Connect the circuit as per the Circuit diagram.2. Initially the starter must be in off position.3. Switch on the D.C. Motor to 220V D.C. Supply by closing the DPST Switch.4. Start the D.C. motor using the three point starter and thereby adjust the speed

to its rated speed using field rheostat.5. Note down the readings of Voltmeter & Ammeters in Table6. Switch off the D.C. Motor to 220V D.C. Supply by opening the DPST Switch.

OBSERVATION TABLE:

At Constant speed of 1500r.p.m.

S.No. Input Voltage V Armature Current Field current

Wc = V Io I2ao Ra Watts = ________ Watts

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CALCULATION TABLE:

I) For Motor

S.No

.

Input

Voltage

(V)(Volts)

Input

Current(I)

(Amp)

Field

current(Ish)

(Amp)

Armature

Copper

Losses(Watts)

Total

Losses

(Watts)

Input

Power

(Watts)

%

II) For Generator

S.No

.

Output

Voltage

(V)

(Volts)

Output

Current(I)

(Amps)

Field

current (Ish)

(Amps)

Armature

Copper

Losses

(Watts)

Total

Losses

(Watts)

Output

Power

(Watts)

%

MODEL GRAPHS:

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PRECAUTIONS:-

1. The field rheostat of the motor must be kept in minimum before switching on the220V D.C. supply.

2. Ensure that the starter arm is at extreme left position.3. Avoid loose connections4. Note down the readings from the meters without any parallax error

RESULTS:

Constant losses = _________ Watts Current at which Max. occurs for motor = _________ Amps Current at which Max. occurs for generator = _________ Amps Maximum Efficiency for motor = __________ %. Maximum Efficiency for generator = __________ %.

CONCLUSIONS:

VIVA VOCE QUESTIONS:

1) What is the significance of Swinburnes test?2) What are the advantages & disadvantages of this test?3) Why this test is not suitable for D.C series motor?4) What is the purpose of 3 point starter?5) What happens if field is open in D.C motor?6) Why we have to keep the field rheostat in minimum position?

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BRAKE TEST ON A

D.C. SHUNT MOTOR

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Experiment no: 5

BRAKE TEST ON A D.C. SHUNT MOTOR

AIM : To obtain the Performance characteristics curves of a D.C. shunt motor byconducting brake test on it.

NAME PLATE DETAILS:

S.No Type DC Shunt Motor

01 Rating 3.0 HP

02 Volts. 220 V DC

03 Current 12 A

04 Exc. Volts. 220 V DC

05 Exc. Current 0.6A

06 Duty S1

07 Ins. Class B

08 Speed 1500 rpm

APPARATUS:




03 Ammeter (0-20) A M.C. 1

04 Rheostat 300, 2 A Wire Wound 1

05 Tachometer 0-10,000 RPM Analog 1

06 Fuse 12A T.C.C. 2

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THEORY:

It is a simple method of testing low rating DC machines and consists of applyinga brake to a water-cooled drum mounted on the motor shaft.

The four important characteristics curves of a D.C. Shunt Motor, namely,Torque, Speed, Armature Current & efficiency, each plotted against the usefulPower, as shown in the model graph are known as Performance characteristics

A belt is wound round the brake drum and its two ends are attached to two springbalances S1 & S2. The tension of the belt can be adjusted with the help of swivels.

The force acting tangentially on the drum is equal to the difference between thereadings of the two spring balances.

The net force, F applied on the brake drum is 9.81(S1 S2) Newtonswhere , S1 & S2 are the readings of Spring balances 1& 2 in Kg.f.

Shaft torque, T developed by the motor is 9.81 (S1 S2) R Nmwhere, R is the radius of the pulley in meters & N is the speed in rpm

Useful Output Power = (2 N T) / 60 Watts Input Power = V IL Watts, where IL = (Ia + Ish) % Efficiency , = (Output power / Input power) x 100. Speed Regulation = [ (No Load speed ) ( Full load speed )] / Full Load. The size of the motor that can be tested by this method is limited from the

consideration of the heat that can be dissipated at the brake drum

Where the output power exceeds about 2 H.P., or where the test is of longduration, its necessary to use a water cooled brake drum.

PROCEDURE:

1. Connect the circuit as per the Circuit diagram.2. Initially the starter must be in off position.3. Switch on the D.C. Motor to 220V D.C. Supply by closing the DPST Switch.4. Start the D.C. motor using the three point starter and thereby adjust the speed

to its rated speed using field rheostat.5. Note down the readings of Voltmeter & Ammeters in Table under No Load

condition.

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6. Apply the Load on the drum gradually in steps by tightening the belt around it.

At each step, note down the readings of the Ammeters, Voltmeter, two Springbalances and the Tachometer.

7. Pour water in the pulley and cool it often when the motor is loaded.8.When the full load is reached, slowly reduce the load and switch off the

Motor from 220V D.C. Supply by opening the DPST Switch

OBSERVATION TABLE:

S.No. Input

Voltage

(V)

(Volts)

Armature

Current

(Ia)(Amps)

Field

current

(Ish)(Amps)

Spring Balances

Speed (N)

(rpm)S1(kgf) S2(kgf)

CALCULATION TABLE:

Radius of the Brake Drum, R = ______ mts.

S.No.

Input

Voltage

(V)

(Volts)

Input

Current

(IL)(Amps)

Torque

(T)

(Nm)

Output

Power

(Watts)

Input

Power

(Watts)

%

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PRECAUTIONS:

1. The field rheostat of the motor must be kept in minimum before switching ON themotor.

2. Ensure that the starter arm is at extreme left position.3. Avoid loose connections4. Note down the readings from the meters without any parallax error5. Tachometer should be kept horizontal to the shaft while measuring the speed.6. Before switch OFF the motor make sure that there is nos load connected to motor.

MODEL GRAPHS:

RESULTS:

At full load:

i) Torque = __________ Nm.ii) Speed = __________ rpmiii) Armature Current = __________ Aiv) Efficiency = __________ %.v) Speed Regulation = __________%

CONCLUSIONS:


1) What is Speed regulation?2) What are the different types of motor?3) What are the characteristics of D.C shunt motor?4) What is the condition for maximum efficiency?5) What are the different methods to reduce the iron losses?6) What are the application of D.C Shunt Motor?

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BRAKE TEST ON A

3 SQUIRREL CAGE

INDUCTION MOTOR

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Experiment no: 6

BRAKE TEST ON A THREE PHASE

SQUIRREL CAGE INDUCTION MOTOR

AIM : To obtain the Performance characteristics curves of a 3 Squirrel cageInduction motor by conducting brake test on it.

NAME PLATE DETAILS:

S.No TypeSquirrel cage

Induction Motor

01 Rating 3.0 HP / 2.2KW

02 Volts. 3 ,415 A.C.

03 Current 4.7 A

04 Connection Delta

07 Ins. Class B

08 Speed 1400 rpm

APPARATUS:



02 Ammeter (0-10) A M.C. 1

03 Wattmeter 600 V ,10A UPF 2



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THEORY:

It is a simple method of testing 3 Squirrel Cage Induction motor and consists ofapplying a brake to a water-cooled drum mounted on the motor shaft.

The five important characteristics curves of a 3 Squirrel cage Induction motorMotor, namely, Speed, Slip, Armature Current, Efficiency & Power factor, eachplotted against the useful Power, as shown in the model graph are known asPerformance characteristics.

Squirrel cage Induction Motor of low rating can be started by Direct-on-lineStarting method or Auto Transformer Method of Starting

A belt is wound round the brake drum and its two ends are attached to two springbalances S1 & S2. The tension of the belt can be adjusted with the help of swivels.

Shaft torque, T developed by the motor is 9.81 (S1 S2) R Nmwhere, R is the radius of the pulley in meters & N is the speed in rpm

Useful Output Power = (2 N T) / 60 Watts Input Power = W1 +W2 Watts % Efficiency , = (Output power / Input power) x 100. % Slip = [Ns-N/Ns] 100 ,

where, Synchronous Speed of Rotating flux is Ns=120 f / P & N is Rotor Speed

Power Factor, Cos = W1+W23VI

The size of the motor that can be tested by this method is limited from theconsideration of the heat that can be dissipated at the brake drum

PROCEDURE:

1. Connect the circuit as per the Circuit diagram.2. Close the TPST switch to 3AC Supply and apply the voltage gradually to the Stator

of the Induction Motor by means of the Variac.3. At no load, note down the readings of all Meters( Ammeter, Voltmeter, Wattmeters &Tachometer) & Spring balances.

4. Gradually apply the load & for various values of current up to rated current, notedown all Meter readings & Spring balance readings.

5. Now release the load gradually and reduce the applied Voltage to zero using Variac.6. Disconnect the Variac by opening the TPST Switch

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OBSERVATION TABLE:

S.No. Input

Voltage

(V)

(Volts)

Armature

Current

(IL)(Amps)

Wattmeters Spring Balances

Speed (N)

(rpm)W1

(Watts)

W2

(Watts)

S1

(kgf)

S2

(kgf)123....

10

CALCULATION TABLE:

Radius of the Brake Drum, R = ______ mts.

S.No

.

Armature

Current

(IL)(Amps)

Torque

(T)

(Nm)

Output

Power

(Watts)

Input

Power

(Watts)% % Slip Cos

123....

10

PRECAUTIONS:

1. Avoid loose connections2. Initially position of the variable on Variac must be in minimum position.3. Note down the readings from the meters without any parallax error4. Tachometer should be kept horizontal to the shaft while measuring the speed.5. Before switch OFF the motor make sure that there is no load connected to motor.

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MODEL GRAPHS:

RESULTS:

At full load:

i) Speed = __________ rpmii) Slip = __________ %.iii) Armature Current = __________ Aiv) Efficiency = __________ %.v) Power factor = __________

CONCLUSIONS:


1. What are the different types of 3 I.M?2. Explain the Performance Characteristics of 3 I.M.3. Explain the Slip-Torque Characteristics of 3 I.M4. Explain different methods of starting 3 Squirrel cage I.M?

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OC & SC TESTS ON A

1 TRANSFORMER

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Experiment no: 7

OPEN CIRCUIT & SHORT CIRCUIT TESTS

ON A SINGLE PHASE TRANSFORMER

AIM : To predetermine the efficiency, percentage regulation and equivalentcircuit parameters of a given single phase transformer by conductingOpen circuit and Short circuit tests on it.

NAME PLATE DETAILS:

S.No Type Transformer01 Rating 2 KVA, 1

02 L V winding 230 V

03 H V winding 440 V

APPARATUS:


01 Voltmeter (0-230) V M.I 1


03 Ammeter (0-1) A M.I. 1

04 Ammeter (0-5) A M.I. 1

05 Wattmeter 300 V ,10A LPF 1

06 Wattmeter 150 V ,10A UPF 1

07 Variac230V / 0-270V

15AContact 1


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THEORY:

The performance of a transformer can be calculated on the basis of the equivalentcircuit which contains four main parameters, the equivalent resistance R01 as referredto primary (or secondary R02), the equivalent leakage reactance X01 referred toprimary (or secondary X02), the core loss conductance G0 or resistance R0 andmagnetizing reactance X0. These parameters can be easily determined by performingOpen circuit test & Short circuit tests. The losses obtained are used in calculating theefficiency & regulation of the transformer. These tests are very economical andconvenient because they furnish the required information with out actually loadingthe transformer.

OPEN CIRCUIT TEST:

The purpose of this test is to determine no load loss or core loss and no loadcurrent I0 which is helpful in finding R0 and X0.

In this test , the transformer secondary winding (usually HV side) is left open andthe primary winding (LV side) is connected to supply of normal voltage andfrequency.

With normal voltage applied to the primary a small current I0 flows in theprimary which produces magnetic flux in the core. The transformer draws reactivepower from the supply to establish the magnetic flux, active power is alsoabsorbed by the transformer to overcome the core loss due to hysteresis and eddycurrent.

The reactive power at no load is much higher than the active power so the powerfactor is small. The primary no load current I0 is small (usually 2 to 10% of therated load current),therefore copper loss is negligibly small in primary and nil insecondary (it being open).

These core loss under no load condition is same for all loads as the net fluxpassing through the core is approximately the same as at no load.

SHORT CIRCUIT TEST:

This is an economical method to determine equivalent impedance (Z01 or Z02),leakage reactance(X01 or X02), total resistance (R01 or R02) of the transformer andcopper loss at full load (or at any desired load).

In this test the secondary side (usually low voltage winding) is solidly short-circuited. Now the primary is connected to a power source of reduced potential

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(usually 5 to10% of normal voltage) that is enough to cause the rated current toflow in both primary and secondary windings.

Since in this test the applied voltage is a small percentage of normal voltage, themutual flux produced is so small percentage of normal value. Hence the corelosses are very small and the losses now taking place will be full load copperloss.

PROCEDURE:

OC TEST:

1. Connect the circuit as per the circuit diagram.2. The HV side of the transformer is kept open & the normal voltage of 230V is applied

to LV side by adjusting the autotransformer ( variac)3. Note down the readings of ammeter, voltmeter and wattmeter on LV side.

SC TEST:

1. Connect the circuit as per the circuit diagram.2. The LV side of the transformer is short circuited & using the autotransformer adjust

the HV side voltage such that the ammeter reads the full load current of HV winding3. Note down the readings of ammeter, voltmeter and wattmeter on HV side.

OBSERVATION TABLE:

OC TEST:S. No. V0(volts) I0 (amps) W0(watts)

1.

SC TEST:

S. No. Vsc

(volts) Isc

(Amps) Wsc

(watts)

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CALCULATION TABLE:

Efficiency characteristics

S.No Load current

( % of full load )

Core

losses

(W)

Copper

losses

(W)

Total

losses

(W)

Input

(W)Output

(W)

Efficiency

1

2

.

.

10

10

20

.

.

100

Voltage Regulation characteristics

S.No Cos SinLagging p.f Leading p.f

full load half load full load half load

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MODEL CALCULATIONS :

O.C. TEST:

WO = VO IO Cos O c; Cos O = ( WO ) / ( VO IO ) =

I = IO Sin O ; I = IO Cos O

RO = VO /I ; XO = VO /I

S.C TEST :

WSC = ISC2 Re2 ; Re2 = WSC / ISC

2

Ze2 = Vsc / Isc ; Xe2 = ( Ze22 - Re2

2 )

% Efficiency ( ) = [ ( x KVA Cos) / ( x KVA Cos + Wi + x2 W cu )] x 100

%Voltage Regulation = x 100

PRECAUTIONS:

1. Avoid loose connections2. Initially position of the variable on Variac must be in minimum position.3. Note down the readings from the meters without any parallax error4. Apply the reduced voltage (only 5 to 10% of rated voltage) slowly to HV side

during SC test

MODEL GRAPHS:

2 2 2 2

2

I Re Cos I Xe Sin

V

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RESULTS:

The parameters of transformer are:

R0 = ____ ; X0 = ____ ; Re2 = ____ ; Xe2 =____

From the graphs plotted, it has been determined that

(i) Maximum efficiency of ____ % occurs at a load current of ____A(ii) Maximum regulation occurs at ____ Pf.(iii) Zero regulation occurs at _____ pf.

CONCLUSIONS:


1. What is the principle of operation of transformer?2. What are the different types of 1 transformers available?3. Why Copper losses in Open circuit test & Core losses in Short circuit4. are considered negligible?5. What is the advantage of deriving the Equivalent circuit?6. What is the condition for maximum efficiency?7. What are the conditions of power factors for maximum & minimum regulation?

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MAGNETIZATION

CHARACTERISTICS OF A

D.C. SHUNT GENERATOR

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Experiment no: 8

MAGNETIZATION CHARACTERISTICS OF

A D.C. SHUNT GENERATOR

AIM : To obtain the Magnetization Characteristics of aD.C. ShuntGenerator and to determine its Critical field resistance& Critical speed.

NAME PLATE DETAILS:

Type DC Shunt Motor DC Shunt Generator

Ratings 3.0 HP 2 KW

Volts. 220 V DC 220 V DC

Current 12 A 12A

Exc. Volts. 220 V DC 220 V DC

Exc. Current 0.6A 0.7A

Duty S1 S1

Ins. Class B B

Speed 1500 rpm 1500 rpm

APPARATUS:





WOUND1


WOUND2

05 Tachometer 0-10,000 RPM ANALOG 1

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THEORY:

I) Magnetization Characteristics

The magnetization characteristics shows the relation between the no loadgenerated emf in armature, E0 and the field (or) exciting current, If at a givenfixed speed as shown in model graph.

These characteristics are also known as the No load saturation characteristics orOpen circuit characteristics. The shape of these characteristics is practically samefor all generators whether separately excited or self excited

Due to the residual magnetism in the poles, some emf is generated even whenIf= 0 represented by OD**. Hence, the curve starts a little way up.

The slight curvature, DE** at the lower end is due to magnetic inertia. It is seenthat the first part of the curve, EC** is practically straight. This is due to the fact

that at low flux densities, reluctance of iron path being negligible (due to highpermeability), total reluctance is given by the air gap reluctance, which isconstant. Hence, the flux and consequentially the generated emf are directlyproportional to the exciting current.

How ever at high flux densities, where is small, iron path reluctance becomesappreciable and straight relation, CF** between Eo and If no longer holds good,i.e., saturation of poles start.

(** refers to the model graph)

II) Critical resistance

It is that maximum value of the field resistance, above which the machine fails toexcite i.e. there will be no build up of the voltage.

This resistance corresponds to the straight-line position of the magnetizationcharacteristic because the magnetic circuit does not offer any appreciablereluctance to the magnetic flux.

III) Critical speed

It is that speed for which the given shunt field resistance will represent criticalfield resistance

(OR)

It is that minimum value of the speed of the machine below which the machinefails to excite .

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PROCEDURE:

1. Connect the circuit as per the circuit diagram.2. Initially the starter must be in OFF & SPST Switch in open positions.3. Switch on the D.C. Motor to 220V D.C. Supply by closing the DPST Switch.4. Start the D.C. motor using the three point starter and thereby adjust the speed

of it to the rated speed of the D.C.generator using field method of speed control.5. Note down the voltage of the voltmeter which represents the residual voltage of

the generator when SPST switch is in open condition.6. Excite the field winding D.C.generator in steps by decreasing its external

resistance gradually and note down various corresponding readings of ammeterand voltmeter till 1.1 to 1.25 times the rated voltage of the generator is reached,maintaining constant speed .

7. Gradually reduce the field current of generator and make it to zero finally byopening SPST switch. and disconnect the D.C. Motor from the 220V D.C.Supply

.OBSERVATION TABLE:

At constant speed of 1500r.p.m.

S.No. Field current( If) Amps Armature Voltage ( Eo ) Volts

MODEL GRAPHS:

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CALCULATIONS:

TO FIND CRITICAL FIELD RESISTANCE:

1. Plot the magnetization curve.2. Draw the tangent such that it touches most of the linear part of the curve.

This line is the Critical field resistance line.

3. The slope of the above line gives the Critical field resistance.TO FIND CRITICAL SPEED:

1. Draw the constant field resistance line Rf .2. From point A draw a line on to the Critical field resistance line.

Now the Critical speed, Nc= (AB /AC) N, where N is the rated speed ofD.C. generator i.e., 1500 r.p.m.

PRECAUTIONS:-

5. The field rheostat of the motor must be kept in minimum & for the generator inmaximum positions before switching on the D.C. supply.

6. Ensure that the starter arm is at extreme left position.7. Avoid loose connections8. Note down the readings form the meters without any parallax error

RESULT:

Critical field resistance = ________ ohms. Critical speed = ________ r.p.m.

CONCLUSIONS:


1. What are Magnetization Characteristics?2. What do you mean by Critical field resistance?3.

What do you mean by Critical speed?4. How do you obtain the O.C.C at any other speed other than rated speed?

5. What are the different types of Generators?6. What are the applications of D.C Shunt Generators?

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REGULATION OF

3 ALTERNATOR

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Experiment no: 9

REGULATION OF 3ALTERNATOR

AIM : To predict the regulation of 3 alternator usingSynchronous impedance method

NAME PLATE DETAILS:

Type DC Shunt Motor Alternator

Ratings 3.0 HP 3 KVA

Volts. 220 V DC 3 ,415V AC

Current 12 A 4.2 AExc. Volts. 220 V DC 220 V DC

Exc. Current 0.6A 1.4A

Duty S1 S1

Ins. Class B B

Speed 1500 rpm 1500 rpm

APPARATUS:



02 Ammeter (0-10) A M.I. 1

03 Rheostat 300, 2 A Wire Wound 1


05 Switch -- Triple pole 1

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THEORY:

Voltage regulation is defined as the rise in terminal voltage, when the full load ata particular power factor is removed keeping the field excitation and speed constantdivided by the rated terminal voltage.

% regulation = (EO-V) / V X100

For small rating machines regulation can be found by direct load where as in thecase of large machines the cost of finding regulation by direct loading becomesexpensive. Hence an indirect method is used for obtaining the regulation of an alternatoris Synchronous impedance or EMF method

In order to calculate the regulation by this method, it requires

a) Armature or stator resistance Ra.b) Open circuit or No-load characteristics.c) Short circuit characteristics.

OD = Eo

Eo = (OB2 + BD2)1/2

OB = OA + AB= V Cos + IRa

BD = BC + CD= V Sin + IXs

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For lag pf, Eo = [ (V Cos + IRa)

2 + (V Sin + IXs)2 ]1/2

For lead pf, Eo = [ (V Cos + IRa)

2 + (V Sin IaXs)2 ]1/2

For UPF, Eo = [ (V + IaRa)2 + (IaXs)2 ]1/2

whereV= Rated voltage, Cos - power factor

% Regulation = (Eo- V) / V X 100

PROCEDURE:OC test:

1) Connect the circuit as per the circuit diagram.2) Initially the starter must be in OFF & SPST Switch in open positions.3) Switch on the D.C. Motor to 220V D.C. Supply by closing the DPST Switch.4) Start the D.C. motor using the three point starter and thereby adjust the speed

of it to the rated speed of the Alternator using field method of speed control.5) With the TPST Switch open , Switch ON the Excitation unit of Alternator & by

varying the field current tabulate the corresponding open circuit voltage readings.

SC test:

6) Conduct the Short circuit test by closing the TPST switch and adjust thealternator field current for which the armature current corresponds to its ratedvalue & tabulate it.

7) Gradually reduce the field current of alternator and disconnect the D.C. Motorfrom the 220V D.C. Supply.

OBSERVATION TABLE:-

OPEN CIRCUIT TEST SHORT CIRCUIT TEST

Ra =____

S. No. VOC

(volts)If (Amps)

S.NoIsc (amps) If(amps)

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MODEL GRAPHS:

CALCULATIONS:

Z = [Open cicuit Voltage per Phase] / [Short circuit current],at constant field current

where Short circuit current can be taken as rated armature current, Ia

Z = Eo per phase / Ia , at constant If

Xs = (Z2 Ra2 )

CALCULATION TABLE:

Voltage Regulation characteristics

S.No Cos SinLagging p.f Leading p.f

full load half load full load half load

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PRECAUTIONS:-

1. The field rheostat of the motor must be kept in minimum position beforeswitching on the D.C. supply.

2. Ensure that the starter arm is at extreme left position.3. Avoid loose connections4. Note down the readings form the meters without any parallax error5. Speed must be maintained constant through out the experiment.6. Make ensure that the terminal voltage is reduced to zero before doing short

circuit test .

RESULTS:

Voltage regulation at full load 0.8 pf. lag is _____& 0.8 pf. lead is ______ Maximum positive voltage regulation occurs at pf of _______ Zero voltage regulation occurs at pf of _______

CONCLUSIONS:


1. What is the principle of operation of an alternator?2. What are the different types of alternators available?3. What are the different factors affecting the Voltage regulation?4. Explain the variations of terminal voltage under different power5. factor conditions.6. Why the synchronous impedance method is termed as Pessimistic method

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Experiment no: 10

DETERMINATION OF TWO PORT

NETWORK PARAMETERS

AIM : To determine Z & Y parameters of a given two port Network.

APPARATUS:

S.No Specification Range Type Quantity

1 Voltmeter (0-300)V PMMC 2

2 Ammeter (0-5)A PMMC 2

3Rheostat (50 , 5A) Wire Wound 3

4 Switches ------ DPDT 2

5 Fuses 5ATin Coated

Copper2

6 Connecting Wires 1 Square mmInsulatedcopper

As perRequirement

THEORY:

A network containing two pairs of terminals is called as two port network.

Normally one pair of terminals coming together to supply power or to withdraw power orto measure the parameters, are called asport. To achieve simplicity, the whole network isshown with a single block.

A typical two port network is as shown below in fig (a)

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OPEN CIRCUIT IMPEDANCE PARAMETERS (Z-parameters):

Z-parameters can be defined by the following equations

V1 = Z11I1 + Z12I2 (1)

V2 = Z21I1 + Z22I2 (2)

Matrix form :

If port 2-21 is open circuited, i.e. I2 = 0 then

Z11 = V1/I1 & Z21 = V2/I1

If port 1-11 is open circuited, i.e. I1 = 0, then

Z12 = V1/I2 & Z22 = V2/I2.

Here,

Z11 is the driving point impedance at port 1-11 with 2-21 open circuited. It can also be

called as open circuit input impedance.

Z21 is the transfer impedance at port 1-11 with 2-21 open circuited. It can also be

called as open circuit forward transfer impedance.

Z12 is the transfer impedance at port 2-21 with 1-11 open circuited. It can also be

called as open circuit reverse transfer impedance and

Z22 is the driving point impedance at port 2-21

with 1-11

open circuited. It can alsobe called as open circuit output impedance.

Z-parameter representation for a two port network, shown above, will be asshown below in fig (b)

I1

I2

Z11V1

V2=

Z21 Z22

Z12

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If the network is

a) Reciprocal then V1/I2 (where I1 = 0) = V2/I1 (where I2 = 0) i.e. Z12 = Z21

b) Symmetrical then V1/I1 (where I2 = 0) = V2/I2 (where I1 = 0) i.e. Z11 = Z22

SHORT CIRCUIT ADMITTANCE PARAMETERS (Y-parameters):

Y-parameters can be defined by the following equations

I1 = Y11V1 + Y12V2 ________________ (1)

I2 = Y21V1 + Y22V2 ________________ (2)

In matrix form

Y11

=Y21 Y22

Y12 V1

V2

I1

I2

1

21

I2I1

Z11

+

_

+

_

Z22

Z12I2 Z21I1

2

11

V2V1

Fig (b) Open circuit impedance parametric representation of a two port net work.

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If port 2-21 is short circuited, i.e V2 = 0 then

Y11 = I1/V1 & Y21 = I2/V1

If port 1-11 is short circuited, i.e V1 = 0 then

Y12 = I1/V2 & Y22 = I2/V2

Here, Y11 is the short circuit driving point admittance at port 1-11 with 2-21

short circuited. It will also be called as short circuit input admittance.

Y21 is the Transfer admittance at port 1-11 with 2-21 short circuited. It will

also be called as short circuit forward transfer admittance.

Y12 is the Transfer admittance at port 2-21 with 1-11 short circuited. It will

also be called as short circuit reverse transfer admittance and

Y22 is the driving point admittance at port 2-21 with 1-11 short circuited. It

can also be called as short circuit output admittance.

Y-parameter representation for a two port network, shown above, will be asshown below

If the network is

a) Reciprocal then I2/V1 (where V2 = 0) = I1/V2 (where V1 = 0) i.e. Y21 = Y12

b) Symmetrical then I1/ V1 (where V2 = 0) = I2/ V2 (where V1 = 0) i.e. Y11 = Y22

1

1

I2I1

Y11 Y22Y12V2 Y21V1

2

11

V2V1

Fig(c) Short circuit admittance parameter representation of a two port net work.

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PROCEDURE:-

1. Connect the circuit as per circuit diagram.2. With the Switches S2 open , S3 close to 11' and S4 open , note down the

corresponding readings of voltmeter and ammeter in S.No 1 in Tabular form afterclosing the Switch S1 to supply mains

3. With the Switches S1 open ,S4 close to 33' and S3 open , note down the

corresponding readings of voltmeter and ammeter in S.No 2 in Tabular afterclosing the Switch S2 to supply mains

4. With the Switches S2 open ,S3 close to 11' and S4 close to 44' , note down thecorresponding readings of voltmeter and ammeter in S.No 3 in Tabular afterclosing the Switch S1 to supply mains

5. With the Switches S1 open ,S3 close to 22' and S4 close to 33 ' , note down thecorresponding readings of voltmeter and ammeter in S.No 4 in Tabular afterclosing the Switch S2 to supply mains

OBSERVATION TABLE:-

S.NO Test Condition V1 (V) I1 (A) V2 (V) I2 (A) Parmeters

1Port 2 Open(I2 = 0) and

Port-1 Active

Z11 = V1/I1=

Z21 = V2/I1=

2Port 1 Open(I1=0) and

Port-2 Active

Z12 = V1/I2=

Z22 = V2/I2 =

3Port 2 Short (4 - 4)

(V2=0) andport-1 active

Y11 = I1/V1=

Y21 = I2/V1=

4Port 1 Short (2 - 2)

(V1=0) andPort-2 active

Y22 = I2/V2=

Y12 = I1/V2 =

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PRECAUTIONS:

1. Note down the readings of voltmeter and ammeter without parallax error.2. The current through a particular element should be maintained below its current

rating.3. The conditions of switches should be thoroughly checked before making the

circuit live

RESULTS:

The values of Z parameters are

Z11 = ________; Z12 = ________; Z21 = ________ ; Z22 = ________

The values of Y parameters are

Y11 = ________; Y12 = ________; Y21 = ________; Y22 = ________

CONCLUSIONS :

VIVA QUESTIONS:

1) What is the significance of the two port parameters?2) How you know the admittance parameters from impedance parameters?3) What are the application of Z& Y parameters?4) What is the condition for reciprocal network?5) Wh t i th diti f t i l t k?

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