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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:
S. No Name of Apparatus Type Range Quantity
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:
S. No Name of Apparatus Type Range Quantity
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:
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 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:
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 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 =
TABLE 2 (For Circuit 2)
S.No Voltmeter (Volts) Ammeter (Amps) Rth = VS/IS()
1. VS = IS = Rth =
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TABLE 3 (For Circuit 3)
S.No Voltmeter (Volts) Ammeter (Amps)
1.Vth = IL =
TABLE 4 (For Circuit 4)
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
03 Ammeter (0-2) 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:
S.No Apparatus Required Rating Type Qty.
01 Voltmeter (0-300) V M.C 1
02 Ammeter (0-1) A M.C. 1
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:
VIVA VOCE QUESTIONS:
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:
S.No Apparatus Required Rating Type Qty.
01 Voltmeter (0-600) V M.C 1
02 Ammeter (0-10) A M.C. 1
03 Wattmeter 600 V ,10A UPF 2
05 Tachometer 0-10,000 RPM Analog 1
06 Fuse 10A T.C.C. 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:
VIVA VOCE QUESTIONS:
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:
S.No Apparatus Required Rating Type Qty.
01 Voltmeter (0-230) V M.I 1
02 Voltmeter (0-75) 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
06 Fuse 10A T.C.C. 2
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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:
VIVA VOCE QUESTIONS:
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:
S.No Apparatus Required Rating Type Qty.
01 Voltmeter (0-300) V M.C 1
02 Ammeter (0-1) A M.C. 1
03 Rheostat 270, 2 AWIRE
WOUND1
04 Rheostat 300, 2 AWIRE
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:
VIVA VOCE QUESTIONS:
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:
S.No Apparatus Required Rating Type Qty.
01 Voltmeter (0-600) V M.I 1
02 Ammeter (0-10) A M.I. 1
03 Rheostat 300, 2 A Wire Wound 1
04 Tachometer 0-10,000 RPM Analog 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:
VIVA VOCE QUESTIONS:
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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