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- 1. 1 | P a g e Basic electrical labarotary experiment
- 2. 2 | P a g e 3 phase POWER MEASURMENT BY 2 WATTMETER METHOD
- 3. 3 | P a g e AIM OF THE EXPERIMENT: To MeasureThree Phase Power By 2 Watt Meter Method. And calculate the percentage error. APPARATUS REQUIRED: THEORY: A watt meter is an instrument for measuring power directly in a circuit. It has two coils which when connected in series give the lower current range of the instrument and when connected in parallel gives the double the range. The pressure circuit has a coil of high resistant. The current coil is connected in series with the circuit in which power is measured and pressure coil across the circuit. Two wattmeter’scan be used to measure power in a three phase 3-wire circuit, by making the connections as shown in below. The load may be balanced or unbalanced. The current coils are connected in series with two phases and the pressure coils between either phase and the third. If one of the wattmeters tends to read negative, the current coil is reversed, but reading of this instrument must be regarded as negative the total power consumed= w1+w2(alzebric sum). Here current through the current coil W1=IR and W2=IB While potential difference across voltage coil of W1=VRY and W2=VBY According to the phasor diagram given below W1=VRYIRcos(30+ ) and W2=VBYIB cos(30- ) Here total power drawn by 3 phase load is W1+W2=VRYIRcos(30+ ) + VBYIB cos(30- ) Here VRY=VBY=VL Then W1+W2= 3 VL I cos (after solving) which is 3 phase power And W2-W1=VLI sin Now tan = Thus pf angle = SLNO NAME OF ITEM SPECIFICATION QUANTITY 1. Watt meter 5A/600V 2 2. Volt meter MI(0-300V) 1 3. Ammeter MI(0-5A) 3 4. Rheostat 50Ω/5A 3 5. 3 phase variac 0-100V 1
- 4. 4 | P a g e CIRCUIT DIAGRAM: Precautions: 1. Don’t switch on power supply without concerning respected teachers. 2. 3 Auto transformer must be kept at minimum potential point before starting. 3. Resistant value of all rheostat should be kept at maximum postion. PROCEDURE: 1. Connect as shown in fig. 2. Keep the rheostat resistance in maximum and slowely increase the output voltage of the variac so that current in each line is about 4A or slightly less. 3. Then vary the resistant of the rehoastats so the load is deliberately unbalanced ,i.e the current in each line becomes different. Corresponding phase voltages across the rehoastat are VRN,VYN ,VBNrespectively. 4. Take reading of W1,W2 ,IR, IB,IY and VRN,VYN,VBN. 5. Decrease the output of the variacabd repeat the above procedure again.
- 5. 5 | P a g e OBSERVATION: SLNO W1 W2 IR IY IB VRN VYN VBN 1 2 ………. 9 10 VERIFICATION: SLNO W1+W2=P1 VRN×IR VYN×IY VBN×IB ∑VI=P2 1 2 ………… 9 10 CONCLUSION: To be written by student. DISCUSSION: 1. Compare W1+W2& ∑VI , comment on the discrepancy if any. 2. Explain how will find the multiplying factor of the watt meter. 3. Verify theoritrically how the two wattmeter method gives the power measurement under both balanced and unbalanced condition. 4. Discuss whether this method is suitable for three star and delta system and also four – wire unbalanced load. REFERENCES: 1. Refer CIRCUIT THEORY BY A. CHAKRABARTI (fifth edition) page-284.
- 6. 6 | P a g e 1 phase Power measurement by 3 AMMETER AND 3 VOLTMETER METHOD
- 7. 7 | P a g e AIM OF THE EXPERIMENT: To measure 1 phase power by 3-ammeter and 3-voltmeter method. APPARATUS REQUIRED: CIRCUIT DIAGRAM: SLNO NAME OF ITEM SPECIFICATION QUANTITY 1. Volt meter MI(0-300V) 3 2. Ammeter MI(0-10A) 3 3. Rheostat 27Ω/10A 1 4. Inductor 1
- 8. 8 | P a g e THEORY: The circuit to be used for measurement of power in an A.C. circuit using three ammeter is shown in fig 1. We know in a D.C. circuit the power is given by the product of voltage and current, whereas ,in A.C. circuit it is given by the product of voltage, current and power factor. For this reason, it is not possible to find power in an A.C. circuit simply from the readings of a voltmeter and ammeter. In A.C. circuits power is normally measured by wattmeter. However, this method demonstrates that the power in a single –phase A.C. circuits can be measured by using three ammeters and 3 voltmeter method. 3 AMMETER METHOD: FIG-1 is used for measurement of power by 3 ammeter method here from the circuit it concludes that I1 is the summation of I2& I3vectorically. Here in this circuit a fixed resister is used which value is known and I2 is the current in that path. And I3 is the current flowing through load. So the phasor diagram of above circuit will be as shown below Here Then I2 2 +I3 2 +2I2I3cos =I1 2 Power factore=Cos = Power=VI COS = POWER DRAWN BY LOAD= Similarly in case of 3 Voltmeter method POWER = VI COS = POWER= ) Precautions: 1. Don’t switch on power supply without concerning respected teachers. 2. Auto transformer must be kept at minimum potential point. 3. Input voltage in case of Run-1 and Run-2 is always be constant through out the experimental reading.
- 9. 9 | P a g e PROCEDURE: RUN-1 : Measurement of power by three-ammeter method: 1. Connect as per the circuit diagram fig-1. 2. Give supply voltage 60V for measurement. If this voltage will more than 60V then the inductor will make sound. 3. Vary the load impendence insteps and note down the reading of the meter in each case. Here in each case I2 and V is constant in all over the experiment. RUN-2: measurement of power by three-voltmeter method: 1. Connect as per the circuit diagram(fig-2) 2. Give supply voltage 120V for measurement. If this voltage will more than 120V then the inductor will make sound. 3. Vary the load impendence (i.e inductor) in steps and note down the reading of the meters in each case.Here V1 is constant in all over the experiment. OBSERVATION AND CALCULATION: RUN-1: SLNO I1 in Amp I2 in Amp I3 in Amp V in Volt POWER P= POWERFACTOR COS = 1 2 4 5 6 7 8 9 10 RUN-2: SLNO V1 in Volt V2 in Volt V3inVolt I in Amp POWER P= POWERFACTOR COS = 1 2 3 4 5 6 7 8 9 10
- 10. 10 | P a g e CONCLUSION: To be written by student. DISCUSSIONS: 1. If the load is capacitive in nature, whether there will be any change in the expression of power factor in both case.
- 11. 11 | P a g e TESTING OF 1 ENERGY METER AT (0.5,0.866 AND 1.0 PF)
- 12. 12 | P a g e AIM OF THE EXPERIMENT: Testing of single 1 energy meter in different power factor 0.5 ,0.866 and unity power factor. APPARATUS REQUIRED: SLNO NAME OF APPARATUS RANGE AND TYPE QUANTITY 1 Voltmeter 0-300(MI) 3 2 Ammeter 0-5/10A(MI) 1 3 1 variac 0-300V(MI) 2 4 1 energy meter 1500 RPK 1 5 Voltmeter 0-600V(MI) 1 6 Tipple Pole Iron Clad Switch 1 THEORY: 1. ENERGY METER: it is used to calculate energy (kilo watt hour) at each instant of time so this type of instrument is indicating type instrument. 2. WHY TESTING: In all supply meters a meter constant is made in them. This constant is expressed in revolutions per kilowatt hour (RPK) usually. Full load current and line voltage for which the meter is intended, are also stated. From these data the number of revolutions per minute which the meter should make, when tested with a certain fraction of its full load can be calculated. The number of revoltions per minute which it actually does make when tested at this load is then observed and the error is calculated. 3. TESTING OF 1 ENERGY METER: for testing of 1∅ energy meter in different power factor is done by help of two phase that is 2400 displaced. Here R phase and B phase are 2400 displaced. In the arrangement given in circuit diagram pressure coil of energy meter is always taking voltage from R phase only and current coil of energy meter has the provision for taking current from R phase and B phase simultaneously or individually. When VRN =0 and VBN has some voltage then current coil of energy meter will take current from B phase. So power factor angle=600 as shown below phasor diagram. And Power factor=cos600 =0.5
- 13. 13 | P a g e When VBN has equal voltage withVRN then current coil of energy meter will take current from B phase and R phase equally. So power factor angle=300 as shown below phasor diagram. Power factor=cos300 =0.866 When VBN =0 and VRN has some voltage then current coil of energy meter will take current from R phase. So power factor angle=00 as shown below phasor diagram. And Power factor=cos00 =1
- 14. 14 | P a g e CIRCUIT DIAGRAM Precautions: 1. Do not switch on power supply without concerning respected teachers. 2. Auto transformer must be kept at minimum potential point. 3. See the revolution of energy meter if it rotates reversely then change the terminal of pressure coil. PROCEDURE: 1. Make connections according to circuit diagram. 2. For testing at loads at different power factor vary the auto transformer positions. Take reading of all voltmeters and ammeters. Calculate the power factor by drawing the vector diagrams.
- 15. 15 | P a g e OBSERVATION: SLNO COS V1 VRN VBN VPC I R.P.M R.P.H VPCICCOS = (P1) =P2 PERCENT AGE ERROR= 1 2 3 4 5 6 7 8 9 REPORT: 1. Find error for each setting of auto transformer (each power factor). 2. Plot a curve between error vs power factors and find error at p.f 0.5,0.866 and 1.0 CONCLUSION: To be written by student. DISCUSSION AND QUESTIONS: Discuss as the result of the experiment question 1. How do you vary the pf using two auto-transformers. 2. What is the basic difference in testing DC and AC energy meters. 3. How many terminals does an energy meter has. 4. Why energymeter is tested at different powerfactor. REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS: 1.www.nptel.iitm.ac.in 2. www.electronics-tutorials.ws/dccircuits 3 www.openbookproject.net 4.www.mhhe.com 5.www.opamp-electronics.com
- 16. 16 | P a g e VOLTAGE- CURRENT RELATIONSHIPS AND LOCUS DIAGRAM OF A SERIES R-L CIRCUIT
- 17. 17 | P a g e AIM OF THE EXPERIMENT: To represent the voltage and current relationship of a series R-L circuit and draw the locus diagram for varying resistance and fixed inductance. APPARATUS REQUIRED: SLNO NAME OF THE APPARATUS RANGE AND TYPE QUANTITY 1 AUTO TRANSFORMER 0-230V(AC) 1 2 VOLTMETER 0-300V MI TYPE 3 3 AMMETER 0-5/10A 1 4 LAMPLOAD 3KW 1 5 WATTMETER LPF 1 6 CHOKE 0.32H 1 CIRCUIT DIAGRAM: THEORY: In a series RL circuit, if the inductance l is kept constant and the resistance R be varied and if a phasor diagram be drawn with the applied voltage V1phasor as the reference(V1 is drawn along x axis) then the locus of the tips of the current phasor is given by the following equation. IX 2 + (IY+ )2 = )2 Where XL= inductive reactance in ohms(of the choke coil) Ix= in phase component of current. Iy= quadrature component of current This is a semi circle with as diameter.which is known as current locus. If VR and VL be the resistive and inductive drops respectively, the VR 2 +VL 2 =V1 2 So that the locus of the tip of the VRphasor is a semicircle with V1phasor as the diameter.which is known as voltage locus.
- 18. 18 | P a g e Precautions: 1. Don’t switch on power supply without concerning respected teachers. 2. Auto transformer must be kept at minimum potential point. 3. Primary voltage that is V1 should always be constant through out the experiment. PROCEDURE: Make connections as shown in the circuit diagram. Adjust the value of the supply voltage V1 to a definite value with the help of variac. Take readings of ammeter, voltmeter and wattmeters, change the current in the circuit by putting on more lamps and take several readings till all the lamps are on. The applied voltage V1 must be maintained constant for each reading. OBSERVATIONS: SLNO V1 I V2 V3 W LOAD 1 2 3 4 5 6 7 8 CALCULATIONS: Make calculations in tabular form shown below: RL= R= VR=I(R+RL) VRL= VL= XL= = 1 2 3 4 5 6 7 8 GRAPHS: 1. Draw the two theoretical semi-circles a. Voltage drops locus and b. For current locus i.e semicircles with V1 as diameter(V1 is along Y-axis) and another semicircle with phasor ( which is at 900 lagging w.r.t v1phasor) as diameter. 2. From the calculations made above draw the phasorsVR , VL. See if the tips of the voltage drop phasor and current phasor lie on the theoretical locus diagram.
- 19. 19 | P a g e CONCLUSION: To be written by student. DISCUSSION question: 1. Discuss why a low power factor watt meter was used by you. 2. What do you understand by the term power factor in reference to a.c. circuits ?. 3. What is the importance of power factor ? CURRENT LOCUS VOLTAGE LOCUS
- 20. 20 | P a g e OPEN CIRCUIT AND SHORT CIRCUIT TEST ON 1 TRANSFORMER
- 21. 21 | P a g e AIM OF THE EXPERIMENT: To perform the open circuit and short circuit test on a single phase transformer and to draw the equivalent circuit after determining its constants. APPARATUS REQUIRED: SLNO NAME OF THE EXPERIMENT RANGE AND TYPE QUANTITY 1 Single phase transformer 3KVA,230/230V,50 HZ 1 2 Wattmeter 2.5/5A LPF 1 3 Wattmeter 10/20A UPF 1 4 Ammeter 0-1A 1 5 Ammeter 0-20A 1 6 Voltmeter 0-300V 1 7 Voltmeter 0-30/75V 1 8 1 variac 0-300V,15A 1 THEORY: The performance of a transformer can be calculated on the basis of its equivalent circuit which contains four main parameters, the equivalent resistance R01 as referred to primary( or secondary R02), the equivalent leakage reactance X01 as referred to primary, the core-loss conductance G0 and the magnetizing susceptance B0. These constants or parameters can be easily determined by two test i.e. Open circuit test and short circuit test. These are very economical and convenient, because they furnish the required information without actually loading the transformer. In fact, the testing of very large a.c machinery consists of running two test similar to the open and short circuit test of a transformer. The purpose of this test is to determine no load loss or core loss and no load I0 which is helpful in finding X0 and R0. One winding of the transformer whichever is convenient but usually high voltage winding is left open and the other is connected to its supply of normal voltage and frequency. A wattmeter(W), Voltmeter (V) and ammeter (A) are connected in the low voltage winding i.e. primary winding in the present case. With normal voltage applied to the primary, normal flux will be setup in the core, hence normal iron losses will occur which are recorded by the wattmeter. As the primary no load current I0 is small, Cu loss is negligibly small in primary and nill in secondary. Hence, the wattmeter reading represents practically the core loss under no load condition. For short circuit test, one winding usually the low voltage winding, is solidly short- circuited by a thick conductor ( or through an ammeter which may serve the additional purpose of indicating rated load current). A low voltage ( usually 5 to 10% of normal primary voltage) at correct frequency (though for Cu losses it is not essential) is applied to the primary and is cautiously increased ill full- load current are flowing both in primary and secondary( as indicated by the respective ammeters). Since, in this test, the applied voltage is a small percentage of the normal voltage, the mutual flux ø produced is also a small percentage of its normal value. Hence, core losses are very small with the result that the wattmeter reading represents the full load Cu loss or i2 R loss for the whole transformer i.e. both primary Cu loss and secondary Cu loss. The equivalent circuit of the
- 22. 22 | P a g e transformer under short- circuit condition. If Vsc is the voltage required to circulate rated load currents, then Z01= Vsc/I1 a two winding transformer can be represented by means of an equivalent circuit as shown below Precautions: 1. Don’t switch on power supply without concerning respected teachers. 2. Auto transformer must be kept at minimum potential point. Before switch on the experiment. CIRCUIT DIAGRAM:
- 23. 23 | P a g e OPEN CIRCUIT TEST PROCEDURE: 1. Connect circuit as shown in the circuit diagram.open circuit the secondary and apply full load voltage to the primary through a variac. The cupper loss is negligible since there is only no load current is flowing.hence power consumed are the core losses of the core. 2. Note voltmeter and ammeter and wattmeter reading. Observation table: SLNO V I0 W IW= IM= COS CALCULATION: See the no load phasor diagram below W=VI0COS IW= , IM= , , SHORT CIRCUIT TEST PROCEDURE: 1. Connect as shown in the circuit diagram. Short circuit the secondary and apply a low voltage to the primary through a auto transformer. The iron losses are negligible since the flux will be very low on account of the primary and secondary. 2. Increase the voltage gradually till full load current flows in the primary. Note voltmeter and ammeter and wattmeter reading. Observation: SL.NO. V I Wc
- 24. 24 | P a g e Calculations: Let the total equivalent resistance of primary and secondary referred to primary side be R1 ohms and the total equivalent leakage reactance referred to primary side be X1 ohms. Wc = I2 R1 Hence R1 = Wc / I2 Also V /I = Z1 and X1 = ohms. Conclusion: 1. Now draw the equivalent circuit. 2. Plot a graph of copper loss versus load current (short circuit current). What is the shape of the curve? 3. Determine the regulation of the transformer at various loads for an assumed load power factor of 0.8 lagging. Regulation percent = X 100 Where Vo = secondary no load voltage. Vt = secondary full load voltage. 4. Plot a curve or regulation versus load current Discussion: 1. Why iron is chosen as the material for the core of the transformer? Why not we use aluminium? 2. What is normally the efficiency of a transformer to be?
- 25. 25 | P a g e LOAD TEST OF 1 TRANSFORM ER
- 26. 26 | P a g e AIM OF THE EXPERIMENT: To study load test on single phase transformer and determine efficiency and voltage regulation of a single phase transformer. APPARATUS REQUIRED: SLNO ITEM NAME RANGE/TYPE QUANTITY 1 1 Transformer 230/230,3KVA 1 2 1 Autotransformer 0-270V 1 3 Wattmeter 150/300/600V,10/20A dynamometer type,UPF 2 4 Ammeter 0-5/10/20 ,moving iron(MI) 2 5 Voltmeter 0-150/300/600V (MI) 2 6 lamp load 3 KW 1 CIRCUIT DIAGRAM: THEORY: When the secondary winding of a transformer is completed through a load and a voltage Vis applied to the primary winding. The transformer is said to be operate under load condition. Under this condition transformer cupper loss increases with increase in current .due to that efficiency changes. At half load efficiencies reaches to maximum it gradually decreases when further load increases. This is because at half load cu loss is less and iron loss is also less but below half load iron loss will more this result efficiency is less.This situation happens when transformer is distribution transformer.When load increases beyond half load cupper loss will increase which again reduces the efficiency.
- 27. 27 | P a g e Precautions: 3. Do not switch on supply without concerning respected teachers. 4. Auto transformer must be kept at minimum potential point. 5. Primary voltage that is V1 should always be constant throughout the experiment. PROCEDURE: 1. Make the connection as shown in the diagram keeping the auto-transformer in zero position and all switches and load is in off position. 2. Switch on the AC supply and then vary the voltage up to rated voltage of the transformer. 3. Now, start loading the transformer by putting on load switch on. So that a suitable current is obtained. 4. And then change the load and note down the reading of all the instrument. 5. Here the primary voltage of the transformer should always be in rated voltage irrespective of load. SAMPLE CALCULATION: Percentage Efficiency= =
- 28. 28 | P a g e Voltage regulation= Here V0=V1 And VL=V2 OBSERVATION TABLE: SLNO LOAD V1 I1 W1 V2 I2 W2 ×100 VOLTAGE REGULATION= CONCLUSION: To be written by student. DISCUSSION QUESTION:
- 29. 29 | P a g e THEVeNIN AND SUPERPOSITION THEOREM VERIFICATION EXPERIMENT
- 30. 30 | P a g e AIM OF THE EXPERIMENT: To verify thievenin and superposition theorem . APPARATUS REQUIRED: SLNO NAME OF ITEM SPECIFICATION QUANTITY 1. Rheostat 27Ω/10A 3 2. Rheostat 10Ω/8A 1 3. Volt meter MC(0-15V) 1 4. Ammeter MC(0-1A) 3 5. Dc source 0-20V 2 Vefification of thevnine theorem RL+ Rin THEORY OF THEVENIN THEOREM: In a passive ,linear,bilateralnetwork,the current in any element is given by the ratio of the open circuit voltage(obtained by open circuiting the element under consideration) to the sum of the resistance of the element and internal resistant,where the internal resistant is obtained by looking across to terminals of the element with the element removed and with all sources assumed to be dead. Then IL=Voc Where IL is the current in the element RL Voc=open circuit voltage across the element RL with RL removed. Rin=internal resistant across the terminal to which RL is connected with (RL removed and all sources made dead). This is thevenine’s theorem. In this experiment, we find out the current in a resistor element RL in the following network, using thevenine’s theorem. CIRCUIT DIAGRAM:
- 31. 31 | P a g e PROCEDURE: 1. Connect as in figure.A 2. When the D.P.D.T switch is closed on connecting terminals 1-1’ to 2-2, the ammeter A reads IL’ amperes. Which is the current in RL found experimentally. 3. Make use of the thevenine’s theorem to find IL as defined earlier. 4. Throw the D.P.D.T switch connecting terminals 1-1’ and 3-3’ in figure-A the voltmeter reads Voc volts, the required value. 5. To find the value of Rin and RL (defined earlier) connect as shown in figure-B a. The D.P.D.T switch is closed connecting 1-1 and 2-2 (figure-B) and the voltmeter respectively.V1/I1 gives the value of RL. b. Close the D.P.D.T switch connecting 1-1 and 3-3(fig-B) and the current I2 and voltage V2 are read on the ammeter and voltmeter respectively .V2/I2 gives the value of Rin. OBSERVATION: SLNO IL ’ VOC I1 V1 I2 V2 ERROR= RESULT: The percentage error is found to be__%. DISCUSSION: The % error is found to be in the range within 10%.The percentage error is due to observational errors, tolerance errors, calibration of instruments.Morover,it can be seen that thevenintheorm can’t be applied to network only containing dependent sources DISCUSSION QUESTION: 1. What are the advantages and disadvantages of using theveninTheorm? 2. Why Thevenintheorm not applied to non-linear circuis? 3. Can Thevenintheorm be applied to circuit having A.C sources?Ifyes,then what will be the difference. 4. How Thevnintheorm be applied to network containing both independent and dependent sources?
- 32. 32 | P a g e THEORY OF SUPERPOSITION THEOREM Superposition theorem A current in a particular branch due to many sources acting in a circuit can v=be obtaoined by adding the current due to the individual sources acting one at a time with the remaining sources dead. EXPLANATION Consider a simple circuit: To calculate current I1,I2 or I3 using superposition theorem, ignoring internal resistantces of the two sources. 1. First Vasources acting with Vb source dead. Corresponding circuit diagram is : Vb source is replaced by a zero resistance path between point C and O. The currents are I1a, I2a and I3a is calculated 2. Next VbactsVa is dead. Source Vahas been replaced by zero resistance path between the points A and O. The currents are I1b, I2b and I3b is calculated.
- 33. 33 | P a g e The currents are given by : I1=I1a+I1b I2=I2a+I2b I3=I3a+I3b PROCEDURE: 1. Connect the circuit as shown below in fig-1. 2. Keep the rheostats to give certain ohmic values R1,R2 and R3. 3. Note the readings of the three ammeters,connected to read I1,I2, and I3 in fig-1 4. Disconnect the voltage source VB,(VA sources acting) and connect the points A and O with a wire.Note the ammeter readings. 5. Disconnect the source VA,(VB source acting ) and connect the points A and O with a wire . Note the ammeter readings. 6. Add the currents obtained in step (4) and (5) and compare with the currents obtained in step(3) , to verify the theorem . 7. Vary R1 ,R2 and R3 suitably repeat step (3) to (6) take no. Of reading. OBSERVATION TABLE: SLNO SOURCE ACTING CURRENT IN AMPS I1 I2 I3 I1a I2a I3a I1b I2b I3b 1 VA 2 VB 3 VA AND VB DISCUSSION QUESTION: 1.What are the advantages and disadvantages of using superposition Theorm? 2.Why superposition theorm not applied to non-linear circuis? 3.Can superposition theorm be applied o circuit having A.C sources?Ifyes,then what will be the difference. 4.How superposition theorm be applied to network containing both independent and dependent sources? REFERENCES: Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 4. A Textbook of Electrical Technology by B.L Thereja. 5. Electrical Science by J. B. Gupta URLS: 1.www.brighthub.com 2.www.allaboutcircuits.com 3. www.howstuffworks.com 4. www.nptel.iitm.ac.in
- 34. 34 | P a g e Magnetization charterstic of separetly excited dc generator
- 35. 35 | P a g e AIM OF THE EXPERIMENT To find Magnetization characteristic of a separately excited DC generator. Apparatus required: SLNO NAME OF APPARATUS RANGE QUANTITY 1 moving coil type ammeter 0-3A 1 2 moving coil type voltmeter 0-300V 1 3 rheostats 0-300ohms.2Amp 2 4 Rheostat 50 Ohms,5Amp 1 5 Tachometer 0-10000RPM 1 6 A DC Generator Coupled to DC motor Set 7.5KW,1500 RPM,230V,31.2Amp 1 Theory: Magnetization curve is also called the open circuit characteristics (o.c.c) .It is the plot between no load or open circuit armature terminal voltage (the induce e.m.f) versus the field current at a constant speed. Hence the name o.c.c.It is also called magnetization characteristic because the induce emf α flux and flux density and the field current is α H and hence the curve is a B-H curve. PRECAUTION: 1. Do not switch on the power supply without concerning respected teacher. 2. Do not make rheostat position of armature to minimum and field rheostat position to maximum at the time of starting. Circuit Diagram
- 36. 36 | P a g e PROCEDURE: Make connection as shown in the diagram. Run the motor and bring it up to rated speed. With zero excitation in the field circuit, take the voltmeter reading. This represents the residual magnetism. Increase the excitation slowly and take reading of ammeter and voltmeter after adjusting the speed constant. While making the field current adjustments do not retrace. Continue till the induce voltage is more than the rated value. 2. Reduce the excitation and take reading of ammeter and voltmeter at constant speed –Do not retrace while reducing excitation. Run II: - Keep the excitation of the generator constant at a partial value and vary the speed. Observations:- Run I: - Speed constant SL NO. If increasing V If decreasing V Run II:- Excitation constant SL NO. N V GRAPHS: 1. Plot the curves of induced voltage versus field current for increasing and decreasing excitation. 2. Plot the curve of induce voltage vs. speed taking care to choose the scale such as point (0, 0) is included in the graph paper. DISCUSSION QUESTION: 1. The shape of the open circuit characteristic curve? What do you mean by saturation? 2. Can you obtain the o.c.c from the experimentally determined o.c.c for a different speed? If so, plot the o.c.c for a speed which is half the value at which you performed the experiment. 3. Why does the induced voltage Vs speed curve pass through (0, 0) but not the induce voltage Vs excitation curve? 4. Explain the effect of variation of the field resistance. 5. Discuss effect of variation of speed on build up voltage. 6. List the condition to be satisfied for voltage build up of a DC generator.
- 37. 37 | P a g e Speed control of DC Motor using armature voltage and field current method
- 38. 38 | P a g e AIM OF THE EXPERIMENT To observe Speed control of DC Motor using (A) Variation of armature circuit resistance. (B) Variation of field circuit resistance. Apparatus required: SLNO NAME OF APPARATUS RANGE QUANTITY 1 moving coil type ammeter 0-5A 1 2 moving coil type ammeter 0-3A 1 2 moving coil type voltmeter 0-300V 1 3 Rheostats 0-300ohms.2Amp 1 4 Rheostat 50 Ohms,5Amp 1 5 Tachometer 0-10000RPM 1 6 DC shunt Motor With Starter 7.5HP,1200RPM,220V,30Amp 1 7 S.P.S.T switch 1 Theory: If V is the applied voltage across the motor terminals, Ebis the back emfdeveloped,then V = Eb+ IaRa. Where Iaand Ra the current and resistance in the armature circuit respectively. But, Eb= = K .Hence V =K + IaRa .i.e. N = K ᶦ This shows that:- i. An increase in the IaRadrop will decrease the value of speed if V remains constant. ii. Speed varies inversely as the field flux and hence varies inversely as the exciting current, if below saturation. Thus, by increasing the resistance in the armature circuit, a motor can be operated at speed below normal. By increasing the resistance in the field circuit, a motor can be operated at speeds above normal. Circuit diagram :
- 39. 39 | P a g e Procedure:- CASE A: Speed control by variation of armature circuit resistance. i. Connect as shown. Make R and Rf zero at the time of starting the motor (which is run on no load) with the motor starter. ii. The motor is on load. Adjust Rf and bring the motor to the rated speed. Note the terminal voltage, V and normal excitation which correspond to the rated speed of motor. Keep Ifconstant throughout, During starting, ammeter A2 may be by passed. iii. With the exciting current kept constant, increase R in steps, at each step note the value of V and the speed in r.p.m. CASE B: - Speed control by variation of field circuit resistance. i. Run the motor as before and bring it to the rated speed at no load. ii.Cut out the field regulating resistance, Rfin steps and note the speed and the field current ,If in each case . Keep the voltage across the armature constant. Take reading for decreasing values of speed by increasing field current. iii. Take reading for increasing values of speed corresponding to same values of If as in 2. iv. Measure Ra after test. Observations:- Case A If = Ra= SL.NO. Speed in R.P.M Voltage across Arm. Case B: - Voltage across armature =............................................. SL.NO. If Speed in R.P.M with decreasing If Speed in R.P.M with increasing If Graphs: Case A:- plot a graph between speed along y-axis and volts across arm. Along x-axis. CaseB: - Plot a graph between speed along y –axis and exciting current If along x-axis. Discussion question: 1.compare the two methods and discuss the advantages and disadvantages of each method. 2. If, by chance, there be a loose contacts in the field circuits rheostat what will be the consequences .What precautions do you take to avoid these consequences. 3. Discuss the graph you obtained.

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