Basic electrical labarotary experiment

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3 phase POWER

MEASURMENT BY 2

WATTMETER METHOD

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AIM OF THE EXPERIMENT:To Measure Three Phase Power By 2 Watt Meter Method. And calculate the percentage error.APPARATUS REQUIRED:SLNO NAME OF ITEM SPECIFICATION QUANTITY1. Watt meter 5A/600V 22. Volt meter MI(0-300V) 13. Ammeter MI(0-5A) 34. Rheostat 50Ω/5A 35. 3 phase variac 0-100V 1

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’s can 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+θ) and W2=VBYIB cos(30-θ)Here total power drawn by 3 phase load is W1+W2=VRYIRcos(30+θ) + VBYIB cos(30-θ) Here VRY=VBY=VL

Then W1+W2=√3 VL I cosθ (after solving) which is 3 phase powerAnd W2-W1=VLI sinθ

Now tanθ=√ 3W 2−W 1

W 2+W 1

Thus pf angle θ=tan−1√3W 2−W 1

W 2+W 1

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VRN

VBN

VYN

600

300

q

VBY

VRY

-VYN

1200

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 ,VBN respectively.

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.

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A

A

A

R

Y

B

W1

W2

V

M L

C V

M

C

L

V

(0-5)A

(0-5)A

(0-5)A

3 ø AUTO TRANSFORMER

(0-230)V

OBSERVATION:SLNO W1 W2 IR IY IB VRN VYN VBN

12……….910

VERIFICATION:SLNO W1+W2=P1 VRN×IR VYN×IY VBN×IB ∑VI=P2

PERCENTANGE ERROR=P2−P1

P2

×100

12…………910

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.

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1 phase Power

measurement by 3

AMMETER AND 3

VOLTMETER METHOD

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AIM OF THE EXPERIMENT:To measure 1 phase power by 3-ammeter and 3-voltmeter method.

APPARATUS REQUIRED:SLNO NAME OF ITEM SPECIFICATION QUANTITY1. Volt meter MI(0-300V) 32. Ammeter MI(0-10A) 33. Rheostat 27Ω/10A 14. Inductor 1

CIRCUIT DIAGRAM:

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A1

230VAC

SUPPLY

AUTO TRANSFORMER

A2

A3

230VAC

SUPPLY

AUTO TRANSFORMER

V2

V3V1

27W/10A

27W/10A

V1

(0-150)V

(0-10)A

(0-10)A(0-10)A

(0-150)V(0-150)V

(0-150)V

FIG-1

FIG-2

I1

I2

I3

q

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 belowHere

I2+I

3= I

1

Then I22+I3

2+2I2I3cosθ=I12

Power factore=Cosθ=I 1

2−I 22−I 3

2

2 I 2 I 3

Power=VI COSθ

=I 2R I 3×I 1

2−I 22−I3

2

2 I2 I3

POWER DRAWN BY LOAD= ( I 12−I 2

2−I 32 )×( R2 )

Similarly in case of 3 Voltmeter method

V 22+V 3

2+2V 2V 3COSθ=V 12

POWERFACTOR=COSθ=V 1

2−V 22−V 3

2

2V 2V 3

POWER = VI COSθ

=V 3×(V 2

R)×

V 12−V 2

2−V 32

2V 2V 3

POWER=¿)

Precautions:

1. Don’t switch on power supply without concerning respected teachers.2. 1∅ 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.

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V2 + V3= V1

V2q

V1

V3

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

AmpI2 in Amp

I3 in Amp

V in Volt POWER P=

( I ¿¿12−I 22−I3

2)R2

¿

POWERFACTOR COS∅=

( I ¿¿12−I 22−I3

2)2 I2 I 3

¿

1245678910

RUN-2:SLNO V1 in

VoltV2 in Volt

V3 inVolt I in Amp POWER P=

(V ¿¿12−V 22−V 3

2)2R

¿

POWERFACTOR COS∅=

(V ¿¿12−V 22−V 3

2)2V 2V 3

¿

12345678910

CONCLUSION:

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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.

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TESTING OF 1∅ ENERGY METER AT (0.5,0.866

AND 1.0 PF)

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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 QUANTITY1 Voltmeter 0-300(MI) 32 Ammeter 0-5/10A(MI) 13 1∅ variac 0-300V(MI) 24 1∅ energy meter 1500 RPK 15 Voltmeter 0-600V(MI) 16 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

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R

B

-BVPC

600

-1200

2400

I

When VBN has equal voltage with VRN 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

R

B

-BVPC

I

300

-1200

2400

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

CIRCUIT DIAGRAM

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R

B

VPC

-1200

2400

I

Precautions:

1. Do not switch on power supply without concerning respected teachers.2. 1∅ 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.

OBSERVATION:

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R

B

N

V V

V

R

N(0-300)V

(0-150)V (0-150)V

V(0-600)V

A (0-10)A

ENERGY METER

C1B1

E1

C2B2

E2

C C

CP

AUTO TRANSFORMER

SL

NO

CO

S

V1 VRN VBN VPC I R.P.M R.P.H VPCICCOS∅=(P1)

R .P . H×10001500

=P2

PERCENTAGE ERROR=(P¿¿1−P2)

P1

×100¿

123456789

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 energy meter 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

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VOLTAGE- CURRENT

RELATIONSHIPS AND LOCUS

DIAGRAM OF A SERIES R-L

CIRCUIT16 | 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 QUANTITY1 AUTO TRANSFORMER 0-230V(AC) 12 VOLTMETER 0-300V MI TYPE 33 AMMETER 0-5/10A 14 LAMPLOAD 3KW 15 WATTMETER LPF 16 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.

IX2+ (IY+

V 1

2 XL)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 V 1

X L as diameter.which is known as current locus.

If VR and VL be the resistive and inductive drops respectively, the VR

2+VL2=V1

2

So that the locus of the tip of the VR phasor is a semicircle with V1phasor as the diameter. which is known as voltage locus.

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230VAC

SUPPLYV1

A

V3

V2

M L

C V

LAMP LOAD

CH

OK

E0.32H

(0-300)V

(0-5)A

(0-300)V

(0-300)VWATT METER

Precautions:

1. Don’t switch on power supply without concerning respected teachers.2. 1∅ 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 LOAD12345678

CALCULATIONS:Make calculations in tabular form shown below:

RL=W

I 2

R=V 2

I

VR=I(R+RL)VRL=

WI

VL=√(V 32−V RL

2 ) XL=V L

I

tanθ=V L

V R

θ=tan−1 V L

V R

1

2

3

4

5

6

7

8

GRAPHS:1. Draw the two theoretical semi-circles

a. Voltage drops locus andb. For current locus i.e semicircles with V1 as diameter(V1 is along Y-axis) and another

semicircle with V 1

X Lphasor ( 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.

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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 ?

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V 1

X L

V1

VR1

VR2

VR3V

R4

VL1

VL2

VL3

VL4

CURRENT LOCUS

VOLTAGE LOCUS

OPEN CIRCUIT

AND SHORT CIRCUIT

TEST ON 1∅

TRANSFORMER

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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 QUANTITY1 Single phase transformer 3KVA,230/230V,50 HZ 12 Wattmeter 2.5/5A LPF 13 Wattmeter 10/20A UPF 14 Ammeter 0-1A 15 Ammeter 0-20A 16 Voltmeter 0-300V 17 Voltmeter 0-30/75V 18 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

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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. 1∅ Auto transformer must be kept at minimum potential point. Before switch on the

experiment.

CIRCUIT DIAGRAM:

OPEN CIRCUIT TEST PROCEDURE:

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A

V230VAC

SUPPLY

AUTO TRANSFORMER

WATT METER

3 KVA230/230V

OPEN

A

V230VAC

SUPPLY

AUTO TRANSFORMER

WATT METER

3 KVA230/230V

M L

C V

(0-1)A

(0-300)V

(0-20)A

(0-30)V

M L

C V

PRIMARY SUPPLY

230VV

1

R0

X0

R01

X01

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=WV

IM=√ I 02−IW

2 COS

∅ 0=WV I 0

CALCULATION:See the no load phasor diagram below

j0

IW

IM

W=VI0COS∅ 0

IW=WV

, IM=√ I 02−IW

2 , R0=VIW

, X 0=VI M

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

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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 = I2R1

Hence R1 = Wc / I2 Also V /I = Z1 and X1 = √Z12−R1

2 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 = V o−V t

V t 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?

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LOAD TEST OF 1∅

TRANSFORMER

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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 QUANTITY1 1∅ Transformer 230/230,3KVA 12 1∅ Autotransformer 0-270V 13 Wattmeter 150/300/600V,10/20A dynamometer type,UPF 24 Ammeter 0-5/10/20 ,moving iron(MI) 25 Voltmeter 0-150/300/600V (MI) 26 lamp load 3 KW 1

CIRCUIT DIAGRAM:

THEORY:When the secondary winding of a transformer is completed through a load and a voltage

V is 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.

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A

V

LO

AD

A

V230VAC

SUPPLY

AUTO TRANSFORMER

WATT METER

3 KVA230/230V

M L

VC

M L

VC

(0-10)A

(0-300)V

(0-10)A

(0-300)V

w2

I2

h=efficiency

=load current

I2= HALF LOAD CURRENT

IRON LOSS=CU LOSS

Precautions:

3. Do not switch on supply without concerning respected teachers.4. 1∅ 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= output powerinput power

×100

=w2

w1

×100

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Voltage regulation=v0−v l

v0

×100

Here V0=V1

And VL=V2

(v¿¿1−v2)v1

×100¿

OBSERVATION TABLE:

SLNO LOAD V1 I1 W1 V2 I2 W2efficiency=

W 2

W 1

×100

VOLTAGE REGULATION=V 1−V 2

V 1

CONCLUSION:To be written by student.

DISCUSSION QUESTION:

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THEVeNIN AND

SUPERPOSITION

THEOREM VERIFICATION EXPERIMENT

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AIM OF THE EXPERIMENT:To verify thievenin and superposition theorem .APPARATUS REQUIRED:SLNO NAME OF ITEM SPECIFICATION QUANTITY1. Rheostat 27Ω/10A 32. Rheostat 10Ω/8A 13. Volt meter MC(0-15V) 14. Ammeter MC(0-1A) 35. 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

V ocRL+R¿

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:

PROCEDURE:1. Connect as in figure.A

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-+6V SO

UR

CE

27 W/10A

V

A

27W/10A 27W/10A

RL

1

1'

2

2'

3

3'

IL'

VOC

1

1'

2

2'

3

3'

-+

6V

A

VRL27 W

/10A

27W/10A 27W/10A

FIG-A

FIG-B

-

+++

- -

+

-

+ -

++ ++

+

- --

-

-

(0-1)A

(0-1)A

(0-10)V

(0-10)V

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 R L=V 1

I 1I2 V2 R¿=

V 2I 2 I L=

V OC

RL+R¿

ERROR=I L' −I L

I L×100

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?

THEORY OF SUPERPOSITION THEOREMSuperposition theorem

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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.EXPLANATIONConsider 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 :

A

A-+

AI1a

I2aI

3a

R1

R3

R2

VA

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 Vbacts Va 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.

A

A

A

-+V

B

I1b

I2bI

3b

R1

R3

R2

The currents are given by :

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I1=I1a+I1b

I2=I2a+I2b

I3=I3a+I3bPROCEDURE:

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-14. 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

SOURCEACTING

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 willbe 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.com2.www.allaboutcircuits.com3. www.howstuffworks.com4. www.nptel.iitm.ac.in

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Magnetization charterstic of

separetly excited dc generator

AIM OF THE EXPERIMENTTo find Magnetization characteristic of a separately excited DC generator.

Apparatus required:

SLNO NAME OF APPARATUS RANGE QUANTITY1 moving coil type ammeter 0-3A 12 moving coil type voltmeter 0-300V 13 rheostats 0-300ohms.2Amp 24 Rheostat 50 Ohms,5Amp 15 Tachometer 0-10000RPM 16 A DC Generator Coupled to DC 7.5KW,1500 1

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motor Set RPM,230V,31.2Amp

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

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V Eg

A

230V DCSupply

Z1

Z2

300W1.7A

50W,5A

A1

A2

+

-

A1

A2

+

-

+

-

(0- 300V) MC

Z1 Z

2

+ -

300W1.7A230V DC

Supply

DPSTS

If

(0-1A) MC

M MDPST

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.

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Speed control of DC Motor

using armature

voltage and field current

method

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AIM OF THE EXPERIMENTTo 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 QUANTITY1 moving coil type ammeter 0-5A 12 moving coil type ammeter 0-3A 12 moving coil type voltmeter 0-300V 13 Rheostats 0-300ohms.2Amp 14 Rheostat 50 Ohms,5Amp 15 Tachometer 0-10000RPM 16 DC shunt Motor With Starter 7.5HP,1200RPM,220V,30Amp 17 S.P.S.T switch 1

Theory:If V is the applied voltage across the motor terminals, Ebis the back emf developed,then

V = Eb+ IaRa. Where Iaand Ra the current and resistance in the armature circuit respectively.

But, Eb = ϕZNP60 A

= KϕN . Hence V =KϕN+ IaRa .i.e. N = K ᶦ (V−IaRa)ϕ

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 :

Procedure:- CASE A:

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A1

A2

230V DC

3AR

f

300W,2A

Z1

Z2

V

A1

A2

50W,5A5AL Z A

300V

Connection diagram

M

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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