ET LabManual

1 IIT BHUBANESWAR, SES, ET Lab.

Indian Institute of Technology Bhubaneswar

School of Electrical Sciences Electrical Technology Laboratory

List of Experiments

1. (a) Characteristics of Incandescent Lamp.

(b) Measurement of power consumption of Fluorescent Lamp.

2. (a) Verification of Superposition theorem.

(b) Verification of Thevinin’s theorem.

3. Study of RLC series circuit. 4. Calibration of Energy meter. 5. Three-phase power measurement using 2-Wattmeter method. 6. (a) Power measurement using 3-Ammeter method.

(b) Power measurement using 3-Voltmeter method.

7. Open Circuit and Short Circuit test on 1-Phase Transformer.

8. Speed control of D.C. Shunt Motor by Field Flux control and

Armature Voltage control method.

9. Open Circuit Characteristic of D.C. Generator.

10. External & Internal Characteristics of D.C. Generator.

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Experiment No- 1(a) CHARACTERISTICS OF INCANDESCENT LAMP

AIM OF THE EXPERIMENT: To obtain the Volt-Ampere (V-I) characteristics of Incandescent Lamp.

APPARATUS REQUIRED:

Instruments/Equipments:

Sl.No Instrument/Equipment Type Specification Quantity

1 Incandescent Lamp Tungsten

Filament

200 Watt, 230 V 1 No

2 Voltmeter MI 0 – 300 V 1 No

3 Ammeter MI 0 - 1000 mA 1 No

4 1- Φ variac Iron core 230 V, 4 A 1 No

5 Connecting Wires Cu 1.5 sq. mm As required

THEORY:

If an electric current is passed through a fine metallic wire heat as well as light energy is radiated when the

temperature is very high. The incandescent lamp consists of an evacuated glass bulb having a fine wire filament. Tungsten

is the most commonly used metal for filament lamps. Resistance of the filament can be calculated by below formula.

𝑹 = 𝑽𝑰

Where, R = Resistance of the filament lamp in Ω.

V = Voltage across lamp in volt.

I = Current through the lamp in ampere.



CIRCUIT DIAGRAM:

V

A

0 – 300 V

0 – 1000 mA

1 Phase Variac

DPSTSWITCH

FUSE230V

1-Phase Supply

LAMP200 W, 230 V

Ph

N

Circuit diagram for V-I characteristic of Incandescent Lamp

PRECAUTION:

1. Connection should be right and tight.

2. Check the circuit connection thoroughly before switching on the supply.

3. Instruments should be connected in proper polarity and range.

4. Do not touch any non-insulated part of any instrument or equipment.

5. Be ensured the zero setting of instrument is on right position. Avoid parallax error.

PROCEDURE:

1. Choose the appropriate ratings of the Ammeters, Voltmeters.

2. Set up the circuit as shown in circuit diagram with the lamps and instruments as indicated.

3. Set the variac at zero output voltage before switching on the power supply.

4. Increase the variac output voltage in steps of 20V to 30 V, until the rated voltage is obtained. At each step, note

the readings of Voltmeter and Ammeter and record them in Table- 1.

5. Repeat step- 4 for decreasing output voltage from rated voltage to zero volts.

OBSERVATION: TABLE- 1

Sl. No. Supply Voltage(Volt)

Current (Amp.)

Inc. Dec. Mean


REPORTS:

a) Plot ‘I’ as a function of voltage ‘V’.

CONCLUSION:

DISCUSSION:

1. Is V-I characteristic a straight line? Justify your answer.

2. Why do the readings differ for increasing and decreasing values of the lamp voltages?

3. State whether a lamp rated 230V, 60W can be used on both ac and dc supply. Give reasons for your answer.

4. Explain the advantages of using tungsten wire as filament material in incandescent lamps.




Experiment No- 1(b) MEASUREMENT OF POWER CONSUMPTION OF FLUORESCENT LAMP

AIM OF THE EXPERIMENT: a) Connection and measurement of power consumption of a Fluorescent lamp.

b) Measurement of its Pick-up and Cut-off Voltage.

APPARATUS REQUIRED:



1 Fluorescent Lamp with Fittings

40 Watt, 230 V 1 No


3 Ammeter MI 0 - 200 mA 1 No


5 Wattmeter LPF 1 A, 300 V 1 No


THEORY:

The fluorescent tube consists of a glass tube. The tube contains argon gas at low pressure and one or two drops of

mercury and inside surface of the tube is coated with a thin layer of fluorescent material in the form of powder. The

coating material used depends upon the colour effect desired may consists of zinc silicate, cadmium silicate or calcium

tungsten. These organic chemicals are known as phosphorus which transforms shot wave invisible radiation into visible

light. A Choke is connected in series with the tube which acts as ballast in running condition and provides a high voltage

impulse or surge for instantaneous time for starting the tube light. The filament which is connected with a starter only to

start the tube light in the other words it is called as starting switch.



CIRCUIT DIAGRAM:

Watt meter

M L

C V

V

A

0 – 300 V

0 – 200 mA

1 phasevariac

DPSTSWITCH

FUSE

230 V 1 phasesupply

FLUORESCENT LAMP 40 W

STARTER

CHOKE

Ph

N

Circuit diagram for Fluorescent Lamp

PRECAUTION:

1. All connections should be tightened.

2. Reading of the meter should be taken correctly.


PROCEDURE: 1. Connect the circuit as shown in circuit diagram.

2. Set the variac to its zero position and switch on the power supply.

3. Gradually vary the variac till the lamp glow. Take the reading of Voltmeter, Ammeter and Wattmeter. This will give

the pick-up voltage of fluorescent lamp.

4. Go on increasing the voltage using variac to maximum 230V and observe the reading of three meters. The

wattmeter reading will give the power consumed by the fluorescent lamp.

5. Gradually go on decreasing the voltage by variac till the fluorescent lamp glows off. This will give the cut-off

voltage of fluorescent lamp.

6. Switch OFF the supply.


OBSERVATION:

TABLE- 2 Sl. No.

Supply voltage in Volts (V)

Current in Amp. (I)

Power consumed in Watts (P)

Pick-up voltage- ________ Cut-off voltage- ________

CALCULATIONS:

Power factor (cosΦ) = Power consumed (P) / (Supply voltage × circuit current).

CONCLUSION:

DISCUSSION:

1. What is the function of Starter in fluorescent lamp?

2. Write the technical specification of fluorescent lamp and indicate what type of Power is being measured by

Wattmeter?

3. What is the function of Choke [Ballast] in fluorescent lamp?




Experiment No- 2(a) VERIFICATION OF SUPERPOSITION THEOREM

AIM OF THE EXPERIMENT: To verify the Superposition theorem.

APPARATUS REQUIRED:



1 Rheostat Tubular 25Ω, 3A 1 No

2 Rheostat Tubular 26Ω, 4 A 1 No

3 Rheostat Tubular 6Ω, 5 A 1 No

4 Ammeter MC 0 - 1A 1 No

4 Multimeter Digital 1 No

5 Battery Dry Cell 12V, 7.2 AH 1 No

6 Rectifier Unit 0-220V, 10 A 1 No


THEORY: Superposition theorem states that “current through or voltage across an element in a linear bilateral network is

equal to the algebraic sum of the currents or voltages produced independently by each source”. This theorem is very

handy tool for solving networks with more than one source. The most obvious advantage of this method is that it does not

require use of mathematical methods like determinants to find required current or voltage. Instead superposition theorem



allows us to calculate cumulative effect of multiple sources on an element, by knowing individual effects on element by

each source. Particular attention must be given to polarity of voltages & direction of currents while applying this theorem.

CIRCUIT DIAGRAM:

A

R1 R2

R3

0 – 1A

V1 V2+

+ +

Verification of Superposition Theorem

PRECAUTION:





5. To avoid parallax error note down readings by standing parallel to the meter.

PROCEDURE:

1. Make a circuit as shown in the circuit diagram. The values of resistors and supply as shown in the circuit are the

suggestions only and any other values may be taken depending upon the availability of components in the

laboratory.

2. Remove source V2 and close the circuit through a link preferably through a wire. Measure I3’due to source V1.

Record V1and I3’ in the observation table.

3. Now put V2 in the circuit and remove V1. Close the circuit through a link in place of V1. Measure I3’’due to source

V2. Record V2 and I3’’ in observation table.

4. Now connect both the sources simultaneously and measure I3 record the result in observation table.

5. Compare the measured and calculated values of currents and draw conclusions.


OBSERVATION:

Table-1

Sl. No.

Voltage Source-1

(V1) in Volt

Voltage Source-2

(V2) in Volt

Current through R3

due to V1 (I3’)

in Amp.

Current through R3 due to V2

(I3”)

in Amp.

Current through R3 due to both

sources (I3) in Amp.

I3’ + I3

” in Amp.

Remark

CALCULATIONS:

CONCLUSION:

DISCUSSION:

1. Can we apply the principle of superposition in a network having more than two sources?

2. Explain why the superposition principle does not work for power in the elements of a circuit?

3. Explain why it is advisable to remove the source and complete the circuit by a link instead of short circuiting the

terminals of the supply while remaining in the circuit.




Experiment No- 2(b) VERIFICATION OF THEVENIN’S THEOREM

AIM OF THE EXPERIMENT: To verify Thevenin’s Theorem.

APPARATUS REQUIRED:


Sl. No. Instrument/Equipment Type Specification Quantity

1 Rheostat Tubular 25Ω, 3 A 1No.

2 Rheostat Tubular 26Ω, 4 A 1 No.

3 Rheostat Tubular 6Ω, 5 A 1 No.

3 Ammeter MC 0 - 1 A 1 No.

4 Multimeter Digital 1 No.

5 Battery Dry Cell 12V, 7.2AH 1 No.

6 Rectifier Unit 0-220 V, 10 A 1 No.


THEORY: Thevenin’s theorem states that “Any two terminals AB of a network composed of linear, passive& active

elements may be replaced by simple equivalent circuit consisting of an equivalent voltage source VOC in series with an

equivalent resistance RTH. The voltage source VOC is equal to the potential difference between two terminals AB caused by

active network with no external resistance connected to these terminals. The series resistance RTH is equivalent resistance

looking back into the terminals AB with all sources within network made inactive.

𝐈𝐋 =𝐕𝐓𝐓

𝐑𝐓𝐓 + 𝐑𝐋

Where IL = Load current, VOC = Open circuit voltage.

RTH = Thevenin’s Resistance and RL = Load Resistance



CIRCUIT DIAGRAM:

RL

RTH

VTH

THEVININ’S EQUIVALENT CURCUIT

A

R1 R2

RL

0 – 1A

V1 V2+

+ +

Circuit diagram for verification of Thevenin’s Theorem

PRECAUTION:



3. Instruments should be connected in proper polarity.

4. Do not touch any non-insulated part of any instrument or equipment

PROCEDURE: 1. Choose the appropriate ratings of the Ammeters, Voltmeters, and Rheostat as per the circuit diagram.

2. Set up the circuit as shown in Figure.

3. Start the rectifier unit by keeping variable knob at exactly zero point. Increase the rectifier output to a required

supply voltage. Note down the reading of load current (IL) and load voltage (VL) Table 1.

4. Measure Thevenin’s voltage and Thevenin’s resistance by a voltmeter and Multimeter by following proper

method.

5. Make Thevenin’s equivalent circuit and find out the load current (IL’). Compare it with IL.

OBSERVATION:

Table-2

Sl. No.

Observed Load

Current (IL)

(in Amp.)

VTH (V)

RTH (From

Multimeter) (Ω)

Computed Load Current

IL=VTH/(RTH+RL) (in Amp.)

Load Current (IL) from equivalent

circuit. (in Amp.)

Remark


CALCULATIONS:

The load current IL can be calculated as

Lth

thL RR

VI+

=

CONCLUSION:

DISCUSSION:

1. What type of ammeter and voltmeter (MI or MC) will you use and why?

2. Can you suggest an alternative procedure for the determination of RTH?

3. Is there any restriction for choice of circuit elements?




Experiment No- 3 STUDY OF RLC SERIES CIRCUIT

AIM OF THE EXPERIMENT: Measurement of current, voltage and power in R-L-C series circuit excited by single phase A.C. supply.

APPARATUS REQUIRED: Instruments/Equipments:


1 Rheostat Tubular 100Ω, 2.5A 1 No

2 Inductor Air core 30 mH, 5 A 1 No

3 Capacitor Oil 70 µF, 660 V 1 No

4 Ammeter MI 0 - 2.5/5 A 1 No

5 Voltmeter MI 0 – 150/300 V 3 Nos

6 Voltmeter MI 0 – 30 V 1 Nos

7 Wattmeter LPF 2.5/5 A, 75/150/300 V 1 No

8 1-Φ variac Iron core 230 V, 4 A 1 No


THEORY:

VI

VR VL VC

R L C

A series R-L-C circuit is shown in the above figure. According to Kirchhoff’s voltage law.

CLR VVVV ++=



IZ

jRIC

L

=

−+= ))1((ω

ω

Where, Z is the impedance of the Circuit. The current though the circuit is given by, ZVI /= In a series R-L-C circuit:

1. If CL ωω /1> then current lags the voltage

2. If CL ωω /1< , then current leads the voltage

3. If CL ωω /1= , then current and voltage are in phase. The phasor diagrams for R-L-C series circuit for:

(a) CL ωω /1=

(b) CL ωω /1>

& (c) CL ωω /1< , is shown below

I

XC

XL

V

IR I

XC

XL

V

IR

XL - XC

I

XC

XL

V

IR

XC - XL

Fig – (a) For unity power factor

Fig - (b) For lagging power factor

Fig – ( c) For leading power factor


CIRCUIT DIAGRAM:

M

C

V

A

0 – 150 V

0 – 5 A

1 phasevariac

DPSTSWITCH

FUSE

230 V 1 phasesupply

VVV

RL C

VR VL VC

0 – 150 V 0 – 30 V 0 – 150 V

2.5/ 5 A, 75/150/ 300 VLPF Watt meter

L

V

Ph

N

Circuit Diagram for RLC series circuit.

PRECAUTION:

1. All connection should be tight. 2. The meter should of proper range. 3. Check zero setting of all meter. 4. While varying the value of Variac note that current in circuit should not exceed safe limit.

PROCEDURE:

1. Connect the circuit as shown in circuit diagram. 2. Set the variac to minimum voltage position. 3. Switch on AC Supply. 4. Set the variac output voltage 90 V. 5. Note down the reading of all meters. 6. Change the setting of Rheostat with fixed position of variac take the reading all the meters. 7. Change the setting of variac. 8. Repeat the step 5 and 6. 9. Record the observation as per table.

OBSERVATION:

S. No. Observation Calculation Remarks

V

Volt

I

Amp

VR

Volt

VL

Volt

VC

Volt

P

Watt

R=

VR/I

XL=

VL/I

XC=

VC/I

Cosφ

P/VI

Z=

V/I

φ


REPORTS:

1. Plot the phasor diagram using the experimental data.

CALCULATIONS:

CONCLUSION:

DISCUSSION:

1. What is power factor in AC circuit and write its importance? 2. What do you mean by Impedance of AC Circuit and what is its units? 3. Define resonance in AC Circuit? What will be consequences when circuit attends resonance?




Experiment No- 4 CALIBRATION OF ENERGY METER

AIM OF THE EXPERIMENT: Connection and testing of a 1-ϕ Energy meter by (a) Short run Test (b) Long Run test.

APPARATUS REQUIRED:



1 Voltmeter MI 0 - 300 V 1 No

2 Ammeter MI 0 – 10 A 1 No

3 Energy meter Induction 230 V, 20 A, 6000 Rev/kWh 1 No

4 Stop watch Standard type 1 No

5 Load box Lamp Load 2 kW, 230 V 1 No


THEORY: Energy meter is an integrating instrument, which is used to measure the consumption of electric energy consumed

by a residence, business or an electrically powered device. It measures energy in kWh (Kilo Watt-Hour) and is an integrating meter. The principle of operation of an energy meter is similar to that of a wattmeter except rotating disc. The number of revolutions made by the disc is counted with the help of a gear train and read on the dial directly as unit (i.e. 1 unit = 1 kWh). Constructionaly it is nearly same with respect to an induction wattmeter, except that the pointer of the wattmeter is replaced by a breaking magnet and a spindle.

A single-phase induction type energy meter consists of the following parts:

1. Moving system 2. Operating mechanism

It consists of – (i) Series magnet

(ii) Shunt magnet

(iii) Breaking magnet

3. Recording mechanism



CIRCUIT DIAGRAM:

V

A

0 – 300 V

DPSTSWITCH

FUSE

230 V 1 phasesupply

CC1 CC2

PC1 PC20-10A

L1 Load (PH)

Load (N)

L2

SINGLE PHASEENERGY METER

2 kW LAMPLOAD BOX

Ph

N

Circuit Diagram for energy meter testing

PRECAUTION:

1. Connection should be right and tight. 2. Check the circuit connection thoroughly before switching on the supply. 3. Instruments should be connected in proper polarity and range. 4. Do not touch any non-insulated part of any instrument or equipment. 5. Be ensured the zero setting of instrument is on right position. Avoid parallax error.

PROCEDURE: 1. Select suitable ranges of the ammeter and voltmeter such that energy meter can be tested over its complete range.

2. Connect the circuit as shown in the circuit diagram.

3. Before switching on the supply, ensure that the Lamp load switches (all) are open (off).

4. Note down the initial reading of the energy meter.

5. Set the desired load by selecting a suitable combination of Switches on the loading rheostat.

6. Switch on the supply and wait for the red indicator of the energy meter disc to come in the front. At this moment start the stopwatch. Note down the voltmeter and ammeter readings.

7. Measure the time (T) for (N) revolutions (say 20 revolutions) switch off the stopwatch immediately. Switch off the supply.

8. By adjusting the loading rheostat take 8 to 10 sets of readings covering the full current range of the energy meter and tabulate the observation in the table.


OBSERVATION:

Sl. No.

Load Voltage

(V)

Load Current

I (A)

Time T (s)

Energy Recorded

by the meter Em

Actual Energy Consumed During N rev. Ea=(V.I.T)/(1000x3600)

Percentage Error (Em-Ea)/Ea x 100

REPORTS:

Plot ‘% Error’ as a function of ‘Load Current’.

CALCULATIONS: For observation no. : ………… Power consumed by the load = VI watt. Time taken to complete 20 revolutions = t secs. Actual energy consumed = VIt/(1000 × 3600) kwh

Energy read by the energy meter = No. of revolutions/Meter constant % Error = [(Actual energy-Measured energy) /Actual Energy]× 100

Note : Slow speed of rotating disc for positive error – Beneficial for consumer Fast speed of rotating disc for negative error – Beneficial for supplier

CONCLUSION:

DISCUSSION:




Experiment No- 5 THREE PHASE POWER MEASUREMENT

AIM OF THE EXPERIMENT: To study about 3-Φ power and to measure the power in a 3-Φ system with balanced and unbalanced loads

separately by Two-Wattmeter Method.

APPARATUS REQUIRED:



1 Rheostat Tubular 100 Ω, 5A 3 Nos



3 Ammeter MI 0 – 2.5/5 A 3 Nos

4 Wattmeter UPF 2.5/5 A, 600 V 2 Nos



THEORY: A wattmeter is an instrument with a potential coil and a current coil so arranged that its deflection is proportional

to VI cosφ, where

V = rms voltage applied across the potential coil.

I = rms current through current coil.

φ = phase angle between V & I.

The 3-phase power can be measured by 3-single phase wattmeter’s having current coils in each line and potential

coils connected across the given line and any common function. Since this common junction is completely arbitrary, it

may be placed on any one of the three lines. The wattmeter connected to that line will show a zero reading.



So according to Blondel’s theorem in an N-wire circuit, only (N-1) number of wattmeter’s are required to measure the

power.

So 2-wattmeter method measures the power of 3-phase star/delta connected balanced/unbalanced loads.

The total power supplied, is given by,W = WA + WB = V3 VLILcosφ (Algebric sum).

R

Y

B

3-phase Load

WA

WB

CURRENT COIL

PRESSURE COIL

SINGLE PHASE WATTMETER

Power factor, cos Φ = 𝑐𝑐𝑐 𝑡𝑡𝑡−1 √3(𝑊𝐴−𝑊𝐵)𝑊𝐴+𝑊𝐵

If one of the wattmeter tends to read negative, the pressure oil is reversed, but the reading of the wattmeter must

be taken on negative.

CIRCUIT DIAGRAM:

V

A

R

Y

B

TPSTSWITCH

3 phaseVariac

FUSEA

A

C V

M L

M L

CV

0 – 5 A

0 – 600 V

0 – 5 A

0 – 5 A

100 , 5 AT

100 , 5 AT

100 , 5 AT

2.5/ 5 A , 600VUPF Wattmeter


R

Y

B

3 phaseSupply

Fig-1 Circuit Diagram for balanced load


V

A

R

Y

B

TPSTSWITCH

3 phaseVariac

FUSEA

A

C V

M L

M L

CV

0 – 5 A

0 – 600 V

0 – 5 A

0 – 5 A



R

Y

B

3 phaseSupply

100 , 5 AT

100 , 5 AT

100 , 5 AT

V

V

V

0 – 150 V

0 – 150 V

0 – 150 VV3

V1

V2

Fig-2 Circuit Diagram for unbalanced load

PRECAUTION:






PROCEDURE:

1.For balanced load:

1. Connect the circuit as shown in Fig.- 1

2. Adjust the rheostats for the maximum resistance.

3. Switch on the supply.

4. Read the meters to obtain VL, I

1, I

2 and I

3. Note the wattmeter reading W

1 and W

2 (Note the multiplying factor on

the wattmeters).

5. Vary the load resistance and obtain at least five sets of observations, the current should not exceed the limit 4 A.

6. Tabulate the readings and check the results by completing the calculations indicated in the table-1.


2.For unbalanced load:

1. Connect the circuit as shown in Fig.-2.

2. Adjust the three rheostats at the maximum values.

3. Switch on the supply and set the autotransformer to 415 V.

4. Take five sets of observation for different rheostat settings such that the reading of I1, I

2 and I

3 in each set is

appreciably different to create unbalanced loading condition. The current should not exceed the limits in each

arm.

5. Tabulate and check the result by completing the computations indicated in Table:-2.

OBSERVATION:

Table-1:

Sl. No.

VL

in (V)

I1 in

(A) I

2In

(A) I

3 in

(A) W

1in

(W) W

2in

(W) Calculated power (Wc) = (V

L/√3) (I

1+I

2+I

3)

(W) = (W1 +W2)

Error = W - Wc Wc *100%

Table- 2:

Sl. No.

VL in (V)

V1 in (V)

V2 in (V)

V3 in (V)

I1 in (A)

I2 in (A)

I3 in (A)

W1in (W)

W2in

(W) Calculated power (Wc) = (V1I1+ V2I2 + V3I3)

(W) = (W1 +W2)

Error = W - Wc Wc *100%

CALCULATIONS:

CONCLUSION:

DISCUSSION:

1. What do you understand by a balanced three-phase load?

2. How would you measure power using a) Three watt meters and b) One wattmeter for balanced/unbalanced loads?

3. Is it possible to measure power factor of the balanced (three –phase load by two-wattmeter method)?




Experiment No- 6(a) POWER MEASUREMENT USING THREE AMMETER METHOD

AIM OF THE EXPERIMENT: To measure the power of a single phase AC circuit by three Ammeter method.

APPARATUS REQUIRED:



1 Rheostat Tubular 100 Ω, 2.5 A 1 No.

2 Rheostat Tubular 26 Ω, 4 A 1 No.

3 Inductance Air core 60 mH, 5 A 1 No.

4 Ammeter MI 0 - 5/10 A 1 No.

5 Ammeter MI 0 – 2.5/5 A 2Nos.

6 Voltmeter MI 0 – 75/150/300 V 1 No.

7 Wattmeter L.P.F. 10A,300V 1 No.

8 1-Φ variac Iron core 230 V, 10 A 1 No.


THEORY:

We know in a dc circuit, the power is given by the product of voltage and current where as in AC 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 AC circuit

simply from the reading of a voltmeter and ammeter. In ac circuits power is normally measured by Wattmeter. However,

this method demonstrates that the power in a single phase ac circuit can also be measured by using 3-ammeters.



The phasor diagram of 3-Ammeter method is shown below

Vθ φ

I2

I1I3

I1

2 = I22 + I3

2 +2I2I3Cos(Φ) Power factor (CosΦ) = (I1

2 -I22 - I3

2 ) / 2I2I3

Power Consumed by the load = I3*V*CosΦ

CIRCUIT DIAGRAM:

V

A

0 – 150V

0 – 5 A

DPSTSWITCH

FUSE

230 V 1 phasesupply

A

A

0 – 10 A

0 – 2 A

60 mH, 5 A

26

100 ,2. 5 AT

,4 AT

I1

I2

I3

Ph

N

Watt meter, 10A, 300V

M L

C V

1Ph Variac

Circuit Diagram for three ammeter method

PRECAUTION:







PROCEDURE: 1. Connect the circuit as per the circuit diagram.

2. Set the 1-Φ variac at minimum position.

3. Switch on the supply and vary the output voltage of the variac in such a way that it should not exceed 120 V.

4. Note down the observations and switch off the power supply.

OBSERVATION:

Observation Calculation

Sl. No.

V in (Volt.)

I1 in (Amp.)

I2 in (Amp.)

I3 in (Amp.)

Wattmeter (W)

CosΦ Cosθ P in Watt (Wc)

Q in Volt-Amp-Reactive

S in volt-Amp

%Error = (W-Wc)/Wc

REPORTS:

1. Choose the proper current and voltage scales and draw the vectors I, V1, V2& V3 , measure the angle Φ & θ

Or

2. By using triangle law find Φ & θ.

CALCULATIONS: %Error = (W-Wc) / Wc W: Wattmeter Meter reading, Wc: Calculated Power

CONCLUSION:

DISCUSSION:

1. Draw the Circuit & Vector diagram to measure the power of a single Φ circuit by 3 Ammeter method

2. Draw and define the power triangle of an AC circuit.




Experiment No- 6(b) POWER MEASUREMENT USING THREE VOLTMETER METHOD

AIM OF THE EXPERIMENT: To measure the power of a single phase AC circuit by three Voltmeter method.

APPARATUS REQUIRED:



1 Rheostat Tubular 100 Ω, 2.5 A 1 No

2 Rheostat Tubular 26 Ω, 4 A 1 No

3 Inductance Air core 60 mH 1 No

4 Ammeter MI 0 - 5 A 1 No

5 Voltmeter MI 0 – 150/300 V 2Nos


7 Wattmeter L.P.F. 10A, 300V 1 No.

8 1-Φvariac Iron core 230 V, 10 A 1 No


THEORY:

Let I be the reference Vector

V2 in phase with I

V3 leads I by angle Φ

∴V1=V2+V3

V12 = V2

2 + V32 +2V2V3Cos(φ)

Power factor of the load (Cosφ) = (V12 -V2

2 - V32 ) / 2V2V3

Power Consumed by the load = V3*I*Cosφ

Φθ

V1

V2

V3

I



Power Factor of entire circuit = cosθ

Active Power Consumed by the Load W= V3ICosΦ watt

Reactive Power Consumed by the load Q= V3IsinΦ VAR

Apparent Power Consumed by the Load S= V3I VA

Power Consumed by entire circuit = V1ICosθ watt

CIRCUIT DIAGRAM:

V

A

0 – 300 V

DPSTSWITCH

FUSE

230 V 1 phasesupply

0 – 5 A

60 mH , 5 A

26Ω, 4APh

N

Watt meter 10A , 300V

M L

C V

1 Ph Variac

VV2

0 – 300 V

100Ω, 2.5A

V

0 – 75 V

Circuit Diagram for three voltmeter method

PRECAUTION: 1. Connection should be right and tight.





PROCEDURE: 1. Connect the circuit as per the circuit diagram.

2. Set the 1-Φ variac at minimum position.

3. Switch on the supply and vary the output voltage of the variac in such a way that it should not exceed 120 V.

4. Note down the observations.

5. Switch off the power supply.


OBSERVATION:

Observation Calculation Sl. No.

I in (Amp.)

V1 in (Volt.)

V2 in (Volt.)

V3 in (Volt.)

Wattmeter (W)

CosΦ Cosθ P in Watt (Wc)

Q in Volt-Amp-Reactive

S in volt-Amp

%Error= (W-Wc)/Wc

REPORTS:

1. Choose the proper current and voltage scales and draw the vectors I, I1, I2 & I3 , measure the angle Φ & θ.

Or

2. By using triangle law find Φ & θ.

CALCULATIONS:

%Error = (W-Wc) / Wc W: Wattmeter Meter reading, Wc: Calculated Power

CONCLUSION:

DISCUSSION:

1. Draw the Circuit & Vector diagram to measure the power of a single Φ circuit by 3 Ammeter method

2. Draw and define the power triangle of an AC circuit.




Experiment No- 7 OPEN-CIRCUIT AND SHORT-CIRCUIT TEST ON TRANSFORMER

AIM OF THE EXPERIMENT: (a) To perform open circuit test on a 1-Φ transformer.

(b) To perform short circuit test on the same transformer.

(c) Calculate the complete parameters of the equivalent circuit of the transformer.

APPARATUS REQUIRED:



1 Voltmeter MI 0 – 150/300 V 2 Nos

2 Voltmeter MI 0 – 30/60 V 1 No



5 Ammeter MI 0 – 10/20 A 1 No

6 Wattmeter LPF 2 A, 150 V 1 No

7 Wattmeter UPF 10 A, 75 V 1 No

8 1-Φvariac Iron core 230 V, 10 A 1 No


Machine specification:

Sl.No Machine Specification Quantity

1. 1-Φ transformer

1-Φ Transformer :-1.5 kVA

230/115 V, 50 Hz

1 No



THEORY:

Open circuit test: - In this test low voltage winding is connected to a supply of normal voltage and frequency and the high voltage

winding is left open. The primary winding draws very low current hardly 3 to 5 percent of full load current when this

condition. As such copper losses in the primary winding will be negligible. Thus mainly iron losses occur in the

transformer under no load on open circuit condition, which are indicated by the wattmeter connected in the circuit.

Hence, total iron losses = W0 (Reading of wattmeter).

Power drawn W0 = Vo I0 cosφ0

Thus no load power factor cosφ0 = W0/V0I0

Core loss component of no load current Iw = I0 cosφ0

Magnetising component of no load current Im = I0 sinφ0

Equivalent resistance representing the core loss R0 = V0/Iw

Magnetising reactance representing the Magnetising current Xm = V0/Im

Short circuit test: -

In this test, low voltage winding is short-circuited and a low voltage hardly 5 to 10 percentage of the rated voltage

of the high voltage side is applied such that rated current flows through the winding. This test is performed at rated current

flowing in both the windings. The iron losses occurring in the transformer under this condition is negligible, because of

very low applied voltage. Hence, the total losses occurring under short circuit are mainly the copper losses of both the

winding, which are indicated by the wattmeter connected in the circuit. Let the various reading be Wsc , Vsc & Isc.

22

2

, RZXIV

Z

IW

R

sc

sc

sc

sc

−==

=

Where,

R = equivalent resistance

X= Equivalent leakage reactance

Z= Equivalent impedance


CIRCUIT DIAGRAM:

A

0 –

150

V

1 phasevariac

DPSTSWITCH

FUSE

230 V 1 phasesupply

V

0 –2 A

M L

C V

LT HT

2 A,150 V LPFWATTMETER

V

0 – 300 V

Ph

N

Fig-1 Circuit Diagram for O.C Test

A

0 –

30 V

1 phasevariac

DPSTSWITCH

FUSE

230 V 1 phasesupply

V

0 – 10 A

M L

C V

LTHT

10 A, 75 V UPFWATTMETER

A

0 – 20 A

Ph

N

Fig- 2 Circuit Diagram for S.C Test

PROCEDURE:

Open circuit test: -

1. Connect the circuit as shown in the Fig-1.

2. Ensure that the setting of the variac is at low output voltage.

3. Switch on the supply and adjust rated voltage across the transformer circuit.

4. Note down the readings of all the meters.

5. Switch-off the a.c. supply.


Short circuit test: -

1. Connect the circuit as shown in Fig-2.

2. Ensure that the setting of the variac is at low output voltage.

3. Switch on the supply and increase the voltage applied slowly, till the current in the windings of the transformer is

full load value.

4. Note down the readings of all the meters.

5. Switch-off the a.c. supply.

OBSERVATION:

Sl.No No load test Short circuit test

Vo Io Wo Vsc Isc Wsc

CALCULATIONS:

Sl.No Cosφo Iw Im Ro Xm R X

CONCLUSION:

DISCUSSION:

1. Why indirect testing of large size transformer is necessary.

2. What type of losses occur in the primary and secondary winding of transformer

3. Which parameters of the equivalent circuit of a transformer can be found through Short-circuit test.

4. Why the HV side of transformer is kept open in Open circuit test?

5. Justify that the power drawn by the transformer under no-load is equal to the iron losses and under short circuit

the full load copper losses.

6. What will happen to the transformer, if a ratted voltage is applied during the short circuit test?




Experiment No- 8 SPEED CONTROL OF DC SHUNT MOTOR

AIM OF THE EXPERIMENT: To perform the speed control of a DC shunt motor by field flux control & armature voltage control method.

APPARATUS REQUIRED:




2 Rheostat Tubular 100 Ω, 5 A 1 No

3 Ammeter MC 0 - 2 A 1 No

4 Voltmeter MC 0 – 300 V 1 No

5 Tachometer Digital 0 – 5000 rpm 1 No




1. D.C. Shunt Motor

D.C. Shunt Motor :- 3.7 kW, 1500 RPM 220 V , 19.7 A , Excitation- 220 V/0 .95A

1 No

THEORY:

If “V” is the applied voltage across the motor terminals, “Eb” is the back e.m.f. developed, then V = Eb + Ia.Ra

Where Ia and Ra are the current and resistance in the armature circuit respectively.

But, Eb = φZNP/60A = KφN

Hence V= KφN + Ia.Ra

i.e N= K (V – Ia.Ra)/ φ



This shows that:

An increase of Ia.Ra drop will decrease the value of speed if “V” remains constant. Speed varies inversely to the

field flux and hence varies inversely as the exciting current. If below saturation, by increasing the resistance in the

armature circuit, motor can be operated at speed below rated speed. By increasing the resistance in the field circuit a

motor can be operated above rated speed.

CIRCUIT DIAGRAM:

L F A

3 point starter

+

-DPST

SWITCH

FUSE

A1

A2

F1

F2

DC SUPPLY220 Volt.

A

M V200 , 1. 5 AT

100 , 5 AT

0 – 2 A

0 – 300 V

+

+

Circuit Diagram for speed control of dc shunt motor.

PRECAUTION:

1. The motor field rheostat should be kept at minimum resistance position.

2. The motor armature rheostat should be kept at maximum resistance position.

3. The motor should be in no load condition throughout the experiment.

4. The motor should run in anticlockwise direction.

PROCEDURE:

1. Connect the machine under test as shown in the circuit diagram.

2. Switch on the D .C supply and start the D.C shunt motor with the help of three point starter by keeping the

external resistance (Field rheostat) in field circuit at its minimum and in the armature circuit at its maximum

position.

Case- A (Armature voltage control)

1. Adjust the field current at normal value (corresponding to normal speed).

2. Keeping the field current constant, vary the voltage across the armature by varying the armature rheostat towards

its maximum value in minimum 5 steps and note down the experimental data’s.


Case- B (Field flux control )

1. Adjust the voltage across the armature at its rated value.

2. Keeping this voltage constant , decrease the field current of the motor by varying the field rheostat towards its

maximum value in minimum of 5 steps and note down the experimental data’s.

OBSERVATION: Case – A

No. Of

Observations

Armature Voltage (Volts)

Speed (RPM) Field Current (Amps)

Case – B No. Of

Observations

Field Current (Amps)

Speed (RPM) Armature Voltage (Volts)

REPORTS:

Draw curves showing :

1. “Speed” vs “Armature voltage” , with field current constant.

2. “Speed” vs “Field current” , with armature voltage constant.

CONCLUSION:

DISCUSSION:

1. Is it possible to obtain the speed higher than the rated speed by armature control discuss ?

2. Why speed control is essential from industrial point of view ?

3. Is it possible to obtain speeds lower than the rated value by using field control ?




Experiment No- 9 OPEN CIRCUIT CHARACTERISTIC OF DC GENERATOR

AIM OF THE EXPERIMENT: a) To conduct open circuit test on a given separately excited dc generator and to plot the characteristic curves.

b) To determine critical resistance of the field circuit and critical speed.

APPARATUS REQUIRED: Instruments/Equipments:




3 Ammeter MC 0 – 2 A 1 No

4 Voltmeter MC 0 – 300 V 2 Nos

8 Tachometer Digital 0 - 5000 rpm 1




1. D.C. Motor coupled with D.C. Generator.

D.C. Shunt Motor :-5 HP , 1500 RPM 220 V , 19.7 A , Excitation- 220 V/0.95 A D.C. Generator :-3.5 kW, 220 V, 16 A, 1500 rpm Excitation- 220 V/0.85 A

1 Set

THEORY: It is also known as Magnetic or Open-Circuit characteristics (O.C.C.). It shows the relation between the no-load

generated e.m.f. of armature “E0”, and the field or exciting current “If”, at (“N”) rated speed of that machine. It is just the

magnetization curve for the material of the electromagnets. Its shape is practically the same for all generators whether

separately excited or self-excited.



CIRCUIT DIAGRAM:

A

L F A

3 point starter

+

-

+

-

DPSTSWITCH

DPSTSWITCH

FUSE

FUSE

A1

A2

F1

F2

F1

F2

200Ω,1.5ADC SUPPLY

220 Volt.

DC SUPPLY220 Volt.

0 – 300 V

0 – 2 A

M GV

+

+

V

0 – 300 V

+

A1

A2

30

0Ω

, 1

.7A

Circuit Diagram for O.C Test on dc generator

PRECAUTION:


2. The generator field rheostat should be kept at maximum resistance position.

3. At the time of starting, the generator should be in no load condition.

4. The generator must rotate in proper direction.

PROCEDURE:

1. Connect the d.c motor and the d.c generator as per the circuit diagram.

2. Set the potential divider feeding the field circuit of the generator for zero output voltage.

3. Switch on the d.c supply to the d.c motor and start it using the three point starter.

4. Adjust the speed of the d.c motor to rated value by varying the resistance in the field circuit.

5. Record the generated e.m.f due to residual magnetism.

6. Switch- on the d.c supply across the field circuit of the generator.

7. Vary the field current of generator in steps and record its value and the corresponding generated e.m.f of the

generator. Observations should be continued upto the generated voltage 25 percent higher than the rated voltage

of the generator.

8. To plot the field resistance line, record the voltage across the field of the generator.

9. Switch off the d.c supply, to stop the motor and also to disconnect the generator field.


OBSERVATION:

Sl. No Vo If

CALCULATIONS:

Calculate the critical resistance and critical speed as follows:

NVVN

IV

R

c

fc

cc

×=

=

1

2

REPORTS:

1. Draw curves showing- Vo Vs If

2. Draw the field resistance line on the magnetization characteristic.

CONCLUSION:

DISCUSSION: 1. Initial portions of Open circuit characteristic [O.C.C.] is almost straight line. Why?

2. What is the significance of residual magnetism?

3. Define critical resistance of the field circuit of D.C. Generator.

4. What is “critical speed” of the generator?

Volta

ge

Field current

OCCAir gap line

O

A

B

C

V1

V2

Rc

Rsh

Vc

Ifc

Nc

N




Experiment No- 10 EXTERNAL & INTERNAL CHARACTERISTICS OF DC GENERATOR

AIM OF THE EXPERIMENT: a) To conduct load test on a given separately excited dc generator and to draw the characteristic curves.

b) To deduce the internal characteristic from the above.

APPARATUS REQUIRED:



1 Rheostat Tubular 200 Ω, 1.5A 1 No

2 Rheostat Tubular 300 Ω, 1.7A 1 No

3 Ammeter MC 0 – 10/20 A 1 No

4 Voltmeter MC 0 - 300 V 1 No

5 Tachometer Digital 0 – 5000 rpm 1 No

6 Load box Resistive 4 kW, 230 V 1 No





1. D.C. Motor coupled with D.C. Generator

D.C. Shunt Motor :-5 HP , 1500 RPM 220 V , 18.6 A , Excitation- 220 V/ 0.95 A D.C. Generator :-2.2 kW, 220 V, 13 A, 1500 rpm Excitation- 220 V/0.85 A

1 Set



THEORY:

The load characteristics on extended characteristics of d.c. generator represents the graphical relationship between

the terminal voltage and the load current, the generator being operated at constant rated speed and with the same

excitation as under no load condition. The nature of this characteristic depends upon the following factors.

1. Voltage drops in the armature winding, Interpol and compensating windings.

2. Voltage drop at the brush contact.

3. Voltage drop due to armature reaction.

External characteristic of the generator indicates the fall in the terminal voltage as the load on the generator increases.

The internal characteristic of shunt generator can be obtained by adding the voltage drop in the armature winding

(IaRa) to the external characteristic plotted experimentally.

CIRCUIT DIAGRAM:

L F A

3 point starter

+

-

+

-

DPSTSWITCH

DPSTSWITCH

FUSE

FUSE

A1

A2

F1

F2

F1

F2

200 , 1.5 AT

300Ω, 1.7A

DC SUPPLY220 Volt.

DC SUPPLY220 Volt.

0 –

300

V

M GV

+

A

ResistiveLoad Box

4 kW

0 – 20 A

+

A1

A2

Fig-1 Circuit Diagram for Load Test on dc generator

PRECAUTION:


2. The generator field rheostat should be kept at maximum resistance position.

3. At the time of starting, the generator should be in no load condition.

4. The generator must rotate in proper direction.


PROCEDURE:

1. Connect the circuit as per the circuit diagram.

2. Set the rheostat so that there is maximum external resistance in the field circuit if the generator.

3. Switch on the d.c supply of the d.c motor and start it with the help of 3-point starter.

4. Adjust the speed of the motor to rated value by varying the resistance in the field and armature circuit of the

motor.

5. Adjust the field current of the generator by its field rheostat so as to obtain rated voltage at no load.

6. Switch on the resistive load & Note down the load current and the terminal voltage.

7. Repeat step-6 for various values of load current; fill the full load current of the generator.

8. Switch off the load on the generator.

9. Switch of the d.c supply to stat the motor.

10. Measure armature resistance (Ra) of the dc generator using multimeter.

OBSERVATION:

Sl. No Load in Watt Terminal Voltage(VL) Load Current(IL) E = VL+IaRa

REPORTS: Draw graphs showing-

1. VL Vs IL

2. E Vs Ia

CALCULATIONS:

E = VL+IaRa

CONCLUSION:

DISCUSSION:

1. Why the voltage drop is so sharp in case of shunt Generator?

2. What should be done if the D.C. shunt Generator fails to build up?

3. What are the reasons of fall of terminal voltage of a D.C. Shunt Generator?

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