47512039 m-i-lab-manual

Gnanamani College of Engineering

NH-7, pachal, Namakkal – 6301018

MEASUREMENTS AND

INSTRUMENTATION LABORATORY

MANUAL

NAME: __________________________________________________

YEAR/SEM: ________________________________ROLL NO______

DEPT.:__________________________________________________

TABLE OF CONTENTSEx.No.

Date Title of the ExperimentPageNo.

Marksawarded

Remarks

Total Marks

Average Marks

Lab Completed Date

Staff Signature :

MEASUREMENTS AND

INSTRUMENTATION LABORATORY

Manual

Third Semester B.E. ( EEE )

Anna University - CoimbatoreBy

Gandhi.R ,ME

Department of EEEGnanamani College of Engineering

NH -7, pachal

Namakkal – 637018

PREFACE

This manual“MEASURMENT &INSTRUMENT” has been

written primarily for Practical for Third semester EEE for the

academic year 2010-2011.

This manual covers all the experiments prescribed by

Anna University – Coimbatore and the experiments are

explained with supportive diagrams and tables.Students can

enter the readings and perform all the calculations in the work

sheets and graph sheets provided in the manual.

We take this opportunity to thank the Management of

Gnanamani College of Engineering and Dr.D.TENSING, Principal,

Gnanamani College of Engineering for the continuous support

and encouragement in completing this work.

Gananamani College of Engineering

NH-7,Pachal,Namakkal-63019

LIST OF EXPERIMENTS

SL.NO

NAME OF THE EXPERIMENTS PAGE.NO

1 STUDY OF DISPLACEMENT TRANSDUCER.

2 PRESSURE TRANSDUCER.

3 AC BRIDGE- SCHERING’S BRIDGE .

4 AC BRIDGE- MAXWELL’S INDUCTANCE,

CAPACITANCE BRIDGE.5 DC BRIDGES -WHEATSTONE BRIDGE

6 DC BRIDGES -KELVINS DOUBLE BRIDGE

7 INSTRUMENTATION AMPLIFIER

8 A/D CONVERTER AND D/A CONVERTER

9 STUDY OF TRANSIENTS

10 CALIBRATION OF SINGLE-PHASE ENERGY METER

11 MEASUREMENT OF THREE-PHASE POWER AND POWER FACTOR

12 CALIBRATION OF CURRENT TRANSFORMER

13 MEASUREMENT OF IRON LOSS

CIRCUIT DIAGRAM

LINEAR VARIABLE DIFFERENRIAL TRANSFORMER

Exp.No: Date:

STUDY OF DISPLACEMENT TRANSDUCER

(Linear Variable differential Transformer)

AIM

To obtain the performance characteristics of Linear Variable differential Transformer (LVDT).

Find the residual voltage and non-electrical quantity displacement in terms of voltage.

REFERENCE1. A.K. Sawhney : A course in Electrical and Electronics Measurements and Instrumentation, Dhanpat Rai & Sons, 1984.2. H.S. Kalsi : Electronic Instrumentation, TMH, 1995.

BASIC KNOWLEDGE REQUIREDPrinciple of working of Linear Variable Differential Transformer and

different transducers.

APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 LVDT Trainer kit - 12 LVDT setup - 13 Multimeter (CRO) Electronic 14 Power Chord - 1

THEORYLinear Variable Differential Transformer (LVDT)

The linear variable differential transformer (LVDT) is a type of electrical transformer used for measuring linear displacement. The transformer has three solenoid coils placed end-to-end around a tube. The centre coil is the primary, and the two outer coils are the secondary. A cylindrical ferromagnetic core, attached to the object whose position is to be measured, slides along the axis of the tube.

An alternating current is driven through the primary, causing a voltage to be induced in each secondary proportional to its mutual inductance with the primary. The frequency is usually in the range of 1 to 10 kHz.

WORK SHEET

As the core moves, these mutual inductances change, causing the voltages induced in the secondary to change. The coils are connected in reverse series, so that the output voltage is the difference (hence "differential") between the two secondary voltages. When the core is in its central position, equidistant between the two secondary, equal but opposite voltages are induced in these two coils; so the output voltage is zero.

When the core is displaced in one direction, the voltage in one coil increases as the other decreases, causing the output voltage to increase from zero to a maximum. This Voltage is in phase with the primary voltage. When the core moves in the other direction, the output voltage also increases from zero to a maximum, but its phase is opposite to that of the primary. The magnitude of the output voltage is proportional to the distance moved by the core (up to its limit of travel). The phase of the voltage indicates the direction of the displacement because the sliding core does not touch the inside of the tube, it can move without friction, making the LVDT a highly reliable device. The absence of any sliding or rotating contacts allows the LVDT to be completely sealed against the environment.

PROCEDURE

1. Make the Connections for the given LVDT kit.

2. Calibrate the LVDT.

3. Place the core of the LVDT to 10 mm by adjusting the micrometer.

4. Gradually increase the micrometer displacement from 10mm to

20mm and note down the forward core displacement from zero mm

to 10mm on the display and measure the secondary output voltage

(mV) across T4 and T7.

5. Similarly, decrease the micrometer displacement from 10mm to zero

mm and note down the reverse core displacement of zero to 10mm

on the display and measure the secondary output voltage (mV)

across T4 and T7.

6. Tabulate the reading of the core displacement, micrometer

displacement and secondary output voltage (mV).

7. Plot the graph between core displacement (mm) along X axis and

secondary output voltage (mV) across Y axis.

8. When the displacement of the core is zero measure the voltage. This

voltage is the residual voltage.

TABULATION

MicrometerDisplacement(m

m)

Core Displacemen

t (mm)

Secondary Output Voltage

(mV)

MODEL GRAPH

Displacement (mm)

O/p

vol

tage

(m

V)

DISCUSSION QUESTIONS

1. Mention some of the transducers.

Variable Resistor, Variable inductor, Variable capacitor, Synchros

& Resolvers

2. State the advantages of LVDT.

The advantages of LVDT are(i) Linearity(ii) Infinite resolution(iii) High output(iv) High sensitivity(v) Ruggedness(vi) Less friction(vii) Less hysterices(viii) Less power consumption

3. State the disadvantages of LVDT?

The disadvantages of LVDT are(i) Large displacements are necessary for appreciable

differential output(ii) They are sensitive to stray magnetic field(iii) Dynamic response is limited by mass of core(iv) Variation in temperature affects the transducer.

Performance Record 05Viva voce 05Total 50

RESULTThus the characteristics of LVDT position sensor with respect to

the secondary output voltage is obtained.Thus, the residual voltage and non-electrical quantity displacement in

terms of voltage are found.

CIRCUIT DIAGRAM

89

+

-

SensorTimerState

+

-

+-

Gain Amplifi

erPSI

Constant

Voltage

Source

To excitation

Instrumentation Amplifier

R6

R5

R3

R2

R1

R4

T2

T3

T4

C3

+

_

Zero

R7

R9

T

1

R8

Exp.No: Date:

PRESSURE TRANSDUCER

AIM To draw the characteristics curve for a given Bourdon tube ie Pressure

Vs output (V or I) and measure the non electrical quantity pressure in terms of voltage (or) current.

REFERENCE

1. A.K. Sawhney: A course in Electrical and Electronics Measurements and Instrumentation, Dhanpat Rai & Sons, 1984.

2. H.S. Kalsi : Electronic Instrumentation, TMH, 1995.

BASIC KNOWLEDGE REQUIRED

Principle of working of pressure transducers, different types of pressure transducers

APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 Pressure Transducer Trainer

kit- 1

2 Multimeter (mV) Electronic 13 Pressure cell Setup - 14 Power Chord - 1

THEORY Pressure Transducer Most pressure measuring devices use elastic members for sensing pressure at the primary stage. These elastic members are of many types and convert the pressure into mechanical displacement, which is later converted into an electrical form using secondary transducers. These devices are many a time known as force summing devices. The commonly used pressure sensitive devices are described below:(i)Bourdon tubes:

Bourdon tubes are made out of an elliptically flattened tube bent in such a way to produce the below mentioned shapes. They are

a) C type b) spiral c) twisted tube and d) helical

Bourdon tube elements have several advantages and these include low cost, simple construction, high pressure range, good accuracy except at low pressure, and improved designs at the pressure for maximum safety. Their greatest advantage is that they easily adapted for designs for obtaining electrical outputs.

89

MODEL GRAPH

89

Test specimen

Active Gauge

Dummy GaugeR1

R2

Strain Gauge

SET UP

BLOCK DIAGRAM

Transducer Bridge

Calibration & Zeroing network

Measure-ment

DC Excitation

Source

Power supply

DC Network

e

Low Pass Filter

PROCEDURE

1. Install the pressure cell setup and interface the 9 pin D connector with

Pressure transducer trainer kit.

2. Connect the Multimeter (in milli volt mode) across T2 and T3 for bridge

voltage measurement.

3. Switch “ON” the module.

4. Initially, open the air release valve and exhaust the tank inlet air and

nullify the bridge voltage by using zero adjustment POT.

5. Now, close the opened air release valve by pressing the pump position,

the pump sucks the air from atmosphere and supply to the cylinder.

Pressure will be developed in the cylinder and now measure the bridge

voltage (mV) across T2 and T3.

6. Gradually increase the pressure by pressing the pump piston and note

down the bridge voltage (mV) for corresponding gauge pressure.

7. Tabulate the readings and plot a graph between gauge Pressure and

bridge voltage (mV).

89

TABULATION

S.No.Gauge Pressure

(psig)Displayed pressure(psi)

MODEL CALCULATION

89

DISCUSSION QUESTION

1. Define transducer?

It is a device which converts a non electrical quantity into an electrical

quantity

2. What is the pressure transducer?It is a device which converts the pressure into mechanical

displacement which is later converted in to electrical quantity using a secondary transducer.

3. Give commonly used pressure sensitive devices?The commonly used pressure sensitive devices are bourdon tubes,

bellows and diaphragms.

Performance 25Record 15Viva voce 10Total 50

RESULT

Thus the characteristics of the pressure cell with respect to bridge voltage

are plotted and the non electrical quantity pressure in terms of voltage or

current is measured.

89

AC BRIDGE- SCHERING’S BRIDGE

89

D AFO

R3

C1 C

2

R4 C

3

RX

Exp.No: Date:

AC BRIDGE- SCHERING’S BRIDGEAIM

To determine the value of the unknown capacitance and loss angle (δ) using low voltage Schering’s bridge.

REFERENCE1. A.K.Sawhney: A course in Electrical and Electronics Measurements and

Instrumentation, Dhanpat Rai & Sons, 1984. 2. H.S.Kalsi : Electronic Instrumentation, TMH, 1985.

BASIC KNOWLEDGE REQUIRED Principle of bridge circuits, loss angle, high voltage Schering Bridge for measurement of capacitance and low voltage Schering bridge for measurement of capacitance.APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 Schering bridge - 12 Decade capacitance

box- 1

3 Multimeter Electronic 14 Patch Chord - 1

5. CRO 1

THEORY

A very important bridge used for the precision measurement of capacitors and their insulating properties is the Schering Bridge. The standard Capacitor C2 is a high quality mica capacitor (low-loss) for general measurements or an air capacitor (having a very stable value and a very small electric field) for insulation measurement. Under balance condition,

R1+[1/jωC1]R4/[1+jωC4R4] = I/jωC2R3

R1+1/jωC1R = R3/jωC2[1+jωC4R4]

R1R4-[jR4/ωC1] = -[jR3/ωC2]+ [R3C4R4/C2]

C1=C2R4/R3

89

Equating real and imaginary terms, R1= R3C4/C2

TABULATION

89

SL.No.

Capacitor

C2 (µf)

R4

(KΩ)C4

(µf)R3

(Ω)

C1

Actual Value (µf)

C1

obtained

Value (µf)

% Error δ

FORMULAE

Two independent balance equations are balanced if C4&R4 are chosen are the variable element. C1= C2 ( R3 / R4) cos2δ Farad Where, R3 – Variable resistance (Ohm) R4 – Standard resistance (Ohm) C1 – unknown Capacitance (Farad) C2 – Standard Capacitance (Farad) Loss angle δ = tan-1(ω C4 R4) C4 – Variable Capacitance (Farad)

% Error = ((Actual Value – Obtained Value) / Actual Value) * 100Dissipation factor D1=tan δ=ωC1R1

=ω[C2R4/R3][R3C4/C2] =ωC4 R4

This bridge is widely used for testing small capacitors at low voltages with very high precision.

89

PROCEDURE1. Connections are made as per the connection diagram shown in fig.2. Connect the unknown capacitance at the C1 (unknown) point.3. Keep R4,R3 in minimum position.4. Connect the CRO across P and Q.5. Switch on the unit.6. Vary resistance R3 to some extent .(above 2K is suggested)7. Choose C2, Such that you can obtain the maximum variation the output.8. Vary the potentiometer R4 such that the amplitude of sine wave

decreases, reaches zero and then it will start increasing, at that point stop the tuning and vary R3 .Here also the amplitude of the sine wave will decrease and at one point it will obtain a minimum of zero amplitude and then it will start increasing, at that point stop the tuning.

9. Repeat the above step such that you will obtain minimum amplitude or zero amplitude.

10. Remove the patching at R3 andR4, find the resistance using the multimeter and note down the reading in the table given and calculate the value of unknown capacitance.

11. One can verify the balancing condition by connecting the bridge output (P&Q) to the input (P&Q) of audio power amplifier and you can hear a minimum noise or no noise .If you vary the potentiometer R4 you can hear a maximum noise.

WORK SHEET

89


1. How can we eliminate the error? Earthed screens are provided in order to avoid errors caused due to inter capacitance between high and low arms of the bridge.

2. Applications of Schering’s bridge? Used in measurement of capacitance, measurement of insulators, insulating coils.

3. What is the use of vibration galvanometer? They are used for power and low audio frequency range.

4. List out commonly used detectors for Ac Bridge. 1. Head phones 2. Vibration galvanometer

`Performance 25

89

Record 15Viva voce 10Total 50

RESULT Thus the value of the unknown capacitance and loss angle (δ) using low voltage Schering’s bridge are determined.

CIRCUIT DIAGRAM

AC BRIDGE-MAXWELL’S INDUCTANCE,CAPACITANCE BRIDGE

89

E

RX

LX

R3

R1

R2

C

D

C

Exp.No: Date:

AC BRIDGE-MAXWELL’S INDUCATNCE,CAPACITANCE BRIDGE

AIM To measure the unknown value of the inductance using Maxwell’s

Inductance Bridge and also to find the Q factor of the coil.

REFERENCE

1. A.K.Sawhney: A course in Electrical and Electronics Measurements and Instrumentation, Dhanpat Rai & Sons,1984.

2. H.S.Kalsi :Electronic Instrumentation,TMH,1995


89

Principle of bridge circuits, low frequency and high frequency inductance measurements.

APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 Maxwells Trainer kit - 12 Unknown

inductance- 1

3 Multimeter Electronic 14 CRO - 1

5 Patch chord - 1

THEORY Maxwell’s bridge measures an unknown inductance in terms of a known capacitor. The use of standard arm offers the advantage of compactness and easy shielding. The capacitor is almost a loss-less component. One arm has a resistance R1 in parallel with C1, and hence it is easier to write the balance equation using the admittance of arm 1 instead of the impedance.The general equation for bridge balance is

Z1Zx = Z2Z3Zx = Z2Z3/ Z1 = Z2Z3Y1

Where, Z1=R1 in parallel with C1 i.e. Y1=1/Z1

Y1=1/R1 + jωC1

TABULATION

S.No

R2

(KΩ)

R3

(KΩ)

RX

(KΩ)

RX (Ω) LX (Ω)

Actual Practical ActualPractica

l

89

MODEL CALCULATION

Z2=R2 & Z3=R3

Zx=Rx in series with Lx=Rx + jωLxFrom equations we have, Rx + jωLx = R2R3(1/R1 + jωC1)

89

Rx + jωLx = R2R3/R1 + jωC1R2R3Equating real terms and imaginary terms we have Rx = R2R3/R1 and Lx=C1R2R3Also Q = ωLx/Rx = (ωC1R2R3 * R1)/R2R3 = ωC1R1 Maxwell’s bridge is limited to the measurement of low Q values (1-10). The measurement is independent of the excitation frequency. The scale of the resistance can be calibrated to read inductance directly. The Maxwell’s bridge using a fixed capacitor has the disadvantage that there is an interaction between the resistance and reactance balances. This can be avoided by varying the capacitances, instead of R2 and R3, to obtain a reactance balance. However, the bridge can be made to read directly in Q. This bridge is particularly suited for inductance measurements, since comparison with a capacitor is more ideal than with another inductance. Commercial bridges measure from 1-1000 H, with + 2% error. (If the Q is very large, R1 becomes excessively large and it is impractical to obtain a satisfactory variable standard resistance in the range of values required)

PROCEDURE

1. Connections are made as per the connection diagram shown in fig.2. Connect the unknown inductance at the Lx (unknown) point.3. Connect the CRO across P and Q.4. Switch on the unit.5. Choose R3, such that you can obtain a maximum variation of output.6. Now set R2 to maximum position.7. Vary the potentiometer R4 such that the amplitude of sine wave will

decrease and at one point it will obtain a minimum of zero amplitude and then it will start increasing at that point stop the tuning and switch OFF the line.

8. Remove the patching at R1 and find the resistance using the multimeter and note down the reading in the table given below and calculate the value of unknown Inductance.

9. One can verify the balancing condition by connecting the bridge output (P&Q) to the input (P&Q) of audio power amplifier and you can hear a minimum noise or no noise .If you vary the potentiometer R1 you can hear a maximum noise

WORK SHEET

89

FORMULAE RX = R2 R3 / R4(Ώ) LX = R2 R3 C4(H)

89

Q factor=ω LX / RX

WhereLX = unknown InductanceRX =Effective resistance of inductance LX

R2, R3, R4 = Known non – Inductance resistanceC4 = Standard capacitance

DISCUSSION QUESTION

1. What are the advantages of Maxwell’s bridge? i) Simple to use. ii) Useful for measurement of a wide range of inductance at power and audio Frequency.

2. What are the disadvantages of Maxwell’s bridge? i) It requires a variable standard capacitor ii) The balancing adjustments becomes difficult

3. List out A.C Bridges i)Maxwell’s inductance bridge ii)Hay’s bridge iii) Schering’s bridge iv) Anderson’s bridge


RESULT

Thus the unknown value of inductance using Maxwell’s Inductance Bridge was determined and the Q factor of the coil was found.

CIRCUIT DIAGRAM

89

Wheat stone’s Bridge

Exp.No: Date:

89

D

P Q

R S

E

DC BRIDGE-Wheat stone BridgeAIM To determine the value of the given low resistance using Wheat stone Bridge

REFERENCE

1. A.K.Sawhney: A course in Electrical and Electronics Measurements and Instrumentation, Dhanpat Rai & Sons, 1984.

2. H.S.Kalsi : Electronic Instrumentation, TMH, 1985.


Principle and operation of bridge circuits

APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 Wheat stone Bridge Trainer

kit- 1

2 Decade resistance box (or) Resistance

- 1

3 Multimeter Electronic 14 Patch chord - 1

Wheat stone’s Bridge:

RX = (RSR1) / R2 (Ω)Where, RS – Standard resistance r - Load resistance RX – unknown resistance

% Error = ((Actual Value – Obtained Value) / Actual Value) * 100

TABULATION89

Wheat stone’s Bridge

SL.No R1 (Ω) R2 (Ω) R3 (Ω)RX (Ω)

% ErrorTheoretical

Practical

89

THEORYThese bridges are used not only for the measurement of

resistance but also used for measurement of various component values like capacitor and inductor etc. Bridge circuit in its simplest form consists of a network of four resistance arms forming a closed circuit. A source of current detector is connected to the two junctions. The bridge circuit uses the comparison measurement methods and operates on null-indication principle. The bridge circuit compares the value of an unknown component with that of an accurately known standard component. Thus the accuracy depends on the bridge component without the null detector. Hence high degree of accuracy can be obtained. In a bridge circuit when no current flows through the null detector which is generally a galvanometer, then the bridge is said to be balanced.Wheatstone bridge

A very important device used in the measurement of medium resistances is the Wheatstone bridge. A Wheatstone bridge has been in use longer than almost any electrical measuring instrument. It is still an accurate and reliable instrument for making comparison measurements and operates upon a null indication principle. The well known expression for the balance of Wheatstone bridge is as follows QR = PS If three of the resistance is known then the fourth may be determined from the eqn, R = S*(P/Q)

Where R is the unknown resistance, S is called the standard arm of the bridge and P and Q are called the ratio arms.

PROCEDURE1. Connection are made as per the circuit diagram2. Connect the decade resistance box at Rx terminal. (Or) connect resistance to be measured at Rx terminal3. Now switch on the unit and vary the resistance at R1 and R3 to get the nearest point of balance.4. Now vary R2 to get exact point of balance.5. Switch off the unit and remove the patching at R2.

6. Now measure the resistance at R2 by using multimeter7. Tabulate the readings and find the value of unknown resistance.

Performance 25

Record 15

Viva voce 10

Total 50

RESULT

89

Thus the value of given resistance was determined using Wheatstone bridge.

CIRCUIT DIAGRAM

KELVINS DOUBLE BRIDGE

89

A

D

++

Rb

RS

c

bR

X

R2

R4

R1

DRB

R3

DRB

aA B

C

EX NO: DATE:

DC BRIDGE-KELVINS DOUBLE BRIDGE

AIM: To find the value of unknown resistance using a Kelvins Double Bridge.

APPARATUS REQUIRED

SL.NO APPARATUS RANGE QUANTITY1 Kelvins BridgeTrainer

kit- 1

2 Multimeter Electronic 13 Unknown resistance - 14 Patch chord - 1

FORMULAE

RX = (RSR1) / R2 + R4 r (R1 / R2 – R3 / R4) / (R3 + R4 + r) (Ω)

Theory

The Kelvin Bridge is a modification of the Wheatstone bridge and

provides greatly increased accuracy in measurement of low value resistances.

An understanding of the Kelvin bridge arrangement may be obtained by a

study of the difficulties that arise in a Wheatstone bridge on account of the

resistance of the leads and the contact resistances while measuring low

valued resistors.

PROCEDURE

1. Connection are made as per the circuit diagram

2. Connect the unknown resistance at Rx terminal.

3. Switch on the unit.

4. Select the range selection switch at the point where the meter reads least

possible value of voltage.

5. Vary the potentiometer (P1) to obtain null balance..

6. Switch off the unit and find the resistance using multimeter at P1.

89

7. Tabulate the reading and find the value of unknown resistance using above

the formula.

TABULATION:

Kelvin’s Double Bridge

SL.No R1 (Ω) R3 (Ω)RX (Ω)

% ErrorTheoretical

Practical

MODEL CALCULATION

89

DISCUSSION QUESTION

1. What are the advantages of bridges?The measurement accuracy is high as the measurement done by

comparing the known & unknown value. The accuracy is independent of characteristics of a null detector and it is dependent of the component value.

2. What is meant by balanced condition for Wheatstone bridge?The bridge is said to be balanced when there is no current flow through

the galvanometer so potential difference across the galvanometer should be zero R1 R4 = R3R2

3. What is the sensitivity of Wheatstone bridge?Sensitivity = Deflection (D)/ Sensitive current ( I)

4. What is meant by Kelvin’s bridge?For measuring the value of resistance below 1Ω the modified form of

Wheatstone bridge is called as Kelvin’s bridge.

5. What is Kelvin double bridge? It consists of another set of arms hence it is called as double bridge.


RESULT

Thus the value of given resistance was determined using Kelvins double bridge.

89

CIRCUIT DIAGRAMINSTRUMENTATION AMPLIFIER

MODEL GRAPH

Gain R1=10

89

V1

+

-

A1

R3

V2

R1

R2

R4

+

+

-

-

A3

A2

V0

o/p

volt

age

I/P voltage

Exp.No: Date:

INSTRUMENTATION AMPLIFIER

AIMTo Study the working of an Instrumentation amplifier.



Principle of working of Instrumentation amplifier.

APPARATUS REQUIREDSL.NO APPARATUS RANGE QUANTITY

1 Instrumentation amplifier Trainer kit

- 1

2 Multimeter Electronic 13 External millivolt

source- 1

THEORYIn a number of industrial and consumer applications, one is

required to measure and control physical quantities. Some typical examples are measurement and control of temperature, humidity, light intensity, water flow etc. These physical quantities are usually measured with the help of transducers. The output of transducers has to be amplified so that it can drive the indicator.This function is performed by an instrumentation amplifier.

Many of the input specification of an Op-amps employed directly determine the input specifications of the instrumentation amplifier.

An analysis of the circuit gives the following equation:Let R1 = R2 = R3 = R4

Considering the basic differential amplifier shown in the figure, the output voltage V0 is given by

V0 = - R2/ R1 V2 + 1/1 + R3/R4 V1 (1+ R2/R1) Or

V0 = R2/ R1 (V2 – 1/1 + R3/R4 (R1/ R2+1) V1V1)

V0 = -R2/R1 V2 + 1/ 1+R3/R4 V1 (1+ R2/R1)

89

V0 = -Rf / Rin (V1 – V2)

TABULATION

Gain R1= 10

S.noInput Voltage

(Vin)Output Voltage

(Vo)Gain=Vo/Vin

89

The Op-amp A1 and A2 have differential input voltage as Zero. For V1 = V2 that is, under common mode condition, the voltage across R will be zero. As no current flows through R and R1 the non-inverting amplifier A1 acts as voltage follower having output V11 and V1. However, If V1 ≠ V2 current flows in R and R2 and (V2-V1).

The gain of an instrumentation amplifier can be varied by changing R1 alone. High gain accuracy can be obtained by using precision metal film resistors for all the resistances.

Because of the large negative feedback used, the amplifier has good linearity typically about 0.01% for a gain less than 10. The output impedance is also low being in the range of milliohms.

The input bias current of the instrumentation amplifier is determined by that of the amplifiers A1 and A2.

Features

The important features of an instrumentation amplifier are1. High gain accuracy and linearity2. High CMRR3. High gain stability with low temperature coefficient.4. Low dc offset5. Low output impedance

The instrumentation amplifier is also called as Data amplifier.

The expression for its voltage gain is generally of the form,A = (V0/V2) – V1

Where V0 = output of the amplifier V2-V1 = differential input is to be amplified.

Requirements of a good instrumentation amplifier

1. Finite, accurate and stable gain2. Easier gain adjustment3. High input impedance4. Low output impedance5. High CMRR6. Low power consumption7. Low thermal and time drift8. High Slew rate

89

WORK SHEET

89

PROCEDURE

1. Switch ON the Instrumentation amplifier unit. Switch SW1 should be

in internal mode.

2. Select the gain of 10 i.e, Switch SW2 should be in R1 mode.

3. Connect the multimeter in millivolt mode across the T1 and T2.

4. Calibrate the unit by using the mV source POT and zero adjustment

POT.

5. When input is zero, display voltages are brought to zero by varying

the zero adjustment POT.

6. After the completion of the calibration, start the experiments.

7. Set the input (say 40 mV) by varying the mV source POT.

8. Measure the output voltage across T5 and GND or from the display.

9. Analyse the output for various input signal.

DesignAn analysis of the circuit gives the following equation:

Let R1 = R2 = R3 = R4 Considering the basic differential amplifier shown below, the

output voltage V0 is given by V0 = - R2/ R1 V2 + 1/1 + R3/R4 V1(1+ R2/R1)

Or,V0 = R2/ R1 (V2 – 1/1 + R3/R4 (R1/ R2+1)V1V1)

V0 = -R2/R1 V2 + 1/ 1+R3/R4 V1 (1+ R2/R1)

89

V0 = -Rf / Rin (V1 – V2)

WORK SHEET

89


1. What is the need of instrumentation amplifier?The low level signal outputs of electrical transducers often

need to be amplified before further processing. This is done by the use of instrumentation amplifier.

2. What are the advantages of instrumentation amplifier?It has low level signal amplification, low noise, low thermal and time Drifts, high common-mode rejection ratio and high slew rate.

3. What are the applications of Op-Amp?The applications are categorized as linear applications,

filter and oscillator applications, comparator and detector applications, special integrated circuit applications and selected system applications.

4. State some linear applications of Op-Amp?In linear circuits, the output signal varies with the input

signal in linear manner. The linear applications are adder, subtractor, Instrumentation amplifier, power amplifier, V-I converter, I-V converter, analog computation, power amplifier etc.


RESULT

Thus an instrumentation amplifier was studied.

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CIRCUIT DIAGRAMA/D CONVERTER

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Control & Timer

SAR

SWITCH TREE

256 R-2R Ladder

Network

VCC

GND

Address latch buffer

8 CHANNEL MUX

ANALOG SWITCHES

Comparator

+

-

STATE OUTPUT LATCH BUFFER

Ref (+))

Ref (-))

SW1SW2SW3

Channel8 Bit Output

Exp.No: Date:

A/D CONVERTER AND D/A CONVERTERAIM

To design and test a four bit A/D converter and D/A converter.


BASIC KNOWLEDGE REQUIREDBasic theory and operation of A/D and D/A and its types

APPARATUS REQUIRED SL.NO APPARATUS RANGE QUANTITY

1 A/D converter, D/A converter Trainer kit

- 1

2 CRO with probe - 13 1 K potentiometer to vary

the input signal- 1

THEORYDIGITAL TO ANALOG CONVERSION

It involves conversion of digital information into equivalent analog information. Digital to analog converter (DAC) acts as a decoding device since it operates on the output of the system. DAC are of two types, Binary weighted resistor type & R-2R ladder type.

R-2R ladder DAC:In this type, the reference voltage is applied to one of the switch

position and the other switch position is connected to ground. The typical values of resistors range from 2.5kΩ to 10kΩ. Let us consider 3 bit R-2R ladder DAC with binary input 001. The output voltage will be VR/ 8, is equivalent to binary input 001.

ANALOG TO DIGITAL CONVERSIONThe analog information is converted into equivalent binary number

in the digital form. Analog to digital converter (ADC) acts as an encoder. The types of ADCs are 1) single slope, 2) Dual slope, 3) successive approximation, 4) Flash type, 5) Delta modulation and 6) Adaptive delta modulation. In this type most frequently used method is successive approximation. Successive approximation:

In this type the basic idea is to adjust the DAC’s input code such that its output is within ±1/2 LSB of the analog input VI. The circuit uses Successive Approximation Register (SAR) to find the required values of each bit by trial and error.

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D/A CONVERTER

89

R

2R

R

Rf

2R2R

2R

-

+

DIGITAL INPUT

LSBMSB

5 6 7 8 9 10 11 12

5K

5K

10K 10K

10V

IOUT

VOUT

GND

3 16

0.1μF

0.1μF

0.01μF

113

10V

V+V-

PROCEDURE

D/A Converter

1. Switch on the Power supply.2. The jumpers J9 to J16 should be in the s/w (right) position.3. The switches sw1 through sw8 are placed approximately to

represent the desired output.4. For example if the input is 4.96v then the switch positions are

as follows.

SW1 SW2 SW3 SW4 SW5 SW6 SW7 SW8 Hex Value

1 1 1 1 1 1 1 0 FEh

5. The output voltage can be observed by using CRO at the terminal pin P2

A/D Converter

1. The power supply is Switched on .2. The Variable terminal of the potentiometer is given to analog

input channel 2.3. To select the analog input channel 2, the channel select switch

position is as follows.

4. The start of conversion (soc) button is pressed. Once to start the Conversion from analog signal to digital form. The LED 9 glows on pressing start of Conversion button.5. The Address latch enable (ALE) button is also pressed once, so as

to enable the digital data to be sent to the output.6. The digital output for the corresponding analog input is displayed

on the LED’s do through D7. For analog input of 4.92V, the digital

output is given as.

D7 D6 D5 D4 D3 D2 D1 D0 Hex Value

1 1 1 1 1 1 0 1 EE

7. The end of conversion is indicated by the LED L10.8. The procedure is repeated for different values of analog voltage.

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SW1 SW2 SW30 1 0

TABULATION

Digital to Analog Conversion

Sl.No

B7 B6 B5 B4 B3 B2 B1 B0Hex

ValueAnalog

O/p

Analog to digital Conversion

Sl.No

Analog I/p

B7 B6 B5 B4 B3 B2 B1 B0Hex

Value

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DISCUSSION QUESTION

1. What are the types of D/A converter?• Binary weighted resistor type • R-2R ladder type.

2. What are the advantages of R-2R ladder D/A converter?

The number of bits can be expanded by adding more section of same R-2R values. It is easier to build accurately as only two precision metal film resistors are required

3. What are the uses of D/A converter?D/A converter are used in computer drives, CRT displays, digital generation of analog of analog waveforms and digital control of automatic process control systems

4. What are the types of A/D conversion?• Successive approximation method• Voltage to time conversion method• Voltage to frequency conversion method• Dual slope integration method

5. What is the use of A/D conversion device?The data to be fed to digital devices normally appears in analog form. Therefore analog to digital conversion devices are used where digital output is needed.


RESULT

Thus the analog output voltage from digital input and digital output from analog input were obtained.

89

Exp.No: Date:

STUDY OF TRANSIENTSAIM

To study the transients of DC circuits and AC circuits.

REFERENCE

1. M.Arumugam and N.Prem Kumar – Electrical circuits Theory, Khanna Publishers, Newdelhi.

2. B.L. Theraja – Fundamentals of Electrical and Electronics, S.Chand and Company Ltd, New delhi.


• Basic concepts of DC and AC circuits

• Basic concepts of RL, RC and RLC transients

THEORY

Transient phenomenon is a periodic function of time and doesn’t last longer. The duration of which they last is very significant as compared with operating time of the system. But they are very important because depending upon the reversibility of the transients, the system may result in blocked condition.

REQUIREMENT OF TRANSIENT IN THE CIRCUIT

1. Either inductor or capacitor or both should be present.2. A sudden change in the parameter as the form should occurs as a fault or any switching operation.a) The following are the simple 3 facts which are the fundamental to the phenomenon of transients in electrical power systems.b) The current can’t change instantaneously through any inductor.c) The voltage across a capacitor can’t change instantaneously.3. The law of conversion of energy should hold good.

DC TRANSIENT RESPONSE OF RLC CIRCUIT

The resistance, inductance and capacitance are connected in series. The capacitor and inductor are initially unchanged and are in series with the resistor. When switch is closed at t=0, we can determine the complete solution for current.

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Applying KVL,

V=iR+Ldi/dt +1/C∫idt

Differentiating above equation

WORK SHEET

89

0=Rdi/dt+Ld²i/dt²+(1/C)i

It is a second order differential equation,

D²+(R/L)D+1/LC=0

The roots are

Dı,D2=-(R/2L)± [(R/2L)²-1/LC]½

By assuming,

Kı=R/2L, K2=[(R/2L)²-1/LC]½

Dı=Kı+K2, D2=Kı-K2

Here,K2 may be positive, negative or zero

K2 is positive when (R/2L)²>1/LC

The roots are real and unequal and give the overdamp response

[D-(Kı+K2)][D-(Kı-K2)]i=0

The solution is i=cıе^(Kı+K2)t+ c2е^(Kı-K2)t

K2 is negative when (R/2L)²<1/LC

The roots are complex conjugate and give the underdamped

response

[Dı-(Kı-jK2)][D2-(Kı-jK2)]=0

Solution is given by

i=е^(Kı*t)[cıcos K2+c2sin K2t]

K2 is zero,where (R/2L)²=1/LC

Solution is given by i=е^Kı+(Cı+C2)t

SINUSOIDAL RESPONSE OF RLC CIRCUIT

Switch ‘S’ is closed at t=0,a sinusoidal voltage V(cosωt+θ) is

applied to RLC series circuit where V=amplitude of the wave,

θ=phase angle.

Applying the KVL,

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V(cosωt+θ)=Rı+Ldi/dt +1/C∫idt

Differentiating the above equation,

Rdi/dt+Ld²i/dt²+i/C=-Vωsin(ωt+θ)

[ D²+(R/L)D +(1/LC)]i=-(Vω/L)(sinωt+θ)

WORK SHEET

89

solving above equation, we get

ip=[Vω²(R/L²)*cos(ωt+θ)]/[(ωR/L)² -(ω²-1/LC)²] + [(ω²-1/LC)

Vmsin(ωt+θ)]/L[((ωR/L)²-(ω²-1/LC)²]

TO FIND M & ø

Msin ø/Mcos ø=tan ø=[ωL-(1/ωC)]/R

ø=–tanˉ¹ [ωL-(1/ωC)]/R

Squaring both equations

M²cos ²ø+ M²sin ²ø=V²/R²[(1/ωc-1/ωL)²]

ip=V/[R²+(1/ωc-ωL)²]½ cos [ωt+θ –tanˉ¹(1/ωc-ωL)/R]

Dı,D2=-(R/2L)± [(R/2L)²-1/LC]½

Dı=Kı+K2, D2=Kı-K2

K2 is positive when (R/2L)²>1/LC

ic= е^(Kı*t)[cıcos K2t+c2sin K2t]+V/[R²+(1/ωc-ωL)²] *cos[ωt+θ –

tanˉ¹(1/ωc-ωL)²]

ic= [е^(Kı*t)]*(cı+c2)t

i= е^(Kı*t)(cı+c2)t+ V/[R²+(1/ωc-ωL)²]½ *cos[ωt+θ +tanˉ¹(1/ωcR-

ω/R)²]


RESULT

Thus the transient in DC and AC circuit for an RLC circuit is

studied.

89

Exp.No: Date:

CALIBRATION OF SINGLE-PHASE ENERGY METER

AIM

To calibrate the given single-phase energy meter by direct loading .


APPARATUS REQUIRED

Sl.No

Apparatus RangeType Quanti

ty

1 Voltmeter (0-300V) MI 1 No.

2 Ammeter (0-10A) MI 1 No.

3 Wattmeter 300V,10A UPF 1 No.

4 Test energy meter1Ø,230V,5-30A

-1 No.

5 Connecting wires - - Require

d

THEORY

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Direct loading

In this method, precision grade indicating instruments are used

as reference standard. These indicating instruments are connected in

the circuit of meters under test. The current and voltage are held

constant during the load test. The number of revolutions made by the

meter disc and the time taken during the test are recorded.

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WORKSHEET

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PRECAUTIONS

i. Auto transformer is kept at minimum position at the time of starting.ii. Rheostat is kept at minimum position in phantom loading.iii. Phase shift transformer is kept at UPF position

PROCEDURE

Direct loading1. Make the circuit connection as per the circuit diagram.

2. Close the DPST switch.

3. Adjust single phase auto transformer till the voltage connected across

the primary winding reads rated primary voltage.

4. Vary the resistive load to vary the load current.

5. Note down readings of time taken for the Energy meter for 5

revolutions, Wattmeter, ammeter and voltmeter.

6. Repeat the same procedure for various load conditions.

7. Calculate percentage error and draw the graph between percentage

error and load current.

FORMULAE

% 100%Actual Trueenergy

ErrorActualenergy

−= ×

100%

3600 1000No. of revolution made by energy meter

Actual energy =meter constant

Wattmeter reading TimetakenTrue energy in kWhr

×= ××

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TABULATION

Direct loading

Sl.No.

Voltage

(Volts)

Current

(Amps)

Wattmeter

Reading (Watts)

Time taken for

5 revolutio

ns

Actual Energ

y

True energ

y

%Error

MODELGRAPH

89


1. What is creeping in energy meter?

It is a slow continuous rotation, when there is no current flows through the current coil and only pressure coil is energized. This behavior is called creeping.

2. What is the provision available in energy meter for adjusting

creeping?

(i)Two diametrically opposite holes are drilled in disc.A small piece of iron is attached to the edge of the disc. The force of attraction exerted by the brake magnet on the iron piece varies the creeping.

3. What are the provisions available in low power factor

measurement energy meter?

(i) Adjustable resistance (ii) shading bands

4. What is calibration and why is it needed for instruments?

It is the procedure for determining the correct value of measured quantity by comparison with the standard one. In order to determine the standards of the instrument , it should be calibrated.


RESULT

Thus the single phase energy meter was calibrated using direct loading method.

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CIRCUIT DIAGRAMStar Connected Load

89

W2

400 Ω / 2A

M L

C V

R

B

440 V

3

50 Hz

AC

Suppl

y

TPSTS

Y

V

A

M L

CV

W1

V400 Ω / 2A

400 Ω / 2A

N

Exp.No: Date:

MEASUREMENT OF THREE-PHASE POWER AND POWER FACTOR

AIMTo measure the power and power factor in three-phase circuit star

connected, Delta connected load & to check the relationship between line and

phase quantity.

REFERENCE

1. A.K. Sawhney : A course in Electrical and Electronics Measurements and

Instrumentation, Dhanpat Rai & Sons, 1984.2. H.S. Kalsi : Electronic Instrumentation, TMH, 1995.


1. Basics of three-phase power and power factor2. Basics of Star and Delta connections.

APPARATUS REQUIRED

Sl.No

Apparatus RangeType Quantit

y

1 Voltmeter (0-600V) MI 1

2 Ammeter (0-10A) MI 1

3 Wattmeter 600V,10A UPF 1

4 Rheostat 5 KW - 1

5 Connecting wires - - Required

THEORY In a three phase, three wire system, we require three elements. But

if we make the common points of the pressure coils coincide with one of the

89

lines, then we require only two elements. Instantaneous power consumed by

load =V1i1+V2i2+V3i3

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WORKSHEET

89

Star Connection Instantaneous reading of wattmeter is P1 and the instantaneous reading ofW2 is P2.Sum of instantaneous readings of two wattmeters =P1+P2.Sum of instantaneous readings of two wattmeter = V1i1+V2i2+V3i3.Therefore, the sum of the two wattmeter reading is equal to the power consumed by the load. This is irrespective of whether the load is balanced or unbalanced.

Delta Connection Here, by means of Kirchoff’s voltage law, sum of instantaneous readings of two wattmeter = V1i1+V2i2+V3i3.Therefore the sum of the two wattmeter readings is equal to the power consumed by the load. This is irrespective of whether the load is balanced or unbalanced.

Total power consumed by load = P1 +P2.

Power factor, cosΦ = cos [tan-1 3 ( P1 -P2)/ ( P1 +P2.)] With unity power factor, P1= P2= (3/2) VI With 0.5 power factor, P1= (3/2) VI, P2=0. With zero power factor, P1= 3 /2VI& P2= (- 3 /2) VIFORMULAE

3 cosL LPower V I φ= (Watts)

where,VL = Line voltage (Volts)

IL = Line current (Amps)cosφ = Power factor

cosφ = / 3 L LPower V I

For Star connected load, For Delta connected load,IL = Iph, IL = 3 Iph,

VL= 3 Vph, VL= Vph,

PROCEDURE1. Make the circuit connection as per the circuit diagram.2. Close the TPST switch.3. Note down the Wattmeter readings W1 and W2.4. know the Multiplication factor, calculate the power. 5. Note down the line voltage and phase voltage using voltmeters.6. Note down the line current and phase current using ammeters.

7. By using the above readings calculate the power and power factor

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TABULATION M.F=

Type of Connectio

n

Voltage (V)Current

(I)Wattmeter

readingPower factor cosφVph

(V)VL

(V)Iph

(A)IL (A)

W1

(W)W2

(W)

Star

Delta

MODEL CALCULATION

89

DISCUSSION QUESTIONS1. What do you mean by power factor?

The cosine of phase angle between the voltage and current is called the power factor

2. What are the methods for measuring power?a. The methods for measuring power are b. (i)Single wattmeter method (ii) Two wattmeter method and (iii) Three Wattmeter method

3. What is the relation between the line and phase quantities in delta connection?

VL=VPH, IL= 3 IPH Where, IL = Line current

V L = Line voltage , VPH = Phase voltage IPH = Phase current

4. What are the advantages of three phase system?i. Three phase transmission line requires less number of conductor material than single phase line for transmitting the same amount of power and voltage.

ii. Parallel operation is simple & power factor is high.iii. Cost wise per unit of output in three phase machine

is very much cheaper.

5. What is energy?Energy is the total power delivered or consumed over a time

interval. Its unit is Kilowatt hour (KWH).Performance 25Record 15Viva voce 10Total 50

RESULT Thus the relationship between phase & line quantities for star and delta connected loads are verified in three phase connection.

CIRCUIT DIAGRAM

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CURRENT TRANSFORMER

Loading Rheostat P

Ammeter

Voltmeter

N Current Transformer

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V

A

V

30 V AC Supply

Exp.No: Date:

CALIBRATION OF CURRENT TRANSFORMER

AIM To study and calibrate current transformer parameters and to draw the curve primary current and Vs Secondary current.

REFERENCE

1. A.K. Sawhney : A course in Electrical and Electronics Measurements and Instrumentation, Dhanpat Rai & Sons, 1984.2. H.S. Kalsi : Electronic Instrumentation, TMH, 1995.


Principle of working of current transformer

APPARATUS REQUIRED

S.No

Name of the Apparatus Range Quantity

1Current Transformer Trainer kit

-1

2 Rheostat 500 Ω,3A 1

3 Loading rheostat 5 kW 1

4 Patch cords - 1

FORMULAERatio error or Current error (%) = 100( KN IS -IP)

IP

KN = Primary winding current

Secondary winding current

Phase Angle error ө = m180 I

pIπ

Im = p

s

I

nI

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TABULAR COLUMN

S No Supply Voltage(V)

Primary Current( IP)

Secondary Current( IS)

Ratio Error KN

Phase angle Error ө

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THEORY A current transformer is an instrument transformer specially designed and assembled to be used in measurement control and protective circuits. Its primary consists of few turns and is connected in series with the circuit whose current is desired to be measured and the secondary is connected to the current measuring instrument. The secondary circuit is closed through the typical low impedance of the instruments connected to it. These are 5 A instruments. The voltage across secondary is the drop through the instruments and loads and usually is only 5 volts.

In ideal CT, the secondary current is inversely proportional to the ratio of turns and opposite in phase to the impresses primary current. The exciting current must be subtracted phasorially from the primary current to find the amount remaining to supply Secondary current. This value will be slightly different from the value that the ratio of turns would indicate and there is slight shift in phase relationship. This results in introduction of ratio and phase angle errors when compared to ideal CT.

PROCEDURE

The calibration of current transformer operation is under two modes (ie.Low voltage and high voltage)

Low Voltage: (ie.30 V)

1. Connect the circuit as per the circuit diagram2. Switch on the kit with rheostat at minimum position.3. Load the CT by using 500 Ω, 3A rheostat.4. Now note down the primary ( IP) and secondary current ( IS) of the transformer.5. Tabulate the readings and calculate calibration parameters.

High Voltage: (ie.230 V)

1. Connect the circuit as per the circuit diagram2. Switch on the kit and switch on the MCB.3. Keep the rheostat in the maximum position.4. Load the CT by using loading rheostat 5k W.5. Now note down the primary ( IP) and secondary current ( IS) of transformer.6. Tabulate the readings and calculate calibration parameters

89

MODEL CALCULATION

89

DISCUSSION QUESTIONS1. Define current transformer. A current transformer (CT) is a type of instrument transformer designed to provide a current in its secondary winding proportional to the alternating current flowing in its primary.2. How is current transformer designed?

The most common design of CT consists of a length of wire wrapped many times around a silicon steel ring passed over the circuit being measured. The CT's primary circuit therefore consists of a single 'turn' of conductor, with a secondary of many hundreds of turns. Common secondaries are 1 or 5 amperes. 3. Mention uses of current transformer.

Current transformers are used extensively for measuring current and monitoring the operation of the power grid. The CT is typically described by its current ratio from primary to secondary. Often, multiple CTs are installed as a "stack" for various uses (for example, protection devices and revenue metering may use separate CTs).4. Mention precautions to be followed while using current transformer.

Care must be taken that the secondary of a current transformer is not disconnected from its load while current is flowing in the primary, as this will produce a dangerously high voltage across the open secondary, and may permanently affect the accuracy of the transformer.5. What is calibration? Mention the need for calibration. Calibration is a measurement process that assigns values to the response of an instrument relative to reference standards or to a designated measurement process. The purpose of calibration is to eliminate or reduce bias in the user's measurement system relative to the reference base. The calibration procedure compares an "unknown" or test item(s) or instrument with reference standards according to a specific algorithm.6. Mention calibrated parameters of current transformer Ratio error and Phase Angle error


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RESULT Thus the given current transformer was calibrated. The Calibrated parameters

are ratio error & phase angle error and the primary Vs Secondary current was drawn.

Exp.No: Date:

MEASUREMENT OF IRON LOSS

AIM: An Ac bridge method is employed for measurement of core losses in ferromagnetic material

THEORY:Any bridge capable of measuring the impedance of any iron core

coil could be used for this.The Maxwell’s wein bridge circuit or Maxwell’s inductance

capacitance bridge is used for the measurement of core loss.La = unknown inductance.Rd = effective resistance of inductance.Ra, Rb, Rc = known iron – inductance resistance.Cb = variable standard capacitor.

Rb (Rd+j wld) [ ] = Ra Rc (or) Z 1 Z4 = Z3 Z2

I+jwCbRb

Ra Rb + j w L d =Ra Rc + j w Ra Rc Cb Rb

In order to use this bridge circuit for the measurement of power loss, the primary of the test frame is connected in bridge. The maximum flux density is calculated by

Eqvg = 4 f B max NA x 10-6

Where Eqvg = average absolute value of abc voltage. F = frequency. Bmax = max flux density N = numbers of turns A = cross sec area of ferromagnetic specimen.

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Balance at fundamental frequency is obtained by adjusting Rb + cb so that the deletor indicates.

If the induction in ferromagnetic specimen is low and if the o/p voltage of power source has a negligible amount of

WORK SHEET

89

Exp.No: Date:

MEASUREMENT OF IRON LOSSAIM:

To measure and study the iron loss of given ring specimen.

APPARATUS REQUIREDS.No

Name of the Apparatus Range Quantity

1Iron loss measurement trainer kit

-1

2 Digital multimeter - 1

3 Microphone - 1

4 Patch chords - 1

THEORY:Any bridge capable of measuring the impedance of any iron core

coil could be used for this.The Maxwell’s wein bridge circuit or Maxwell’s inductance

capacitance bridge is used for the measurement of core loss.La = unknown inductance.Rd = effective resistance of inductance.Ra, Rb, Rc = known iron – inductance resistance.Cb = variable standard capacitor. Rb (Rd+j wld) [ ] = Ra Rc (or) Z 1 Z4 = Z3 Z2

I+jwCbRb

Ra Rb + j w L d =Ra Rc + j w Ra Rc Cb Rb

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In order to use this bridge circuit for the measurement of power loss, the primary of the test frame is connected in bridge. The maximum flux density is calculated by Eqvg = 4 f B max NA x 10-6

Where Eqvg = average absolute value of abc voltage. F = frequency. Bmax = max flux density N = numbers of turns A = cross sec area of ferromagnetic specimen.

Balance at fundamental frequency is obtained by adjusting Rb + cb so that the deletors indicate.

FORMULA:

At balance condition:

Unknown Resistance

RS = Std. R1 x Std.R3 x C

For the value of R3 select Std. Resistance according to the specimen’s inductance. Say low value of inductance in specimen needs a lower value of std.r3. In our unit the Std. resistance R3 values to be chosen are 10Ω/100Ω/1KΩ.

Unknown Resistance

RS = Std. R1 x Std.R3 R3

Where: R2 – Std. Resistance measured by using multimeter across pot2.Iron loss = IL2 x (RS – R W)

Where: I1 – Current floe to the specimen in ampere (A).RS Specimen Resistance.RW Winding Resistance (measure by using multimeter)C – Std. Capacitor (0.1µF)

PROCEDURE:

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1. Connections are made as per the connection diagram fig 1.2. Connect the ring specimen to the bridge arm, for which

measurement to be made.3. Keep the POT 2 in maximum position and switch on the unit.4. The output can be detected either by microphone (or) CRO /

multimeter.5. For detecting the output vary the POT 1 from lower to higher

value. At one stage the output goes to minimum value. (ie, bridge become balanced or current flow through detector is zero (or) minimum.

6. Now note down the Resistance of the POT 1 by using multimeter.7. In this condition note down the A.CX current though ring

specimen (I1), value of POT 1 and the Source current by using milli ammeter (2 or 200mA range selection).

8. Substitute these values in an approximated formula and find out the iron loss of the given ring specimen.

9. Similarly repeat the same procedure for the given three ring specimens.

RESULT:

The iron loss of the given ring specimen is measured.

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TABULAR COLUMN

SL.No

Inductance (LS )in mH Resistance (RS) in ohms

Current (I1) in mA

(AC)

Iron LossTheoretical

valuePractical

value

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PHANTOM LOADING

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CIRCUIT DIAGRAM

SINGLE PHASE ENERGY METER

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47512039 m-i-lab-manual

Education