electrical and electronics lab

8/9/2019 Electrical and Electronics Lab

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ELECTRICAL AND ELECTRONICS LAB

LIST OF EXPERIMENTS

S.N

ONAME OF THE EXPERIMENT

PAGENo.

1 A. THEVENINS THEOREM.NORTANS THEOREM

1

4

2MAXIMUM POWER TRANSFER THEOREM 7

3 A. SUPER POSITION THEOREMB. RECIPROCITY THEOREM

10

13

4SERIES AND PARALLEL RESONANCE 16

5 OPEN CIRCUIT CHARACTERISTICS OF DC SHUNTGENERATOR

21

6SWINBURNES TEST

24

7BRAKE TEST ON DC SHUNT MOTOR

30

8 OPEN CIRCUIT AND SHORT CIRCUIT TESTS ON A SINGLEPHASE TRANSFORMER 33

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THEVENINS THEOREM

Exp. No: 1(A)

AIM:

To verify Thevenins Theorem for given circuit experimentally

APPARATUS:

S. NO EQUIPMENT SPECIFICATION QUANTITY1 VOLTAGE SOURCE 0-30V 12 RESISTORS 2.2K 33 RESISTORS 1K 2

4 Variable Resister 560 15 AMMETER ( DMM ) 0-20mA 1

6 VOLTMETER ( DMM 0-15V 17 CONNECTING WIRES

THEORY:

STATEMENT:

It states that any linear active two terminal network containing resistance and voltage

sources and /or current sources can be replaced by single voltage source Vth in series with a

single resistance Rth. The thevenin equivalent voltage Vth is the open circuit voltage at the

network terminals, and thevenin resistance Rth is the resistance between the network terminals

when all the sources are replaced by their internal resistances.

EXPLANATION:

a) Steps to find the Rth

Open circuit current sources and short circuit voltage sources.

Open circuit the load resistance

Find out the equivalent resistance which is Rth across the open circuit

terminals.

b) Steps to find the Vth


Find out the open circuit voltage Vth across the open circuit terminals by using

mesh analysis.

c) Find out the load current ILby connecting the load resistance to the given circuit.

d) Draw the equivalent circuit. Connect the voltage source Vth in series with Rth and RL .

CIRCUIT DIAGRAM:

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Fig.1

Fig.2

PROCEDURE:

1. Connect the circuit as shown in fig 1.

2. Adjust the input voltage to 4V.

3. Calculate Vth theoretically.

4. Find open circuit voltage Vth across AB using voltmeter with out connecting RL.

5. Connect the load resistance RL and measure load current I L through branch AB using

ammeter.

6. Short the voltage source and open the terminals A and B.

7. Measure the resistance (Rth) seen through the open circuit terminals AB using DMM.

8. Connect Thevenins equivalent circuit (shown in figure 2).

9. Measure the load current ILI and tabulate the readings.

10. Compare IL , ILI and observe both are equal.

11. Repeat the procedure for different voltage values.

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OBSERVATIONS:

S.No Vs (v) R th () Vth(v) IL(mA) IL(mA)

THEORETICAL CALCULATION:

Rth Computation:

Vth Computation:

PRECAUTIONS:

1. Reading must be taken without parallel error.

2. Measuring instruments must be properly calibrated.

3. Avoid loose connections.

4. Selection & Connection must be proper when using DMM.

RESULT:

VIVA-VOCE:

1. State Thevenins theorem.

2. For what type of networks Thevenins theorem is applicable?3. Is Thevenins voltage is open circuit voltage?

4. Advantages of Thevenins theorem.

NORTONS THEOREM

Exp. No: 1(B)

AIM:

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To verify Nortons Theorem for given circuit experimentally

APPARATUS:

S. NO EQUIPMENT SPECIFICATION QUANTITY1 VOLTAGE SOURCE 0-30V 1

2 RESISTORS 2.2K 33 RESISTORS 1K 2

4 VARIABLE RESISTER 560 1

5 AMMETER ( DMM ) 0-20mA 1

6 VOLTMETER ( DMM) 0-15V 1

7 CONNECTING WIRES

THEORY:

STATEMENT:

It states that any linear active two terminal network containing resistance and voltage

sources and /or current sources can be replaced by single current source IN in parallel with a

single resistance RN. The Nortons equivalent current IN is the short circuit current through the

network terminals, and Nortons resistance RN is the resistance between the network terminals

when all the sources are replaced by their internal resistances.

EXPLANATION:

a) steps to find the RN

Open circuit current sources and short circuit voltage sources.


Find out the equivalent resistance RN across the open circuit terminals.

b) Steps to find the IN

Short circuit the load resistance.

Find out the short circuited current IN through the short circuit terminals by using

mesh analysis.

c) Find out the load current IL by connecting the load resistance to the given circuit.

d) Draw the equivalent circuit and connect the current source IN in parallel with RN and

RL.

CIRCUIT DIAGRAM:

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Fig.1

Fig.2

PROCEDURE:

1. Adjust the input voltage to 4V

2. Calculate IN theoretically.

3. Find short circuit current IN through AB using ammeter by shorting terminals AB.

4. Connect the load resistance RL and measure load current I L through branch AB using

ammeter.

5. Short the voltage source and open the terminals A and B.

6. Measure the resistance (RN ) through the open circuit terminals AB using DMM.

7. Connect Nortons equivalent circuit (shown in figure 2).

8. Measure the load current ILI and tabulate the readings.

9. Compare IL , ILI and observe both are equal.

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10. Repeat the procedure for different voltage values.

OBSERVATIONS:

S.No Vs (v) R N () IN(mA) IL(mA) IL(mA)

THEORETICAL CALCULATION:

RN Computation:

IN Computation:

PRECAUTIONS:




4. Selection & Connection must be proper when using DMM.

RESULT:

VIVA-VOCE:

1. State Nortons theorem.

2. For what type of networks Nortons theorem is applicable?

3. Is Nortons current is short-circuit current?

4. Why we need to short voltage sources and open current sources to find RN5. Relation between Thevenins and Nortons theorem

MAXIMUM POWER TRANSFER THEOREM

Exp. No: 2

AIM:

To verify Maximum Power Transfer Theorem for given circuit experimentally

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APPARATUS:

S. NO EQUIPMENT SPECIFICATION QUANTITY1 Voltage source 0-30V 12 Resistors 22 2

3 Resistors 10 24 Variable resister 470 1

5 Ammeter ( DMM ) 0-20mA 1

6 Voltmeter ( DMM ) 0-15V 17 Connecting wires

THEORY:

STATEMENT:

In a linear bilateral network containing an independent voltage source in series with

resistance Rs delivers maximum power to load resistance RL when RL= Rs.

Or

In a linear bilateral network containing an independent current source in parallel with

resistance Rs delivers maximum power to load resistance RL when RL= Rs.

EXPLANATION:

Maximum power P is given as

Pmax =L

th

R

V

4

2

Where

Vth the open circuit voltage which is given by open circuiting the load resistance.

Rth or source resistance Rs is the equivalent resistance which is given by short

circuiting the voltage source and open circuiting the load resistance.

CIRCUIT DIAGRAM:

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PROCEDURE:

1. Connect the circuit as shown in fig.

2. Short circuit the voltage source and measure the resistance (Rs) seen through the

terminals AB.

3. Now connect the voltage source and load resistance (10+ variable resistances) across

AB.

4. Connect ammeter in the branch AB.

5. Now vary the load resistance in steps and note down the corresponding ammeter

reading.

6. Tabulate the reading and find the power dissipated in the resistor using formula.

7. Draw the graph between power and resistance.

8. From the graph find resistance corresponding to max power.

OBSERVATIONS:

S.No. Voltage

V ab

(Volts)

Load Current

I ab

(mA)

Load resistance

(RL)

(ohms)

Power

P = I2 R

(watts)

PRECAUTIONS:


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3. Avoid loose connections

RESULT:

VOVA-VOCE:

1. State maximum power theorem.

2. Define power, and energy.

3. Give the condition for maximum power transfer.

4. What is the draw back of maximum power transfer theorem?

5. What is ratio of load voltage to source voltage?

SUPER POSITION THEOREM

Exp. No: 3(A)

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AIM:

To verify superposition theorem for given circuit experimentally.

APPARATUS:

S. NO EQUIPMENT SPECIFICATION QUANTITY1 VOLTAGE SOURCE 0-30V 1

2 RESISTORS 1K 3

3 AMMETER ( DMM ) 0-20mA 1

4 VOLTMETER ( DMM ) 0-15V 2

5 CONNECTING WIRES

THEORY:

STATEMENT:

In a linear bilateral network containing two or more independent sources , the voltage

across or current through any branch is algebraic sum of individual voltages or currents

produced by each independent source acting separately with all the independent sources set

equal to zero.

EXPLANATION:

1. Select only one source and replace all other sources by there internal resistances. (If the

source is the ideal current source replace it by open ckt. if the source is the ideal voltage

source replaces it by short ckt.)2. Find the current and its direction through the desired branch.

3. Add all the branch currents to obtain the actual branch current

CIRCUIT DIAGRAM:

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PROCEDURE:

1. Connect the circuit as shown in fig.

2. Adjust the V1 = 5V, V2 = 5V.

3. Measure the current through the branch AB using ammeter. I.e., Iamb.

4. Adjust V2 = 0V, and V1 = 5V and measure current (I) in branch AB.

5. Adjust V2 = 5V, and V1 = 0V and measure current (I) in branch AB.

6. Tabulate the readings.

7. Repeat the procedure for different voltage values of V1, V2.

8. Then calculate I = I + I.

9. Compare I ab, I and observe that both should be equal.

OBSERVATIONS:

S. No.V1

(volts)

V2

(volts)

I ab

(Without using

theorem)

I

(V2=0)

(mA)

I

(V1=0)

(mA)

I = I+I

(mA)

1

23

4

5

PRECAUTIONS:



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RESULT:

VIVA-VOCE:

1. State super position theorem.

2. Define linear network.

3. Define bilateral and unilateral networks.

4. State Kirchhoffs laws.

5. Define ohms law.

RECIPROCITY THEOREM

Exp.No:3(B)

AIM: To verify the reciprocity theorem for a given network.

APPARATUS:

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S.NO EQUIPMENT RANGE TYPE QUANTITY1 RPS 0-30V 1

2 Ammeter 0-20 MA DMM 1

3 Resistance 1 K 5

4 Voltmeter 0-15 V DMM 2

5 Connecting wiresTHEORY:

STATEMENT:

It states that in any passive element linear, bilateral single source network if a

voltage source E acting in one branch of a network causes a current I to flow in another branch

the network then the same voltage source E acting in the second branch would cause an

identical current I to flowing the first branch.

CIRCUIT DIAGRAM:

Fig 1

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Fig 2

PROCEDURE:


2. Find the current (I cd) in branch CD using ammeter and note down the values.

3. Now connect the circuit as shown in fig 2.

4. Compare I ad and I cd, and observe that both are equal.

5. Repeat the procedure for different values

OBSERVATIONS:

S.NO Applied Voltage V (volts) Current in branch

CD

Icd(mA)

Current in Branch

AB

I ab (mA)

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PRECAUTIONS:



RESULT:

VIVA-VOCE:

1. State reciprocity theorem?

2. Reciprocity theorem is applicable to__________________ only.

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SERIES AND PARALLEL RESONANCE

Exp. No: 4

AIM:

To obtain frequency response of a series and parallel resonance circuit and also

calculate bandwidth and Q-factor.

APPARATUS:

S. NO EQUIPMENT SPECIFICATIONS QUANTITY1 FUNCTION GENERATOR 1MHz 1

2 CRO 20MHz 1

3 RESISTORS 1K 1

4 CAPACITOR 0.047f 1

5 INDUCTOR 70mH 1

6 CONNECTING WIRES

THEORY:

RESONANCE: A RLC network is said to be in resonance when the applied voltage and

current are in phase and the frequency at which this phenomena occurs is known as resonance

frequency fr. Resonance occurs when inductive reactance is equal to capacitive reactance i.e.

XL = XC .

Resonant frequencyLC

fr2

1=

At resonance the impedance is pure resistance .At resonance frequency the current in

the circuit is maximum which is given by Imax.

BAND WIDTH: A band of frequencies at which the current or voltage is2

1times its

maximum value. At that instant the power delivered to the circuit is half of the power at

resonance. Hence they are called as half power frequencies 1 and 2. The frequency 1 is

termed as lower cut off frequency and 2 is termed as upper cut off frequency. And hence the

difference between the two half power frequencies is known as Band width.

B.W = 2 - 1

QUALITY FACTOR: The ratio between resonance frequency fr to bandwidth B.W. It is also

given as the ratio of capacitor or inductor voltage at resonance to supply voltage.

Q factor =V

VLor

V

VC

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Or Q factor = 2 cycleperdissipatedenergystoredenergymaximum

Or Q factor =B.W

fr

CIRCUIT DIAGRAM:

SERIES RESONANCE:

Fig.1

PARALLE RESONANCE:

Fig.2

PROCEDURE:

Series Resonance:


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2. Feed a sine wave of 5v (peak to peak) amplitude and of 1 KHz frequency from

function generator to the input terminals.

3. Vary the input frequency and observe the output on CRO and also note the output peak-

to-peak amplitude. Tabulate these values.

4. Draw the graph between input frequencies vs. output voltage

5. From the graph observe the resonant frequency.

6. And also calculate Cut-off frequencies f1 and f2, band width and Q-factor.

Parallel Resonance:

1. Connect the circuit as shown in figure 2.

2. Repeat the procedure from steps 2 to 6 for parallel resonance.

3. From the graph observe the resonant frequency i.e. frequency at witch voltage is

maximum

4. By using graph calculate cutoff frequency, f1, f2, bandwidth, and Q-factor.

OBSERVATIONS:

Series Resonance:

SI.N

OFrequency(Hz) Output voltage(V)

1

2

,

,

20

Parallel Resonance

SI.N

OFrequency(Hz) Output voltage(V)

1

2

,

,

20

MODEL GRAPH:

Series Resonance:

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Parallel Resonance:

PRECAUTIONS:



RESULT:

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VIVA-VOCE:

1. Define resonance and give condition for resonance.

2. What is meant by half power band width?

3. Define quality factor and give expression for it?

4. Why at frequency less than resonance frequency the circuit behaves like a capacitive

ckt?

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OPEN CIRCUIT CHARACTERISTICS OF DC SHUNT GENERATOR

Exp. No: 5

AIM:

1. To determine the magnetization characteristics (open circuit characteristics) of DCShunt Generator.

2. To determine the Critical field Resistance.

APPARTUS:

S. No Equipment Type/ Rating Quantity

1 Rheostat Wire Wound / 1.5A/300ohms 2

2 Ammeter Moving Coil 0-2A 1

3 Voltmeter Moving Coil /0-300V 1

4 Tachometer Digital/0-10000 rpm 1

5Connecting

wires- -

NAME PLATE DETAILS:

THEORY:

An electrical generator is a device which converts mechanical energy into electrical

energy in this energy conversion is based on the principle of production of dynamically

induced emf i.e., when ever a moving conductor cuts the magnetic flux lines, an emf is induced

across the conductor terminals.The performance of the generators can be determined by the following characteristics.

They are

open circuit characteristics.(Eo/If)

external characteristics.(V/I)

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internal characteristics.(E/Ia)

An open circuit characteristic is also known as magnetic characteristics. It shows the relation

ship between the no-load generator emf in armature and field current at constant speed.

CIRCUIT DIAGRAM:

PROCEDURE:

1. Connections are made as shown in the circuit diagram

2. Field rheostat of motor is kept minimum resistance position and that of Generator is

kept at maximum position

3. DC supply is switched on and DPST switch is closed

4. Motor is started using the starter slowly cutting the starter Resistance

5. Motor field rheostat is varied to run the at rated speed

6. Generator field rheostat is varied and field current (If) and generated no-load voltage

(Eo) is noted down up to 125% of rated voltage.

7. A Graph is drawn between If Vs Eo.

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OBSERVATION TABLE:

MODEL GRAPHS:

RESULT:

VIVA-VOCE:

S. No. If (A) Eo (V)1.

2.

3.

4.

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1. Why do we use 3-point starter to start DC Motor?

2. What is critical field resistance?

3. At the time of starting of DC Motor the field resistance is kept in minimum

position why?

4. What is critical speed?

SWINBURNES TEST

Exp. No: 6

AIM:

To pre-determine the efficiency of a D.C. shunt machine considering it as a generator

or as a motor by performing Swinburnes test on it.

APPARATUS:

S. No Equipment Type/ Rating Quantity

1 DC voltmeter (M.C) 0-300v 1 NO.

2 DC Ammeter (M.C) 0-20A 1 NO.


4 Tachometer 0-2000RPM 1 NO.

5 Variable Rheostat 300/1.5A 1 NO.

NAME PLATE DETAILS:

THEORY:

Testing of D.C .machines can be divided into three methods:

Direct test

Regenerative test

Indirect. test

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Swinburnes Test is an indirect method of testing a dc machine. In this method, the

constant losses of the DC machine are calculated at no-load. Hence, its efficiency either as

motor or as a generator can be pre-determined. In this method, the power requirement is very

small. Hence, this method can be used to pre-determine the efficiency of higher capacity dc

machines as a motor and as a generator.

Disadvantages:

(i) Efficiency at actual load is not accurately known

(ii) Temperature rise on load is not known and

(iii) Sparking at commutater on load is not known.

Power input at No-load = Constant losses + Armature copper losses (which is negligible)

Power input at No-load = Constant losses

Power input = Va Ia + Vf If

LOSSES IN A DC MACHINE:

The losses in a D.C. machine can be divided as 1) Constant losses 2) Variable losses,

which changes with the load.

CONSTANT LOSSES:

Mechanical Losses: Friction and Wind age losses are called mechanical losses. They depend

upon the speed. A dc shunt machine is basically a constant speed machine both as a generator

and as a motor. Thus, the mechanical losses are constant.

Iron Losses: For a dc shunt machine, the field current hence the flux per pole is constant

(Neglecting the armature reaction which reduces the net flux in the air gap). Hence, hysterics

and eddy current losses (which are also called as iron losses) remain constant.

Field Copper Losses: Under normal operating conditions of a D.C. shunt machine, the field

current remains constant. Thus, power received by the field circuit (which is consumed as field

copper losses) is constant.

Constant losses in a dc shunt machine=Mechanical + losses Iron losses+ Field cu. Losses.

VARIABLE LOSSES:

The power lost in the armature circuit of a dc machine increases with the increase in load.

Thus, the armature copper losses are called as variable losses.

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EFFICIENCY OF A DC MACHINE:

% Efficiency =Inputpower

rOutputpoweX 100

As a generator Input power P in = Pout + Constant losses+ Armature copper losses at a

Given load I2a Ra

Pout = VLIL

Where Ia = IL + If Self excited generator

Ia= IL Separately excited generator

As a motor Input power Pin = VL IL + Vf If

Output power Pout = Pin Constant losses Armature Copper losses.

IL= Ia + If Self excited motor(VfIf is not accounted for Pin)

IL = Ia Separately excited motor>

Note: While calculating the armature copper losses on load condition, the hot resistance of

the armature= 1.2 Ra (Normal Temperature) is considered.

CIRCUIT DIAGRAM:

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PROCEDURE:

1. Connections are made as per the circuit diagram.

2. At the time of starting keep the field rheostat of the motor at minimum position.

3. Motor is started under no load using the 3-point starter.

4. Motor is brought to the rated speed by varying the field rheostat.

5. At the rated speed No load readings of the ammeters and voltmeter are noted down.

OBSERVATIONS:

At no-load (separately excited dc motor):

V

(volts)

IL

(Amps)

IF

(Amps)

Ia

(Amps)

Pin= V IL

(watts)

I2aRa

(watts)

Constant losses =

(Pin - I2aRa)

(watts)

EFFICIENCY AS A MOTOR:

Let us assume that the current drawn by the armature =

Input to the motor = input to the armature + input to the field =

Total losses = constant losses + armature copper losses =

Output = Input total losses =

Efficiency m = InputOutput

X 100 =

EFFICIENCY AS A GENERATOR:

Let us assume that the current delivered by the armature =

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Output =

Total losses =constant losses + armature copper losses =

Input = Output + Losses =

Efficiency g = InputOutput

X 100 =

AS A MOTOR:

S.

No.

IL

(Amps)

Power

Input

(watts)

Copper

Loss

(watts)

Total

Loss

(watts)

Power

input

(watts)

Efficiency

1.

2.

3.4.

5.

AS A GENERATOR:

S.

No.

IL

(Amps)

Power

Input

(watts)

Copper

Loss

(watts)

Total

Loss

(watts)

Power

Output

(watts)

Efficiency

1.

2.3.

4.

5.

CONCLUSION :

1. The power required to conduct the test is very less as compared to the direct loading

test.

2. Constant losses are calculated from this method are used to compute the efficiency of adc machine as a generator and as a motor without actually loading it.

3. Hence, this is an economic method.

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MODEL GRAPH :

RESULT:

VIVA-VOCE:

1. Why do we conduct this test?

2. What are the different types of losses present in a DC Machine?

3. Why armature of a DC Machine is laminated?

4. What are the advantages of Swinburns test?

5. Swinburnes test is applicable to what type of machines and why?

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BRAKE TEST ON A DC SHUNT MOTORExp. No: 7

AIM:

To obtain the performance characteristics of a DC Shunt motor by a load test.

APPARATUS:

S. NO EQUIPMENT TYPE/ RATING QUANTITY

1 DC voltmeter (M.C) 0-300v 1 NO.2 DC Ammeter (M.C) 0-20A 1 NO.


4 Tachometer 0-2000RPM 1 NO.

5 Spring balance 0-10Kg 2 NO.

6 Rheostat 0-200 1 NO.

NAME PLATE DETAILS:

THEORY:

This is a direct method of testing a dc machine. It is a simple method of measuring

motor output, speed and efficiency etc., at different load conditions A rope is would round the

pulley and its two ends are attached to two spring balances S1 and S2 . The tensions provided

by the spring balances S1 and S2 are T1 and T2 the tensions of the rope can be adjusted with the

help of swivels.

The force acting tangentially on the pulley is equal to the difference between the

readings of the two spring balances in kg- force.

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The induced voltage Eb =V-Ia Ra and Eb= KN, Thus, K=Eb /N

V= applied voltage, Ia =armature current, Ra =Armature resistance.

Total power input to the motor P in =Field circuit power + Armature power

= VfIf + Va Ia

CIRCUIT DIAG RAM:

PROCEDURE;-

1. Connections are made as shown in the circuit diagram

2. Before giving supply, the field rheostat is kept in minimum position.

3. During starting load is not applied and by varying the field rheostat reading make the

motor to run at its rated speed and note down the no load values.

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4. The Motor is slowly loaded with the help of spring balances, tightening the rope.

5. At every equal increment of load, ammeter, voltmeter, and spring balances are noted.

6. The efficiency is calculated for different load conditions and maximum efficiency point

is indicated.

OBSERVATION TABLE:

S.No Voltag

e

(V)

Speed

(rpm)

Load

Current

IL(A)

Field

Current

If(A)

W1

(kg

)

W2

(kg

)

Torque

(N-M)

Input

(Watt)

Output

(watt)

=(output/input)*100

MODEL GRAPHS:

RESULT:

VIVA VOCE

1. What are the disadvantages of this test?

2. Why do interpoles in DC machines have tapering shape?

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3. If the field of DC shunt motor gets opened while the motor is running then what

happens?

4. What will happen if the filled of a DC shunt motor is open?

5. What happens when a Dc motor is connected across AC supply?

OPEN CIRCUIT AND SHORT CIRCUIT TESTS

ON A SINGLE PHASE TRANSFORMER

Exp. No: 8

AIM:

1. To predetermine the efficiency and regulation of the given single phase transformer at

different power factors.

2. To draw equivalent circuit referred to the primary.APPARATUS REQUIRED:

S.No: Apparatus Type Rating Quantity

1. Ammeter Moving iron 0-20A 1

2. Voltmeter Moving iron 0-300V 1

3. Wattmeter Dynamometer 5A/150V,LPF 1

4. Wattmeter Dynamometer 20A/150V,UPF 1

NAME PLATE DETAILS:

THEORY:

TRANSFORMER:

Transformer is a static device which transfers electrical power from one circuit to

another circuit without any electrical connection but through magnetic medium. It works on the

principle of Faraday law of electro magnetic induction. It transfers energy without change in

frequency but usually change in voltage and current.

The performance of transformer can be pre determined with reasonable accuracy the

parameters of its equivalent circuit. The parameters and losses in a transformer can be

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determined by conducting two simple tests to estimate the efficiency and voltage regulation

without actually loading the transformers.

They are

Open circuit or no load test.

Short circuit test.

Open circuit test is conducted by opening the secondary (H.V) of the transformer.

By using open circuit test the core loss of the transformer can be determined.

From open circuit test data

RO= Vo / Ic

Xm=Vo / Im

Short circuit test is conducted by short circuiting the secondary (L.V) of the

transformer. By using short circuit test the copper loss of the transformer can be determined.

From short circuit test data:

Zh=Vsc/Isc

RH=Wsc/Isc2

XH=(ZH2

-RH2

)

1/2

CIRCUIT DIAGRAM :

Open Circuit Diagram:

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Short Circuit Diagram:

PROCEDURE:

Open circuit test

1. Connections are made as per the circuit diagram

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2. HV side is kept open and rated voltage is applied to the low voltage winding by

adjusting the autotransformer

3. The meter readings are noted down and tabulated

Short circuit test

1. Connections are made as per the circuit diagram

2. LV side is short circuited and adjusting the autotransformer, rated current is send on the

HV side

3. The meter readings are noted down and tabulated

OBSERVATIONS:

Open circuit test:

Vo

(V)

Io

(A)

Wo

(W)

Cos o=Wo/(Vo*Io)

Ic=Io*

Cos

o(A)

Im=Io*Sin

o(A)

Short circuit test:

Vsc

(V)

Isc

(A)

Wsc

(Watts)

Efficiency curve:

Efficiency at any load (X times full load) and at a given power factor can be calculated as

follows.

Output at X times full load =X*rated KVA*PF

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Iron loss = Wo

Copper loss at x times full load =X2 *full load copper loss

%Efficiency = output*100/(output + losses)

Efficiency at different assumed loads for a given power factor are calculated and tabulated as

follows

S.No. Load(X) Output

(watts)

Iron

loss

(watts)

Copper

loss

(watts)

Input=output + losses

(watts)

%Efficiency

Regulation curve:

Percentage regulation= rCos Xsin

r, percentage resistance=I*R1*100/V

x, percentage reactance=I*X1*100/V

I, rated LV side current

V, rated LV side voltage

Positive sign for lagging power factor and negative sign for leading power factor percentage

regulation for full load for different power factors are calculated and results are tabulated as

shown bellow.

S.No Cos sin %Regulation

38


39/39

PRECAUTIONS:

1. Wattmeter connections must be done as per the rating of the transformer

2. LPF wattmeter to be used for open circuit test

RESULT:

VIVA VOCE:

1. What is the operating principle of transformer?

2. Why OC and SC tests are convenient and very economical?

3. What is the main purpose of OC and SC test?

4. Why iron losses are negligible during SC test of a transformer?

5. What is the primary reason to conduct OC test only on the Low voltage winding of

the transformer?

electrical and electronics lab

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