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EXPERIMENT NO. 1

To verify Kirchoff’s current law and Kirchoff’s voltage law

EXPERIMENT NO. 1

Aim: - To verify Kirchoff’s current law and Kirchoff’s voltage law.

Apparatus: - Trainer kit, connecting leads and digital multimeter.

Theory: -

Kirchoff Law: -

Circuit calculations are made using Ohm's law and Kirchoff's laws which stem from the law of conservation of energy. Among it the Kirchoff's laws makes it possible to find out the branch currents and voltages in circuits and networks. In given board it is made to simply verify these laws.

Kirchoff current law: -

The first law called 'Kirchoff current law' or KCL. It is applied to circuits nodes and states that 'the algebraic sum of the currents meeting at node of an electric circuit is zero' i.e. ΣΙ = 0.

The currents included in above sum differ in sign according to whether they flow from or flow to at meeting node. On these basis of KCL, an equation is written for the currents at any node of the circuit. For given electrical circuit in fig 1, the equation is written as At summing node A = Ι + (-Ι1) + (-Ι2) + (-Ι3) = 0 where the currents with -ve sign indentify as leaving currents and with + sign it is entering current.

It follows that, Ι1 + Ι2 + Ι3 = Ι which verify the KCL, that sum of currents at meeting or leaving node is equal to all branch currents.

Kirchoff voltage law: -

In second law known as 'Kirchoff voltage law' or KVL. It states that the algebric sum of product of current and resistance in each of the conductors in any closed path in a network plus the algebric sum of the e.m.f. in the closed path is zero. i.e. ΣIR+ΣE.M.F =0

The algebric sum of e.m.f around a close circuit is equal to the algebraic sum ofthe voltage drop across the resistances in the circuit' i.e. ΣE = ΣΙR

Circuit diagram: -

Fig1: - Circuit Diagram of KCL

Fig2: - Circuit Diagram of KVL

Procedure: -

KCL:

1. Make the connection according to the ckt diagram 2. Set the three rheostats to their max value. 3. Switch on the power supply 4. Change the setting of the rheostats to get different readings in all the three ammeters. 5. Measure the current in the three ammeters 6. Check that at every time current in the main branch is equal to the sum of currents in the two branches. Repeat the setting of the rheostat 7. Switch off the power supply.

KVL:

1. Connect the circuit as per the circuit diagram 2. Switch on the power supply 3. Note down the readings of the voltmeters 4. Change the value of the rheostat and repeat the step several times and switch off the power supply.

Observation Table: -

KCL S.No. I I1 I2 I = I1 + I2 Error%

KVL

S.No. APPLIEDVOLTAGE

(volts)

V1(volts)

V2(volts)

V3(volts)

V=V1+V2+V3(volts)

Error%

Precautions: -

1. All connections should be tight and correct. 2. Switch off the supply when not in use. 3. Reading should be taken carefully.

Result: -

1. The incoming current is found to be equal to the outgoing current.2. The total input voltage is equal to the total voltage drop in the ckt.

Questions: -

Q.1 What is the statement of Kirchhoff’s first law?A. The sum of the currents entering at any junction is equal to the sum of the currents leaving the junction.

Q.2 According to Kirchhoff’s second law, the algebraic sum of all IR drops and emf’s in any closed loop of a network is equal to…A. It is equal to zero.

Q.3 Kirchoff’s second law is related to what?A. EMF and IR drops.

Q.4 What is the internal resistance of the ideal voltage source?A. Zero

Q.5 What is higher , the terminal voltage or the emf?A. The emf

Q.6 What is he internal resistance of the current source ideally?A. Infinity

Q.7 What is the active network?A. An active network is that which contains one or more than one sources of emf. or current Sources.

Q.8 What is the bilateral network?A. It is the circuit whose properties are same in either direction

Q.9 What is the difference between a node and a branch?A. A node is a junction in the circuit where two or more than two circuit elements are connected together. The part of the network, which lies between two junctions, is called branch.

Q.10 What is the non-linear circuit?A. The circuit whose parameters change with the change in voltage and current is called the non-linear ckt.

EXPERIMENT NO. 2(a)

To verify Thevenin Theorem.

EXPERIMENT NO. 2(a)

Aim: - To verify Thevenin Theorem.

Apparatus: - Trainer kit, connecting leads and digital multimeter.

Theory: -

Thevenin theorem

It is a very powerful theorem and provides mathematical technique for replacing given network in two terminal network, with one single source, and one single resistance. In a linear bilateral network, it says “the current flowing through a linear network across any two terminals is given by

Procedure: - 1. Connect the circuit as shown in given figures.2. Firstly connect 9 volt dc power supply with the network across CD points.3. Connect voltmeter in place of A and B point and note open circuit voltage as Vth.

4. Calculate the open ckt voltage Vth which appears across terminal A and B. Vth = I.RTH. This is called Thevenin’s voltage.5. Calculate IL= Vth/(RL+RTH).

Observation Table: -

S.No. Voltage (V) Vth Rth Observed valueIL

Calculated valueIL’

ErrorIL - IL’/ IL x 100

1.

2.

3.

Precautions: -

1. All connections should be tight and correct. 2. Switch off the supply when not in use.

3. Reading should be taken carefully.

Result: - Thevenin theorem is proved and verified.

Questions: -

Q.1 To what type of circuit Thevenin’s theorem is applicableA. Linear and bilateral

Q.2 What is the use of Thevenin’s theorem?A. To convert the complex ckt into a voltage source and a series resistance

Q.3 How RTH is connected with the ckt?A. In series

Q.4 How is RTH connected with the load resistance?A. In series

Q.5 What modification is done in galvanometer to convert it into a ammeter?A. A large resistance in parallel

Q.6 What modification is done in the galvanometer to convert it into a voltmeter?A. A series resistance

Q.7 Resistance is a n active element or the passive?A. Passive

Q.8 How will you calculate the RTH?A. The resistance between the two terminals

Q.9 In place of current source, what is placed while calculating RTH?A. Replace current source by open ckt

Q.10 In place of voltage source which electrical parameters is placed?A. A short ckt.

EXPERIMENT No. 2 (b)

To verify Norton Theorem

EXPERIMENT NO. 2 (b)

Aim: - To verify Norton Theorem.

Apparatus: - Trainer kit, connecting leads and digital multimeter.

Theory: - Norton Theorem

It is a theorem and provides mathematical technique for replacing given network in two terminal networks with one single current source and one single resistance. In a linear bilateral network it says “In any two terminal network containing voltage source and resistances when viewed from its terminals via equivalent to a constant current source and a parallel resistance equal to internal resistance Ri. The constant current source is Isc (short circuited current)". According to the statement 1st removing RL, short circuit current is measured as IN, then the equivalent internal rsistance found by replacing emf source with short circuit and measuring RN at open circuit.

Procedure: -

1. Connect the circuit as shown in given figures.2. Remove the load resistance3. Find the Norton’s resistance RN4. Measure the Norton’s current IN5. Now measure the current in the load resistance directly6. Find out the current in the load7. Using formula find out the current in the load resistance8. Verify that these two are equal.

Observation Table: -

S.No. Voltage(V)

VN RN ISC=VN /RN

Observed valueIL

Calculated value

IL’

ErrorIL - IL’/ IL x 100

1.

2.

3.

Precautions: -

1. All connections should be tight and correct. 2. Switch off the supply when not in use. 3. Reading should be taken carefully.

Result: - The actual current IL’ is equal to the calculated value of load current IL & verified the theorem

Questions: -

Q.1 To what type of network Norton’s theorem applicable?A. Two terminal linear network containing independent voltage and current sources.

Q.2 How is RN connected to IN?A. In the parallel

Q.3 What is placed in place of voltage sources while calculating the RN?A. Their internal resistance replaces these.

Q.4 Give an example of unilateral ckt?A. Diode rectifier

Q.5 What is unilateral ckt?A. Whose characteristics changes with the change in direction of operation

Q.6 Give one example of the bilateral n/w?A. Transmission lines

Q.7 What is the limitation of Ohm’s law?A. Provided physical conditions do not change

Q.8 What is the reason that ground pin are made of greater diameter in the plugs?A. R=ρL/A

Q.9 Where is the voltage divider rule applicable?A. Two resistance in series

Q.10 Where is the current divider rule applicable?A. When there are two resistances in parallel.

EXPERIMENT No. 3

To verify Superposition Theorem

EXPERIMENT NO. 3

Aim: - To verify Superposition Theorem.

Apparatus: - Trainer kit, connecting leads and digital multimeter.

Theory: - Superposition Theorem According to this theorem, " in a linear network of resistances whichcontains more than one source of e.m.f, the current which flows at any pointis the sum of all currents which flows at that point if each source is consideredseparately, whether other source is replaced with their internal resistance".In given network shown in fig 1, the current flows into RL is the sum of the currents flow from two different voltage sources E1 and E2. Replacing source

E2 with short circuit ( assume the internal resistance = 0 ), the network reduce to fig 2. The current flows from E1 through R1 and parallel combination of RL and R2. the equivalent resistance of the circuit is, Req = R1 + R2 // RL. The current flows in the circuit will be I' 1 = E1 / Req

This current again divides in two path at point A, thus current in eachpath is

I' = I' 1 ( R2 / RL + R2 ),I' 2 = I' 1 ( RL / R2 + RL ), since I' 1 = I' + I' 2 …………(1)

Replacing source E1 with short circuit, the network reduce to fig 3. Thecurrent flows from E2 through R2 and a parallel path of RL and R1. The totalresistance of this circuit is Req = R2 + R1 // RL. The current flows in thecircuit is I" 2 = E2 / Req.This current divides at point A in two paths and current in each path is

I" = I" 2 ( R1 / RL + R1 ),I" 1 = I' 2 ( RL / R1 + RL ),since I' 2 = I'' + I' 1 ………(2)

The actual currents are than known by superposition of these two eq (1) & (2) as.

I1 = I' 1 - I'' 1, I2 = I'' 2 - I' 2 and I = I' + I'' ……………………………………... (3)

Procedure: - 1. Connect the circuit as shown in fig.2. Taking both the voltage sources connected, measure the current (I3) IN I-I’

branch using milli ammeter.3. Short circuit the E2 battery and measure current (I’) IN I-I’ branch. 4. Short circuit the E1 battery and measure current (I”) IN I-I’ branch.

Observation Table: -

S.No.

I3 I’3 I”3 I’3 +I”3 Error%=I3-(I’3+I”3)

1.

2.

Result: - From the observation and calculation, superposition theorem has been verified.

EXPERIMENT No. 5

TO STUDY FREQUENCY RESPONSE OF A SERIES R – L – C

CIRCUIT AND DETERMINE RESONANT FREQUENCY AND Q FACTOR

FOR THE VARIOUS VALUES OF R, L AND C

EXPERIMENT No. 5

Aim: - To Study frequency response of a series R – L – C circuit and determine resonant frequency and Q factor for the various values of R, L and C.

Apparatus: - RLC Trainer kit, function generator and connecting leads

Theory: - A resonant circuit, also called a tuned circuit consists of an inductor and a capacitor together with a voltage or current source. It is one of the most important circuits used in electronics. For example, a resonant circuit, in one of its many forms, allows us to select a desired radio or television signal from the vast number of signals that are around us at any time. A network is in resonance when the voltage and current at the network input terminals are in phase and the input impedance of the network is purely resistive.

Circuit Diagram: -

Procedure: -

L.C.R SERIES RESONANCE CIRCUIT1. Connect voltmeter to 'V' sockets and milliameter to mA sockets.2. Connect output terminals on Function Generator to sockets marked signal output.3. Keep Function Generator selector switch at sine wave.4. Keep dial at minimum i.e. 100 c/s and with amplitude control set voltmeter Reading 3 to 4 volt.

5. Note Frequency from the dial and current from the milliameter.6. Change the frequency from the dial in steps of 100 or so and note the Corresponding current. Keeps Voltmeter reading constant by adjusting with Amplitude control knob.7. Plot a graph between frequency & Current in mA.8. Repeat the Exp. with same Procedure with the different R&L.

Observation Table: -

S.No. Frequency(in KHZ) Current(in mA)

1.

2.

3.4.5.6.

7.8.9.

RESULT: - 1) Inductance is calculated at different points from values of

capacitance considering the case of resonance.2) A graph is plotted between capacitance and corresponding currents.

Theoretically graph look like as shown in fig.

EXPERIMENT No. 6

TO STUDY FREQUENCY RESPONSE OF A PARALLEL R – L – C

CIRCUIT AND DETERMINE RESONANT FREQUENCY AND Q FACTOR

FOR THE VARIOUS VALUES OF R, L AND C

EXPERIMENT No. 6

Aim: - To Study frequency response of a parallel R – L – C circuit and determine resonant frequency and Q factor for the various values of R, L and C.

Apparatus: - RLC Trainer kit, function generator and connecting leads.

Theory: - A resonant circuit, also called a tuned circuit consists of an inductor and a capacitor together with a voltage or current source. It is one of the most important circuits used in electronics. For example, a resonant circuit, in one of its many forms, allows us to select a desired radio or television signal from the vast number of signals that are around us at any time. A network is in resonance when the voltage and current at the network input terminals are in phase and the input impedance of the network is purely resistive.

Circuit Diagram: -

Procedure: -

L.C.R PARALLL RESONANCE CIRCUIT

1. Connect voltmeter to 'V' sockets and milliammeter to mA sockets and Function generator to Signal input Sockets.

2. Keep Function Generator selector switch at Sine Wave.3. Keep dial at minimum i.e. 100 c/s and with amplitude control set voltmeter

reading 3 to 4 volt.4. Note Frequency from the dial and current from the milliameter.5. Change the frequency from the dial in steps of 100 or so and note the

corresponding current. Keeps voltmeter reading constant by adjusting with amplitude control knob.

6. Plot a graph between frequency & Current in mA.7. Repeat the Exp. with same Procedure with the different R&L.

RESULT: - 1) Inductance is calculated at different points from values of

capacitance considering the case of resonance.2) A graph is plotted between capacitance and corresponding currents.

Theoretically graph look like as shown in fig

EXPERIMENT NO. 7

To perform open and short circuit tests on 1-phase transformer and to calculate efficiency

EXPERIMENT NO. 7

Aim: - To perform open and short circuit tests on 1-phase transformer and to calculate efficiency.

Equipments Needed: -

S.NO. NAME TYPE RANGE QUANTITY1) Ammeter MI 0-2A 012) Wattmeter Dynamometer 2.5A, 200V 013) Voltmeter MI 0-300 V 014) Single phase

variac------------- 230/270 V,

15A01

5) Ammeter MI 0-15A 016) Wattmeter Dynamometer 15A, 75V 017) Voltmeter MI 0-30 V 01

Theory: -

OPEN CIRCUIT TEST: - In this test low voltage winding (primary) is connected to a supply of normal voltage and frequency and the high voltage winding (secondary) is left open as shown in fig (1). The primary winding draws very low current hardly 3 to 5 % of full load current under this condition. As such copper losses in primary winding will be negligible. Thus mainly iron losses occur in transformer under no load or open circuit condition, which are indicated by wattmeter connected in the circuit.Hence, total iron losses = W0 (Reading of wattmeter) From the observation of this test, the parameters R0 and Xm of parallel branch of equivalent circuit can also be calculated, following the steps given below: Power drawn, W0 = V0 I0 COS

Thus, no power factor, COS = W0 / V0 I0

Core loss component of no load current, Iw =I0 COSQ0 And, magnetizing component of load current, Im = I0 SIN Equivalent resistance representing the core loss, R0 = V / Iw

Magnetizing reactance representing Magnetizing current, Xm = V / Im

SHORT CIRCUIT TEST: - In this test, low voltage winding is short circuited and a low voltage hardly 5 to 8 % of rated voltage of high voltage winding is applied to this winding. This test is performed at rated current flowing in both winding. Iron losses occurring in the transformer under short circuit are mainly copper losses of both winding, which are indicated by wattmeter connected in circuit as shown in fig (2) Thus total full load copper losses = Wsc

The equivalent resistance referred Req, and reactance Xeq referred to a particular winding can also be calculated from observation of this test, following the steps given below.Equivalent resistance referred to H.V. winding, Req = Wsc / square of Isc

Also, equivalent impedance referred to H.V. winding, Zeq = Vsc / Isc

Thus, equivalent reactance referred to H.V. winding, Xeq = Under root of (square root of Zeq -square root of Req)PERFORMANCE CALCULATIONS : Complete performance of the transformer can be calculated based on above observations of open - circuit and short - circuit test following the steps given by,

1. Efficiency at different loads :

(a) EFFICIENCY AT FULL LOAD: Total losses at full load = W0 + Wsc

Let the full load output power of transformer in KVA be P0.Then % efficiency at full load, ηf = (P0 * 1000 * COSQ / P0 * 1000 * COSQ + W0

+ Wsc) * 100

2. Equivalent Circuit: All the parameters of approximate equivalent circuit has been calculated above. Thus an approximate equivalent circuit of the transformer can be drawn with these values of parameters marked on it. The equivalent circuit can be solved easily for estimating the performance like terminal voltage across secondary etc.

3. Regulation : - Regulation of transformer can now be calculated based on the parameters of the equivalent circuit using the approximate formula given below.

% Regulation = (I (Req COS + Xeq SIN ) / V) * 100 Where, I = rated current of winding, referred to which Req and Xeq have been calculated. V = Voltage of that winding. COS = Power factor at which regulation is to be calculated.

CIRCUIT DIAGRAM: -

Fig (1)CIRCUIT DIAGRAM OF OPN CIRCUIT TEST IN SINGLE PHASE TRANSFORMER

Fig (2)CIRCUIT DIAGRAM OF SHORT CIRCUIT TEST IN SINGLE PHASE TRANSFORMER

PROCEDURE: -(A) Open Circuit Test: -

1. Connect the circuit as per in fig (1)2. Ensure that the setting of the variac is at null position.3. Switch on the supply and adjust rated voltage across the transformer

circuit. 4. Record no load current, voltage and no load power, corresponding to rated

voltage of transformer winding.5. Switch off the ac supply.

(B) Short Circuit Test : 1. Connected the circuit as per fig (2) for conducting short circuit test.2. Adjust the setting of the variac, so that the output voltage is zero.3. Switch on the ac supply to the circuit.4. Increase the voltage applied slowly, till the current in the winding of

transformer is full load rated value.5. Record, short circuit current, corresponding applied voltage and power

with full load current flowing under short circuit conditions.

6. Switch off the ac supply.

OBSERVATION TABLE: -

FOR OPEN CIRCUIT TEST: -S.NO. V0 (V0LTS) IO(AMPS) W0(Watts)

1.

2.3.

4.

Sample Calculation for Open Circuit Test : -

The total iron loss = Wo

No load power factor = Wo/VoIo

Core loss component of the current = w =IoCosφMagnetizing component of current=Iu=IoSinφ Ro=V1/Iw Xo=V1/Iu

FOR SHORT CIRCUIT TEST: -

S.NO. Vsc (V0LTS) Isc (AMPS) Wsc (Watts)

1.2.3.4.

Sample Calculation for Short Circuit Test : -Total copper loss = Wsc R = Wsc/ Isc2 Z = Vsc/Isc X = √Z2-R2

Precautions: -1 All connections should be tight.2 The circuit should be according to circuit diagram.

3 The power should be on when the circuit is checked completely.

Result: - The shunt parameters and core loss can be find out with open circuit test. and the short circuit test is performed to find the copper losses and series parameters like the resistance and reactance.

EXPERIMENT NO. 8

NO LOAD AND BLOCK ROTOR TEST ON 3- PHASE INDUCTION MOTOR

EXPERIMENT NO. 8

Aim: - NO LOAD AND BLOCK ROTOR TEST ON 3- PHASE INDUCTION MOTOR.Equipments Needed: - Panel for three phase induction motor, three phase induction motor and connecting leads.

Theory: -Light running or No Load Test: -In this test, the motor is made to run without load that is no load conditions the speed of the motor is very close to the synchronous speed but less than the synchronous speed. The rated voltage is applied to the stator. The total input line current and total input power is measured. Two wattmeter methods are to calculate the total input power.

As the motor is on No Load the power factor is very low which is less than 0.5 and one of the two wattmeter’s read negative. It is necessary to reverse the connections of the current coil or pressure coil connections of such a wattmeter to read positive reading. This reading must be taken negative for the further calculations. The power input ‘w’ consists of following losses; stator copper losses, stator core losses iron losses friction and windage losses.The no load rotor current is very small and hence the rotor copper loss is negligiblysmall the rotor frequency is‘s’ time the supply frequency and on no load it is verysmall. Rotor core losses are proportional to this frequency and hence negligibly small. Under no load I is very small and in many practical cases it is also neglected thus ‘w’ consists of stator iron loss and friction and windage loss which are constant for the entire load. Hence ‘w’ is said to give fixed losses of the motor.

Block rotor test: -In this test the rotor is locked and not allowed to rotate. Thus the slip is 1. Thesituation is exactly similar to the short circuit test on transformer. If under shortcircuit condition the primary is excited with rated voltage a large short circuit current can flow which is dangerous to the winding point of view so similar to thetransformer short circuit test the reduced voltage (about 10 to 5 % of full load voltage) is just enough such that the stator carries the rated current is applied. During this test the stator carried the rated current, hence the stator copper loss is also dominated similarly the rotor also carried the short circuit current to produce the rotor losses the voltage is reduced the iron loss which is proportional to voltage is small enough the motor is at standstill hence mechanical loss that is friction and windage loss is absent.

Circuit Diagram: -

Procedure: -NO LOAD TEST –

1. Initially TPST switch is kept open.2. Autotransformer must be kept at minimum potential position.3. The machine must be started at no load.

BLOCKED ROTOR TEST: -1. Initially the TPST switch is kept open.2. Autotransformer must be kept at minimum potential position.3. The machine should be started on full load.

Observation Table: - NO LOAD TEST: S.No Voltage

Voc(Volts)

CurrentIOC

(Amps)

Wattmeter readings (W1)

Wattmeter readings (W2)

W0=W1+W2 WATTS

Observed Observed Calculate total power drawn

1234

Voc= open circuit voltageIoc = open circuit current

BLOCKED ROTOR TEST:S.No Voltage

VSC(Volts)

CurrentISC

(Amps)

Wattmeter readings (W1)

Wattmeter readings (W2)

W0=W1+W2 WATTS

Observed Observed Calculate total power drawn

12345

Vsc = short circuit voltageIsc = short circuit current

Precautions: -1. Connect the circuit as per the circuit diagram.2. The autotransformer should be kept in minimum voltage position.3. All connections should be neat and tight.4. Do not run the motor in these conditions for more than 1 minute. It can be

dangerous for the motor.5. Connecting leads should be perfectly insulated.

Result: -The total power drawn by the motor is equal to the sum of two wattmeter’s readings.

EXPERIMENT NO. 9

To measure power in 3 phase system by two wattmeter method

EXPERIMENT NO. 9

Aim: - To measure power in 3 phase system by two wattmeter method.

Equipments Needed: -

S.NO. NAME TYPE RANGE QUANTITY1) Ammeter MI 0-10A 012) Wattmeter Dynamometer 5A, 500V 023) Voltmeter MI 0-500 V 014) Three phase

variac------------- 470V, 15A 01

5) RESISTIVE LOAD

3- PHASE 440V, 10A 01

Theory: -The load in a 3-phsase system may be connected in star or delta. A balanced load is that in which the loading in each phase is exactly the same. The power in a 3-phase system is always equal to the sum of the powers in each of the three phases. Consider the circuit of Fig. which shows a three-phase star connected balanced load supplied by a three phase balanced supply. There are two Wattmeter’s, W1 and W2, connected in the circuit as shown in Fig. The common point lies on the line corresponding to phase b. Hence, there is no Wattmeter connected in this phase. The current coil of W1 measures Ia, while the potential coil measures Vab. The current coil of W2 measures Ic, while the potential coil measures Vbc. Thus, the two Wattmeter readings are

‘Three single-phase Wattmeter’s can effectively measure the active power consumed by a three-phase load. This is done by connecting their current coils in series with each phase, and their potential coils in parallel from each phase to the neutral, and finally adding all the three Wattmeter readings. The neutral point represents any common point in the circuit, where all the three Wattmeter potential coils are connected,

Procedure:1. Connect the circuit as shown in FIG.2. Switch on the supply taking care that the variac is kept on its zero output

position and adjust the variac to give 230 V line voltage.3. Switch on a resistive load and take the readings at power factor =1. Keeping the load current to about 3 A. Record V1, IL, W1and W2.4. In each case record V1, IL, W1and W2.5. Tabulate the readings and calculations as shown below

Observation Table: -

S.NO. V1 (V0LTS) IL(AMPS) W1(Watts) W2(Watts) P=W1 +W2

1.

2.3.

4.

Result: -The total power drawn in three phase system in resistive load is measured two wattmeter method.

EXPERIMENT NO. 10

To study the following electrical appliances.1.Thermostat2.Fluorescent tube light3.Compact fluorescent lamps4.Incandescent lamp5.Fan6.S ingle phase water pump7.Batteries8.Electric fuses & MCB9.Solar cell

Experiment No. 10

Aim: - To study the following electrical appliances.10.Thermostat11.Fluorescent tube light12.Compact fluorescent lamps13.Incandescent lamp14.Fan15.S ingle phase water pump16.Batteries17.Electric fuses & MCB18.Solar cell

Apparatus: - 1. Thermostat: -

A thermostat is the component of a control system which regulates the temperature of a system so that the system's temperature is maintained near a desired set point temperature. The thermostat does this by switching heating or cooling devices on or off, or regulating the flow of a heat transfer fluid as needed, to maintain the correct temperature. The name is derived from the Greek words thermos "hot" and stators "a standing".A thermostat may be a control unit for a heating or cooling system or a component part of a heater or air conditioner. Thermostats can be constructed in many ways and may use a variety of sensors to measure the temperature. The output of the sensor then controls the heating or cooling apparatus.This circuit is intended to control a heating system or central heating plan, keeping constant indoor temperature in spite of wide range changes in the outdoor one. Two sensors are needed: one placed outdoors, in order to sense the external temperature; the other placed on the water-pipe returning from heating system circuit, short before its input to the boiler.The Relay contact wiring must be connected to the boiler's start-stop control input. This circuit, though simple, has proven very reliable:

2. Fluorescent tube light : - A fluorescent tube is a gas discharge lamp that uses electricity to excite mercury vapor. The excited mercury atoms produce short-wave ultraviolet light that then causes a phosphor to fluoresce, producing visible light. A fluorescent lamp converts electrical power into useful light more efficiently than an incandescent lamp. Lower energy cost typically offsets the higher initial cost of the lamp. The lamp fixture is more costly because it requires a ballast to regulate the current through the lamp. While larger fluorescent lamps have been mostly used in commercial or institutional buildings, the compact fluorescent lamp is now available in the same popular sizes as incandescent and is used as an energy-saving alternative in homes.

3. Compact fluorescent lamps: -

4. Incandescent lamp : -

The incandescent light bulb, incandescent lamp or incandescent light globe produces light by heating a metal filament wire to a high temperature until it glows. The hot filament is protected from oxidation in the air with a glass enclosure that is filled with inert gas or evacuated. In a halogen lamp, filament evaporation is prevented by a chemical process that redeposit metal vapor onto the filament, extending its life. The light bulb is supplied with electrical current by feed-through terminals or wires embedded in the glass. Most bulbs are used in a socket which provides mechanical support and electrical connections.Incandescent bulbs are manufactured in a wide range of sizes, light output, and voltage ratings, from 1.5 volts to about 300 volts. They require no external regulating equipment, have low manufacturing costs, and work equally well on either alternating current or direct current. As a result, the incandescent lamp is widely used in household and commercial lighting, for portable lighting such as table lamps, car headlamps, and flashlights, and for decorative and advertising lighting.

5. ceiling fan : - A ceiling fan is a fan, usually electrically powered, suspended from the ceiling of a room, that uses hub-mounted rotating paddles to circulate air. A ceiling fan rotates much more slowly than an electric desk fan; it cools people effectively by introducing slow movement into the otherwise still, hot air of a room, inducing evaporative cooling. Fans never actually cool air, unlike air-conditioning equipment, but use significantly less power (cooling air is thermodynamically expensive).

Parts of a ceiling fan

An electric motor (see Types of ceiling fans below for descriptions) One to six paddles (called "blades"); usually made of wood, MDF, metal,

or plastic; which mount under, on top of, or on the side of the motor. The majority of residential ceiling fans have either four or five blades while most industrial ceiling fans have three. However, a very few specialized art fans (fans made more for artistic appearance than functionality) have other numbers of blades, such as one, or eight or more.

Metal arms, called blade irons (alternately blade brackets, blade arms, blade holders, or flanges), which connect the blades to the motor.

6. S ingle phase water pump : -

A pump is a device used to move fluids, such as liquids, gases or slurries.A pump displaces a volume by physical or mechanical action. Pumps fall into three major groups: direct lift, displacement, and gravity pumps. Their names describe the method for moving a fluid.The following types of pumps:piston pumprotary gear pumpHydraulic pumpvane pump

1. piston pump : -A piston pump is a type of positive displacement pump where the high-pressure seal reciprocates with the piston. Piston pumps can be used to move liquids or compress gases.

2. Rotary-type positive displacement: internal gear, screw, shuttle block, flexible vane or sliding vane, circumferential piston, helical twisted roots (e.g. the Wendelkolben pump) or liquid ring vacuum pumps.

3. Hydraulic pumps are used in hydraulic drive systems and can be hydrostatic or hydrodynamic.Hydrostatic pumps are positive displacement pumps while hydrodynamic pumps can be fixed displacement pumps, in which the displacement (flow through the pump per rotation of the pump) cannot be adjusted, or variable displacement pumps, which have a more complicated construction that allows the displacement to be adjusted

4. A rotary vane pump is a positive-displacement pump that consists of vanes mounted to a rotor that rotates inside of a cavity. In some cases these vanes can be variable length and/or tensioned to maintain contact with the walls as the pump rotates

7. Batteries: -

An electrical battery is one or more electrochemical cells that convert stored chemical energy into electrical energy.

There are two types of batteries: primary batteries (disposable batteries), which are designed to be used once and discarded, and secondary batteries (rechargeable batteries), which are designed to be recharged and used multiple times. Batteries come in many sizes, from miniature cells used to power hearing aids and

wristwatches to battery banks the size of rooms that provide standby power for telephone exchanges and computer data centers.

Battery cell typesThere are many general types of electrochemical cells,The variation includes galvanic cells, electrolytic cells, fuel cells, flow cells and voltaic piles.

8. Electric fuses & MCB: -

Fuse: - In electronics and electrical engineering, a fuse (from the French fusée, Italian. fuso, "spindle) is a type of low resistance resistor that acts as a sacrificial device to provide over current protection, of either the load or source circuit. Its essential component is a metal wire or strip that melts when too much current flows, which interrupts the circuit in which it is connected. Short circuit, overloading, mismatched loads or device failure are the prime reasons for excessive current.A fuse interrupts excessive current (blows) so that further damage by overheating or fire is prevented. Wiring regulations often define a maximum fuse current rating for particular circuits.The fuse element is made of zinc, copper, silver, aluminum, or alloys to provide stable and predictable characteristics. The fuse ideally would carry its rated current indefinitely, and melt quickly on a small excess. The element must not be damaged by minor harmless surges of current, and must not oxidize or change its behavior after possibly years of service.

KIT KAT FUSE: -

MCB: - Miniature circuit breakers cut off a circuit, or provide circuit interruption, when overload conditions are detected in a controlled circuit. Miniature circuit breakers are most commonly used in residential electrical units and in commercial buildings to control power sent from an electrical supply to the protected circuits.Miniature circuit breakers contain two contacts that close and open via a spring-based mechanism. A thermal magnetic trip device in the circuit breaker trips the spring when current levels increase above a certain limit, which in turn causes the breaker contacts to open to divert the current.

9. Solar cell : -

A solar cell (also called photovoltaic cell or photoelectric cell) is a solid state electrical device that converts the energy of light directly into electricity by the photovoltaic effect.Photovoltaics is the field of technology and research related to the practical application of photovoltaic cells in producing electricity from light, though it is often used specifically to refer to the generation of electricity from sunlight.Cells are described as photovoltaic cells when the light source is not necessarily sunlight. These are used for detecting light or other electromagnetic radiation near the visible range, for example infrared detectors, or measurement of light intensity.The solar cell works in three steps:

1. Photons in sunlight hit the solar panel and are absorbed by semiconducting materials, such as silicon.

2. Electrons (negatively charged) are knocked loose from their atoms, causing an electric potential difference. Current starts flowing through the material to cancel the potential and this electricity is captured. Due to the special composition of solar cells, the electrons are only allowed to move in a single direction.

3. An array of solar cells converts solar energy into a usable amount of direct current (DC) electricity.