Electrical Engineering Interview Questions And Answers

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 1)

2022-02-12T17:06:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 1)

Question1: Is it possible to start up a 36w or 40w 220V AC/DC tube light without the help of Coil or Choke?

Answer 1: It is not possible without the help of electric coil or choke. Because, the coil or choke ionize the particles in the tube light. In normal voltage it is not possible to ionize the particle.

Question2: What is the difference between three point and four point starters?

Answer 2: In the four point starter the line and Shunt are connected separately. In the three point starter the line and shunt are connected in same. This is one of the advantages of three point starter.

Question 3: What types of resistance is used in bulbs or tube lights? What will happen if a 220volt AC operated bulb or device run by the 220volt DC?

Answer 3: The AC operated Bulb or device has a very low resistance. But, the devices whose are designed for AC offers high impedance to the supply. When the DC voltage is applied to the AC devices, the passing DC current is very high due to the low resistance of the device. This high voltage may damage the device.

Question 4: What do you mean by electric traction?

Answer 4: Electric Traction means running a machine by using electric energy or power. Electrical energy, energies the traction system. Basically DC motors are used in traction system. Now electric traction system widely used in trams, trolleys, industries, railways .Bullet train is also running by the magnetic traction, magnetism is setup by the electric energy in the bullet train.

Question 5: Why the capacitor works doesn’t work on DC?

Answer 5: The capacitor normally gives an infinite resistance to DC component. But, it allows AC power component to pass through the devices. These resistances of the capacitor blocks work of DC component.

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Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 6)

2021-08-19T13:36:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 6)

In this topic we know about different types of motors starting torque and current, rheostat position, role of armature reaction, stepper motor, KVAR.

Question: In the synchronous motor, DC motor or Induction motor which types of the motor has the high starting torque and starting current? What are the load criteria for those motors on the starting condition?

Answer: From these motors the starting torque is present at the DC motor.

The load criteria for the starting conditions are as follows,

· The DC series motor can not start without the presence of the load.

· The synchronous motor and Induction motor can not start at the presence of load.

Question: In the motors starting purpose why the armature rheostats at maximum position while the field rheostat is kept in minimum position?

Answer: in the time of starting a motor the field rheostat is kept in minimum state for the high starting torque.

In the time of starting a motor the armature rheostats is used to control or reduce the high starting current. This high starting current can damage the motors starting systems.

Question: What are the types of power in electrical power?

Answer: There are normally three types of power are counted in electrical power. They are,

· Apparent power

Active power
Reactive power

Question: What does the KVAR means?

Answer: The KVAR indicates the electrical power. KVAR means

“Kilo Volt Amperes with Reactive components”

Question: In the magnetic fluxes, what is the role of armature reaction?

Answer: The armature flux plays an important role for the running condition. This armature flux can oppose the main flux or it may support the main flux for better running condition. This effect of supporting and opposing of main flux to armature flux is called armature reaction.

Question: What do you understand by the two phase motor? Give us one example.

Answer: The motor which have a phase split in the running and starting winding is called as a two phase motor.

Example: Servo motor is one types of two phase motor. In the servo motor there is phase difference of 90 degree between control winding and auxiliary winding.

Question: Define stepper motor. What is the use of stepper motor?

Answer: The motor which work or act on the applied input pulse in it, is called as stepper motor.

This stepper motor is under the category of synchronous motor, which does not fully depend of complete cycle. It likes to works in either direction related to steps. for this purpose it mainly used in automation parts.

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Electrical engineering interview

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 7)

2021-06-24T21:52:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 7)

In this topic we know about the Power Factor, the leading and lagging condition for the power factor, the disadvantage of the low Power Factor and the main causes of the low power factor.

Question: What is Power Factor?

Answer: In an a.c. circuit the cosine of angle between voltage and current is called as power factor.

Fig: Power factor angles

Question: What are the leading and lagging conditions in the power factor improvement?

Answer: Normally there are two types of circuit, capacitive and inductive circuits.

· In the capacitive circuits, current leads the voltage and power factor is called as leading.

· In the inductive circuits, current lags behinds the voltage and power factor is called as lagging.

Question: What are the disadvantages of low power factor?

Answer: The main disadvantages of low power factor are as follows,

1. Large size of the Conductor.

2. Large Kilo Volt Ampere rating of the electrical machinery or equipment.

3. Poor or unaccepted voltage regulations.

4. A big amount of Copper loss.

5. The handling capacity of the system reduced.

Question: What are the main causes of the low power factor?

Answer: In the supply system normally the power factor is lower then 0.8 for the overall load. Low Power factor is very undesirable, because it hampers the economic parameters for the power generation and distribution. Now, some short important causes are as follows,

· In the low lagging factor some electrical equipment such as electric discharge lamps, Arc lamps and the industrial heating furnaces are running.

· The power factor is also depending on the load conditions. The load is varying time to time. The supply voltage is increasing in the low load period which increases the amount of the magnetization current. This magnetizing current is resulting in the decreased power factor.

· The single phase and three phase ac motors have low lagging power factor and they are induction type also. The power factor is varying for these types of motor. It can run from 0.2 to 0.9 values depends on load. In the heavy load the power factor can be 0.8 or 0.9 and in the light load the power factor can be 0.2 to 0.3.

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Electrical engineering interview

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 8)

2020-10-31T10:26:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 8)

In this topic we know about the substation and its some requirements for setup.

Question: What do you mean by substation? Give some details.

Answer: The electric power is now generated in the form of alternating current. The electric power is produced in the power station, which is generally very far from the consumer premises. So, the generation, transmission and distribution are facing much difficulty. It is distributed to the consumer or user by the help of a very large interconnected network. In the network it is very important to change some characteristics for transmitting and distributing the electric power properly. Such as, ac to dc conversion (by using thyratron and mercury arc rectifier), voltage up or down (If the generating voltage is 11kv then transfer it to the high voltage range of 220kv), frequency change, power factor correction and improvement. This change needs a setup of some apparatus properly, in short this types of setup is called substation.

The assembly of electrical apparatus whose purpose is changing the some major characteristics of the electric supply system is called substation.

The substation is very important for achieving the continuity of supply to the consumers. In some cases the ac is need to convert in dc for running the dc motor, traction, electroplating processes this type of conversion is performed by the sub-station (ignitron).

SUB STATION

Question: What are the basic requirements for setup a substation? Or which conditions are considerable for setup a substation?

Answer: The continuity of electric energy supply system is solely depending on substation. So, enough care should be taken when designing and set upping a sub station in a suitable place.

Some points are as follows,

· The substation must have the chance of easy operation and maintenance.

· The place must be proper for setup a substation. It must be setup near the centre of the load.

· The substation must have enough safety apparatus or safety setup. There must be enough chance for running the maintenance and clearing process. Also there should be setup for fire protection.

· The system must be reliable. When designing protected gears and other instruments must be included. The construction process also maintained properly for reliable operation.

· The capital cost must be as low as possible.

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Electrical engineering interview

Job interview Topics for Electrical and electronic Engineers (Part-2).

2018-01-07T13:40:00.000+06:00

Electrical Engineering interview question and answers:Job interview or viva Topics for Electronics and Electrical Engineers (Part-2).

In this topic we discuss about Load Curve,Demand Factor and Load Factor.

4. Load Curve:

The curve which shows the variation of load on the electrical power station with respect to time is known as load variation curve or simply load curve.

5. Importance of Load Curve:

The Daily Load Curve gives the information of load on the power station during different running hours of the day.
The number of unit’s generation per day is found from the area under the daily Load Curve.
Average load is found from the Load Curve.

Average load= [Area (KWh) under daily load curve/24 hours]

The maximum demand of the station on that day is found from the highest point of the daily Load Curve.
The size and the number of generating units can be determined from the load curve.
This Load Curve helps to determine the operation schedule of the station. In that case when all the units or the less units needs to running is found.

6. Demand Factor:

The ratio of maximum demand and connected load on the power station is called as demand factor.

Demand Factor= [Maximum Demand/Connected Load]

In the power station Maximum demand < Connected Load.
Demand Factor < 1
Demand Factor is very important in determining the capacity of the plant equipment.

7. Load Factor:

The ratio of Average load and the Maximum demand in a given period of time in a power plant is known as load factor.

Load factor = [Average load/ Maximum demand]

In a power plant Average load < Maximum demand
Load factor<1.

The Load factor is very important for determining the cost of per unit generation of power.

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Electrical engineering interview

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 9)

2016-09-03T21:54:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 9)

In this topic we will know about motor protection

In the nameplate of the motor we see IP XY . i.e. IP 55
Example: With the IP rating IP 23,

2 describes the level of protection from solid objects and
3 describes the level of protection from liquids.

IP stands for

Ingress Protection Ratings

IP rating normally has two (or three) numbers:

Protection from solid objects or materials
Protection from liquids (water)
Protection against mechanical impact

specifying the environmental protection the enclosure provides.

Ingress Protection (IP) ratings are Introduced by the European Committee for Electro Technical Standardization .

NEMA IEC 60529 Degrees of Protection Provided by Enclosures - IP Code,

IP First number - Indicates the Protection against solid objects. Which we indicates as X in the example

0 No special protection

1 Protected against solid objects over 50 mm, e.g. accidental touch by persons hands.

2 Protected against solid objects over 12 mm, e.g. persons fingers.

3 Protected against solid objects over 2.5 mm (tools and wires).

4 Protected against solid objects over 1 mm (tools, wires, and small wires).

5 Protected against dust limited ingress (no harmful deposit).

6 Totally protected against dust.

IP Second number -Indicates the Protection against liquids.Which we indicates as Y in the example

0 No protection.

1 Protection against vertically falling drops of water e.g. condensation.

2 Protection against direct sprays of water up to 15^o from the vertical.

3 Protected against direct sprays of water up to 60^o from the vertical.

4 Protection against water sprayed from all directions - limited ingress permitted.

5 Protected against low pressure jets of water from all directions - limited ingress.

6 Protected against temporary flooding of water, e.g. for use on ship decks - limited ingress permitted.

7 Protected against the effect of immersion between 15 cm and 1 m.

8 Protects against long periods of immersion under pressure.

Example - IP 35

An electrical socket rated IP22 is protected against

Protected against solid objects over 2.5 mm (tools and wires).
Protected against low pressure jets of water from all directions - limited ingress.

IP stands for

Ingress Protection Ratings

IP Third number - Indicates the Protection against mechanical impacts .

Its not used in normal cases.

the third number is not available in IEC 60529

0	No protection.
1	Protects against impact of 0.225 joule (e.g. 150 g weight falling from 15 cm height).
2	Protected against impact of 0.375 joule (e.g. 250 g weight falling from 15 cm height).
3	Protected against impact of 0.5 joule (e.g. 250 g weight falling from 20 cm height).
4	Protected against impact of 2.0 joule (e.g. 500 g weight falling from 40 cm height).
5	Protected against impact of 6.0 joule (e.g. 1.5 kg weight falling from 40 cm height).
6	Protected against impact of 20.0 joule (e.g. 5 kg weight falling from 40 cm height).

Electrical engineeringinterview

Job interview or viva Topics for Electrical Engineers (Part-6)

2015-03-01T19:21:00.000+06:00

Electrical engineering interview question and answers:

Job interview or viva Topics for Electrical Engineers (Part-6)

21. Swing equation:

During the transient state of the synchronous machine, the relation between the accelerating power and angular acceleration is called swing equation.

M (d²δ/dt²) = Ps – Pe

Here,

M = angular momentum of the rotor,

Ps = mechanical power input,

Pe = electrical power output,

δ= load angle,

t = time,

22. Swing Curve:

The graph which shows the relation between δ versus t is called swing curve. These graphs provide the stability information. This swing curve information is useful for understanding the adequacy of relay protection in the power station. This information helps to solve the faults before one or more machines become unstable.

23. Unsymmetrical faults:

The unbalanced conditions of the system are called unsymmetrical faults. If an unbalanced connection is attached at any point of the balanced system, then this point is called the fault point of the system. This unsymmetrical fault has normally two types:

1. Shunt fault.

2. Series fault.

24. Shunt Fault:

An unbalanced shunt fault is an unbalanced condition between phases and ground or, an unbalance between phases.

25 Classification of shunt fault:

In a 3-phase system the shunt fault are as follows,

· Single line to ground fault (LG).

· Line to line fault (LL).

· Double line to ground fault (LLG).

· Three phase short Circuit fault (LLL).

· Three phase to ground fault (LLLG).

Here,

· LG, LL, LLG ---- unsymmetrical faults.

· LLL, LLLG ---- symmetrical faults.

26. Series fault:

The unbalance in the line impedances is known as series fault.

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Job interview and viva Topics for Electrical and electronics Engineers (Part-4).

2013-05-11T13:32:00.000+06:00

Electrical engineering interview topic:Job interview and viva Topics for Electrical and electronics Engineers (Part-4).

In this topic we know about Power System Stability,Steady State Stability,Static Stability,Dynamic Stability and Steady State Stability Criterion

13. Power System Stability:

In the large power system there might be some disturbance always. The system has the tendency for the system to develop forces for bringing the system in a normal or stable condition. The ability of the system to obtain the normal or stable condition after getting disturbed is called Power System Stability.

The stability has mainly two types:

1. Steady State Stability.

2. Transient Stability.

14. Steady State Stability:

The ability of being stable after a small disturbance in the system is called Steady State Stability.

The Steady State Stability is concerned with the very small power changes .which are changed gradually.

The Steady State Stability has two types:

1. Static Stability

2. Dynamic Stability.

· Static Stability:

In the Static Stability there is a large duration phenomenon with small disturbance. This is the inherent stability that can achieve with out the help of automatic control devices.

· Dynamic Stability:

In the Dynamic Stability there is a small duration phenomenon with large disturbance. This Dynamic Stability refers to artificial stability. This artificial stability is given to an inherently unstable system by the help of automatic control devices.

15. Steady State Stability Criterion:

The rate (dP/dδ) of the differential power increased is obtained per differential load angle increased. This are called as electrical stiffness or the synchronizing power coefficient of a synchronous machine. This is considered as the measure of the stability of the system.

The Steady State Stability has basically three Criterion or state.

When,

· (dP/dδ) > 0 ; Direct State Synchronous Stability.

· (dP/dδ) = 0 ; Steady State Stability.

· (dP/dδ) < 0 ; The System is unstable.
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Determine the efficiency & characteristic of a Shunt wound DC motor.

2012-12-12T22:56:00.000+06:00

Objective:-To Determine the efficiency & characteristic of a Shunt wound DC motor.

Theory:- The field coil and the armature windings are connected in shunt or parallel across the power source. The armature winding consists of relatively few turns of heavy gauge wire. The voltage across two windings is the same but the armature draws considerably more current than the field coil. Torque is caused by the interaction of the current caring armature winding with the magnetic field produced by the field coil. If the DC line voltage is constant, the armature voltage and the field strength will be constant. The speed regulation is quite good; the speed is a function of armature current and is not precisely constant. As the armature rotates within the magnetic field, an EMF is induced in its wining. This EMF is in the direction opposite to the source EMF and is called the counter EMF, which varies with rotational speed. Finally, the current flow through the armature winding is a result of the difference between source EMF and CEMF. When the load increases, the motor tends to slow down and less CEMF is induced, which in turn increases the armature current providing more torque for the increased load. Motor speed is increased by inserting resistance into the field coil circuit, which weakens the magnetic field. Therefore, the speed can be increased from “basic” or full-load, full-field value to some maximum speed set by the electrical and mechanical limitations of the motor.

The shunt motor is constant speed, medium starting torque & high running torque.

Equipments Required:-

1. Power supply

2. DC Motor/generator

3. DC Voltmeter/Ammeter

4. Electrodynometer (fixed 240V ac)

5. Connection Leads

6. Tachometer

7. Timing belt

Connection Diagram:-

Table:

E (volts)	I (amps)	SPEED (r/min)	TORQUE (N.m)
240			0
240			0.3
240			0.6
240			0.9
240			1.2

Characteristic Curve:

Calculation:

Regulation =

P_out =

Discussion:

Job interview or viva Topics for Electrical Engineers (Part-5).

2012-05-30T13:36:00.000+06:00

Electrical Engineering interview question and answers:Job interview or viva Topics for Electronics and Electrical Engineers (Part-5).

In this topic we will describe Transient Stability,Stability Limit,Steady State Stability Limit,Transient Stability Limit,Power-angle Equation.

16. Transient Stability:

In the system the ability of maintaining synchronous operation to reach a stable condition or state or the one close to it after a large disturbance is called Transient Stability.

For the Transient Stability there are some important assumptions. Such as,

· System resistance(R) may be neglected in comparison with reactance(X).

· The System may be reduced in form of two-machine equivalent system.

· Each machine may be considered to supply an infinite bus.

· Direct Axis Reactance= Quadrature Axis Reactance.

· The rotor shape can be cylindrical.

· For Machine representation Direct Axis Reactance can be used.

· For few seconds after occurrence of disturbance, the input power of the shaft may be assumed constant.

17. Stability Limit:

In the system, when the maximum amount of power can be transferred between source and loads in a network, without the loss of synchronism is called Stability Limit.

18. Steady State Stability Limit:

Under Steady state condition, when the load is gradually increased but there is no unstable condition in the system and the maximum power can transferred is called Steady State Stability Limit.

19. Transient Stability Limit:

The amount of maximum power that can be transferred in the system without making the system unstable when a sudden or large disturbance occurs. This is called Transient Stability Limit.

20. Power-angle Equation:

Here, Pe= Power, δ= load angle.

Pe=Pemax Sin δ

The maximum steady state power transferred, when δ = 90°.

The value of Pemax often called pull-out power or steady state limit.

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Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 5)

2012-05-20T13:00:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 5)

In this topic we will find some question about unit cell in crystallization, circuit breaker and its properties, advantage of circuit breaker over fuse and switches, Arc voltage, re-striking voltage, recovery voltage.

Question: What is unit cell in crystallization?

Answer: The small cell which is most convenient in the crystal and carries all the properties and nature of the crystal is called unit cell. In the three dimensional system the repetitions of unit cell creates a whole structure of crystal.

Question: What is circuit breaker? What are the properties of circuit breaker?

Answer: A circuit breaker is a piece of equipment which can save devices from faults by using moving or fixed contacts, called as electrodes.

The main the properties of circuit breaker are,

· Make a circuit by remote control or manually under fault conditions.

· Under a fault condition break a circuit automatically.

· Make or break a circuit by remote control or manually under normal conditions.

Question: What is the advantage of circuit breaker over fuse and switches? What are the disadvantage of fuse and switches?

Answer: The fuse and switches are connected in series with the circuits. So the problems are as follows,

· The switches take quite sometimes for replacement and restore the power supply to the users, when it blows up.

· The fuse can not successfully block or interrupt a heavy current due to its series connectivity, which may damage the high capacity circuits.

Due to this disadvantages circuit breaker used over fuse and switches.

Question: What is Arc voltage?

Answer: During the arcing period the voltage that appears across the contacts of the circuit breaker is called arc voltage.

Question: What is re-striking voltage?

Answer: During the arcing period this is the transient voltage, which appears across the contacts at or near current zero is called as restriking voltage.

Question: What is recovery voltage?

Answer: After final arc extinction it is the normal arc frequency (50 Hz) root mean square (r.m.s) voltage that appears across the circuit breaker is called recovery voltage. The system voltage and recovery voltage are approximately equal.

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Electrical engineering interview

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 4)

2012-05-09T20:25:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 4)

In this Part we find some question about Crystallization

Question: What do you understand by a crystal?

Answer: A three dimensional periodic arrangement of molecules, atoms, or ions is called as crystal. In the crystal structure degree of symmetry and periodicity is most important. For the crystal structure the orientation of neighbors and their numbers are clearly mentioned.

Question: What is crystallization?

Answer: The process by which crystals of a substance are formed another phase of that substance is called as crystallization. Crystallization needs (10^3- 10^4) number of atoms or molecule of the substance. The nucleation process for the crystallization is increasing by the atomic diffusion from the vapor or the melt.

Question: What is crystalline solid? What are the most important properties of crystalline solid?

Answer: Crystalline Solid is a solid in which atom or molecules are arranged in a periodic arrangement to form a long range order.

Or,

A Crystalline solid is a solid in which atoms bond with each other in a regular pattern to form periodic collection.

The most important properties of crystalline solid are,

· Periodicity

· Long range order

Question: Describe the Crystalline solid in terms of lattice and basis. What are the lattice parameters? When we have the chance to find the actual crystal structure?

Answer: For describing the Crystalline solid,

Lattice: A lattice is an infinite periodic array of geometric points in space without atom.

Or,

A lattice is a regular array of points in space with a discernible periodicity. In three dimensional spaces there are possibilities of 14 distinct lattices. When at each lattice points an atom or molecule is placed, the resulting regular structure is called a crystal structure.

The lattice has two parameters mainly. They are,

1. The angle between the sides of unit cell

2. The lengths of the side of the unit cell.

Basis: The identical group of atom or molecules that are placed at each lattice points is called basis.

Or,

A molecule, a collection of molecules or an atoms placement at each lattice point to generate the crystal structure of a substance truly is known as basis.

When we place identical group of atoms or molecules at each lattice points then we find the actual crystal structure.

Question: who are the inventors of transistors? When they get Nobel Prize?

Answer: The inventors are,

1. William Shockley

2. John Bardeen

3. Walter Brattain

They get Nobel prizes in 1956.

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Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 3)

2012-04-27T18:09:00.000+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 3)

In this topic we know about ,electric supply system,electric supply system classification and principle components,components are used for dc to ac convert or ac to dc convert

Question 10: What is electric supply system?

Answer 10: Electric supply system: The process of transporting or conveyance of electrical energy or power from a power station to the users premises (home, industry etc) is called as electric supply system.

Question 11: What are the electric supply system classification and principle components?

Answer 11: The electric supply system can be classified into two types:

і) AC or DC supplies system.

ii) Underground or Overhead Supply system.

In the electric supply system the 3 phase 3 wire system is used for AC energy generation and transmission. In the distribution system three phase four wires is used now a days. This system is now used all over the world.

In the underground system tunnels are used and in overhead system poles are used.

Question 12: Which components are used for dc to ac convert or ac to dc convert? What is the operation range of these devices? What is present day’s trend for generation and transmission?

Answer 12: The high voltage dc transmission is much better than the high voltage ac transmission. So, now the electrical engineers grow interest for transmission of dc in place of ac.

The use of Mercury arc rectifiers and thyratron gives opportunities of converting a.c into d.c and the conversion of dc to ac to the electrical engineers.

Figure: One line diagram for the high voltage dc transmission

In the given one line diagram the electrical power is generated as ac and stepped up to high voltage by the help of stepper transformer. Then this high voltage energy is transferred to the Mercury arc rectifiers, which change ac to dc. After that time this high voltage dc is passed through the thyratron, this thyratron convert dc to ac again. After that this high voltage is stepped down to low voltage in receiving transformer.

The operation range of such devices is up to 30MW at 400KV in one single unit.

The present trend is using ac for generation and distribution. And the dc is used for high voltage transmission.

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Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 2)

2012-04-21T21:46:00.002+06:00

Electronics and Electrical Engineering interview Question and Answers for fresh engineers (Part 2)

In this topic we will find some question about transmission voltage, regulator difference, motor type of fans, unit of electrical energy

Question 6: What are the advantages of high transmission voltage? What are the limitations of the high transmission voltage?

Answer 6: Advantage of high transmission voltage:

1. It reduces the volume of conductor material.

2. It increases the transmission efficiency.

3. It decreases the line drop percentage.

Limitations of high transmission voltage:

1.The insulation cost of the conductor increased.

2.The other terminal operators, switchgear and transformers cost is increasing.

Question 7: Which is better between electrical rheostat regulator and electronic regulator for fans? Which is the basic component difference for them?

Answer 7: The electronic regulator is better then electrical rheostat regulator. In electronic regulator the power loss is very low. The electronic regulator only transfers the necessary power for the necessary speed of the fans. But in electrical rheostat regulator the power loss is same for every speed, so no power is saved.

The electronic regulator use TRIAC for speed control by varying the firing angle speed.

The electrical rheostat regulator use control resistance, which is increased or decreased by the variation of speed.

Question 8: In our normal household fans (such as exhaust fan, pedestal fan, ceiling fan, table fan etc) which types of motor is used? What is the rotor type and starting type normally?

Answer 8: In our normal household fans single phase induction motor is normally used.

The rotors are mostly squirrel cage rotor. The fans are normally capacitor start and run.

Question 9: Which is the unit of electrical energy?

Answer 9: Answer: the unit of electrical energy is joule or watt-second

Electrical energy in joule or watt-second

= voltage in volts* current in amperes* time in seconds.

Joule or watt-second is very small for the measuring of electrical energy in the practical field purpose.

1 watt-hour = 1watt * 1 hour

= 1watt * 3600 seconds

= 3600 watt-sec

1 kilowatt-hour = 1kW * 1hr

=1000watt* 3600 sec

--------------------------------------------------------------------------------------------------------

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Curriculum vitae writing tips

2012-04-12T21:47:00.002+06:00

Curriculum vitae

I will leave it to you to determine on the design and content of your

Curriculum vitae. There are several sources which can advise you on these

Aspects. Here are some transient pointers:

Use the best quality paper you'll afford.
Use solely white or off-white paper.
Don’t use a flowery exhausting cowl or binder along with fancy hard cover.
Use a straightforward font like Arial and use identical font in your Covering letter.
Never use a generic covering letter. Be specific in its content by.Bringing out one or 2 components of how you match what they're looking for. It’s not simply a wrapper for your CV!
Your CV must get past that 1st thirty seconds once the recruiter has Picked it up. Imagine yours is simply one among-st fifty or sixty or additional Applications received. you wish your professionalism to square and not stand out as a result of you've used bright yellow paper!
Pointing your achievements clearly in your CV; not merely list your past employers and job titles.
accommodates any directions within the adverts like supplying your current salary – you'll want to try and do this inside your covering letter.

Imagine your delight, constant reader, when 2 weeks later you receive

on beautiful headed paper an invitation to attend an interview for this

Potentially wonderful new job! That’s when your exertions very

Begins. In Chapter a pair of I’ll explore what you would like to be brooding about

before you truly attend the interview.

Job interview Topics for Electrical and electronic Engineers (Part-3).

2012-04-06T12:42:00.000+06:00

Electrical engineering interview topic:

Job interview Topics for Electrical and electronic Engineers (Part-3).

8. Diversity Factor:

The ratio of Sum of individual maximum demand and Maximum demand in power station is called Diversity Factor.

Diversity Factor= [Sum of individual maximum demand/ Maximum demand in power station]

Sum of individual maximum demand <Maximum demand in power station.
Diversity Factor> 1

9. Load Duration Curve:

In the Load Duration Curve the load elements are shown with respect to time. When the Load elements are arranged in the descending magnitude order in the Curve, thus the curve is known as Load Duration Curve.

10. Plant Use Factor:

The ratio of Station Output in KWh with respect to Plant Capacity and hours of use is called Plant Use Factor.

Plant Use Factor= [Station Output in KWh/ Plant Capacity * hours of use]

11. Node Equation:

In the electrical circuit when a junction is formed with two or more pure elements (The elements can be Resistive, Inductive, Capacitive or an ideal source of current or voltage) and are connected to each other at their terminals are called nodes. The process of calculating this nodes and check their working basis is known as node equation. In the Node Equation usually the major nodes are considered. In the nodes, where more then two pure elements are connected in a junction is known as major nodes.

In the Node Equation there are two types of admittance. They are Self Admittance and Mutual Admittance. Some times the Self Admittance is called Driving point & Mutual Admittance is called transfer admittance.

12. Plant Capacity Factor:

The ratio of Actual Energy Produced and the Maximum Energy that could have been produced in the power station is called Plant Capacity Factor.

Plant Capacity Factor= [Actual Energy Produced/ Maximum Energy that could have been produced in the power station]

= [Average Demand/Plant Capacity]

----------------------------------------------------------------------

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Job interview Topics for Electrical and Electronics Engineers (Part-1)

2012-04-02T12:27:00.000+06:00

Electrical Engineering interview question and answers:

Job interview Topics for Electrical and Electronics Engineers (Part-1)

1.Per Unit (pu) Quantities:

The numerical per unit value of any quantity is defined as the ratio of its actual value to another arbitrarily chosen value of quantity of the same dimensions, assumed as the base or reference. It is basically used in one line diagram.

Per unit (pu) quantities|vivasolve.blogspot.com

View more PowerPoint from boromama

Per unit value ≈ (Actual value/ Base value of the same direction)

2. Advantage of Per Unit Representation:

The Per Unit impedance value for apparatus of like rating lies within a very small range.
The impedance of machines is specified by the manufactures of the machine in terms of per unit values.
Per Unit value provides more meaningful information .The chance of confusion between line and phase voltage values on a 3-phase system is reduced.
The computational disturbance or effect in power system is highly reduced with the use of Per Unit quantities.
The Per Unit impedance which is referred on any side of a single phase transformer is the same.
The Per Unit impedance which is referring to the either side of a three-phase transformer is the same regardless of the three-phase connection .Whether the connections are delta-delta, Y-Y, or delta-Y.

3. One –line Diagram:

In the Electrical Circuits there are various components in a circuit diagram. Sometimes, it is needed to simplified electrical circuit diagram by omitting the completed circuit through the neutral and by indicating the component parts by standard symbols, in place of their equivalent circuits. Circuit parameters are not shown and a transmission line is represented by a single line between its two ends.

----------------------------------------------------------------------

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Job interview or viva Topics for Electrical and Electronics Engineers (Part-9).

2012-03-12T23:59:00.000+06:00

Electrical Engineering interview question and answers:

Job interview or viva Topics for Electrical and Electronics Engineers (Part-9).

Importance of electrical energy:

The convert process of electrical energy into any form gives the present advancement of human civilization. This advancement of science and technology give a pride to the electrical energy. The survival and advancement of industry and social structure is now completely based on the low cost, uninterrupted power supply of electrical energy. The following reasons are give electrical energy a major position then other energies,

Easy and reliable control:

The electrical machines have simple and convenient operation along with start and control. Such as, the motor operation can be easily controlled by using turn on and turn off switch. Also the speed of the motor can be varied to any desired range.

Flexibility:

The most important feature of electrical energy is its greater flexibility. This electrical energy can be transported from one place to another with the help of conductor.

Convenient form:

The electrical energy has a wide variety of convenient form. The electrical energy can be converted in most of the known forms. Such as, the electrical energy can be converted into light and heat. The only thing is needed to pass the electrical energy through of high resistance. The more the resistance the more heat can be produced. The electrical energy can be converted into mechanical form when it passes through a motor. Not only this forms, it can be used in various forms.

High efficiency in transmission:

the consumers of the electrical energy normally use this energy from a very distance place from the production center. This electrical energy can be passed through the overhead conductors, known as electrical wire. This electrical energy can be passed through this wire with a very high efficiency and conveniently.

Cleanliness and cheapness:

The electrical energy is normally free from smoke, poisonous gas, fume etc. which helps to make the environment clean. The electrical energy is much cheaper then other energy. If only transmission is considered, once it was setup the only cost is maintenance.

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vivasolve.blogspot.com

Job interview or viva Topics for Electrical and electronic Engineers (Part-8).

2012-03-09T11:33:00.001+06:00

Job interview or viva Topics for Electrical and electronic Engineers (Part-8).

28. Booster transformer: The transformer which boosts up the power flow in the system is normally mentioned as booster transformer. These types of transformer are used for control the voltage of a transmission line which is very far away from the main transformer.

The booster transformer has normally two types. They are:

1. in phase booster transformer.
2. Quadrature booster transformer.

29. In phase booster transformer: In this types of transformer setup an amount of voltage is injected in the system, the additional voltage which is given can be in phase or in phase opposition to the existing line to earth voltage. This in phase booster transformer control the reactive volt amperes transferred in the interconnected system. It use on load tap changing transformers compensates in conjunction, to control the voltage drop.

30. Quadrature booster transformer: These types of transformer leads the system by injecting voltage by 90° .This controls the original or real power flow in the system. The line impedance is negative. The injection of series in phase results in the flow of a quadrature reactive current (KVA) . While the injection voltage is in quadrature the current flows in a direction lagging the injected voltage by nearly 90° . For that the resulting current is in phase with the phase voltage of the line. The tap changing procedure of the voltage of these types of transformer is very slow. This process normally takes 15-20 seconds foe tap changing. This is, not suitable for the rapidly varying load.

38.Write a program to show the sum of even numbers by using pointers.

2012-03-08T22:51:00.002+06:00

38.Write a program to show the sum of even numbers by using pointers.

#include
#include
int main()
{
clrscr();
int numbers[50],*ptr;
int n,i;
cout<<"\nEnter the count\n"; cin>>n;
cout<<"\Enter the numbers one by one\n"; for(i=0;i>numbers[i];
ptr=numbers;
int sum=0;
for(i=0;i {
if(*ptr%2==0)
sum=sum+*ptr;
ptr++;
}
cout<<"\n\nSum of even numbers: "< getch();
return 0;
}

Output:

Enter the count
5
Enter the numbers one by one
10
16
23
45
34

Sum of even numbers: 60

37.Create a class box whose function is passed three double value each of which represents the length of one single of the box compute the volume of the box.

2012-03-08T22:50:00.001+06:00

37.Create a class box whose function is passed three double value each of which represents the length of one single of the box compute the volume of the box.

#include
#include
class box
{
double l,w,h,v;
public:
box(double length,double width,double height)
{l=length ;w=width; h=height;}
void volume(void);
};
void box::volume(void)
{
v=l*w*h;
cout<<"Volume of the box: "< }
int main()
{
clrscr();
box b(3.0,2.5,3.5);x
b.volume();
getch();
return 0;
}

Output:

Volume of the box: 26.25

36.Write a program to show the hexadecimal of n.

2012-03-08T22:49:00.003+06:00

36.Write a program to show the hexadecimal of n.

#include
#include
int main()
{
clrscr();
int n;
cout<<"Enter value of n: "; cin>>n;
if(n<=9)
cout<<"Hexadecimal value:"< else if(n==10)
cout<<"Hexadecimal value: A";
else if(n==11)
cout<<"Hexadecimal value: B";
else if(n==12)
cout<<"Hexadecimal value: c";
else if(n==13)
cout<<"Hexadecimal value: D";
else if(n==14)
cout<<"Hexadecimal value: E";
else if(n==15)
cout<<"Hexadecimal value: F";
getch();
return 0;
}

Output:

Enter value of n: 12
Hexadecimal value: c

Enter value of n: 2
Hexadecimal value:2

35.Write a program to show how the unary minus operator is overloaded .

2012-03-08T22:49:00.001+06:00

35.Write a program to show how the unary minus operator is overloaded .

#include
#include
class space
{
int x;
int y;
int z;
public:
void getdata(int a,int b,int c);
void display(void);
void operator-();
};
void space:: getdata(int a,int b,int c)
{
x=a;
y=b;
z=c;
}
void space::display(void)
{
cout< cout< cout< }
void space::operator-()
{
x=-x;
y=-y;
z=-z;
}
int main()
{
clrscr();
space s;
s.getdata(10,-20,30);
cout<<"s: ";
s.display();
-s;
cout<<"s: ";
s.display();
getch();
return 0;
}

Output:

s: 10 -20 30
s: -10 20 -30

34.Write a program to show the total marks of a particular student of a subject using inheritance .

2012-03-08T22:48:00.000+06:00

34.Write a program to show the total marks of a particular student of a subject using inheritance .

#include
#include
class student
{
char name[30];
int roll_no;
public:
void getdata(void);
void putdata(void);
};
void student:: getdata(void)
{
cout<<"Input:\n"; cout<<"Enter student name: "; cin>>name;
cout<<"Enter roll number: "; cin>>roll_no;
}
void student::putdata(void)
{
cout<<"\n\nOutput:\n\n";
cout<<"Student Name: "< <<"Roll Number: "< }
class attendance
{
public:
int att_mark;
void get_att_mark(int);
void put_att_mark(void);
};
void attendance::get_att_mark(int a)
{
att_mark=a;
}
void attendance::put_att_mark(void)
{
cout<<"Attendance Mark: "< }
class final_exam:public student
{
public:
float part_a;
float part_b;
void get_mark(float,float);
void put_mark(void);
};
void final_exam::get_mark(float x,float y)
{
part_a=x;
part_b=y;
}
void final_exam::put_mark()
{
cout<<"\nMarka of Part A: "< <<"Marks of Part B: "< }
class class_test
{
public:
int ct1,ct2,ct3;
void get_ct_mark(int,int,int);
void put_ct_mark(void);
};
void class_test::get_ct_mark(int p,int q,int r)
{
ct1=p;
ct2=q;
ct3=r;
}
void class_test::put_ct_mark()
{
cout<<"Marks of Class Test 1: "< <<"Marks of Class Test 2: "< <<"Marks of Class Test 3: "< }
class result:public attendance,public final_exam,public class_test
{
float total;
public:
void display(void);
};
void result::display(void)
{
total=att_mark+part_a+part_b+ct1+ct2+ct3;
putdata();
put_att_mark();
put_mark();
put_ct_mark();
cout<<"\nTotal= "< }
int main()
{
clrscr();
result s;
s.getdata();
s.get_att_mark(27);
s.get_mark(90.0,85.5);
s.get_ct_mark(19,20,20);
s.display();
getch();
return 0;
}

Output:

Input:
Enter student name: MOHIN
Enter roll number: 1354

Output:

Student Name: MOHIN
Roll Number: 1354
Attendance Mark: 27

Marka of Part A: 90
Marks of Part B: 85.5

Marks of Class Test 1: 19
Marks of Class Test 2: 20
Marks of Class Test 3: 20

Total= 261.5

33.Write a program by inheritance and show a= 5 b= 10 c= 50 a= 5 b= 20 c= 100

2012-03-08T22:45:00.007+06:00

33.Write a program by inheritance and show

a= 5
b= 10
c= 50

a= 5
b= 20
c= 100

#include
#include
class B
{
int a;
public:
int b;
void get_ab();
int get_a(void);
void show_a(void);
};
class D:public B
{
int c;
public:
void mul(void);
void display(void);
};
void B::get_ab(void)
{
a=5;b=10;
}
int B::get_a()
{
return a;
}
void B::show_a()
{
cout<<"a= "< }
void D::mul()
{
c=b*get_a();
}
void D:: display()
{
cout<<"a= "< cout<<"b= "< cout<<"c= "< }
int main()
{
clrscr();
D d;
d.get_ab();
d.mul();
d.show_a();
d.display();
d.b=20;
d.mul();
d.display();
getch();
return 0;
}

Output:

a= 5
b= 10
c= 50

a= 5
b= 20
c= 100

0	No special protection
1	Protected against solid objects over 50 mm, e.g. accidental touch by persons hands.
2	Protected against solid objects over 12 mm, e.g. persons fingers.
3	Protected against solid objects over 2.5 mm (tools and wires).
4	Protected against solid objects over 1 mm (tools, wires, and small wires).
5	Protected against dust limited ingress (no harmful deposit).
6	Totally protected against dust.

0	No protection.
1	Protection against vertically falling drops of water e.g. condensation.
2	Protection against direct sprays of water up to 15^o from the vertical.
3	Protected against direct sprays of water up to 60^o from the vertical.
4	Protection against water sprayed from all directions - limited ingress permitted.
5	Protected against low pressure jets of water from all directions - limited ingress.
6	Protected against temporary flooding of water, e.g. for use on ship decks - limited ingress permitted.
7	Protected against the effect of immersion between 15 cm and 1 m.
8	Protects against long periods of immersion under pressure.