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Monday, 22 July 2019

BECIL(Broadcast Engineering Consultant India Limited) JOBS

BECIL Recruitment 2019 notification released. Broadcast Engineering Consultant India Limited (BECIL) invites online applications from eligible candidates for deployment in Noida Metro Rail Corporation purely on a contract basis.



You can check BECIL Vacancy 2019 details listed below.

Interested candidates may apply for 199 posts for Station Controller /Train Operator, Customer Relations Assistant, Junior Engineer (Electrical, Mechanical, Electronics, Civil), Maintainer, Accounts Assistant & Office Assistant from 22nd July 2019.

BECIL Recruitment 2019: Important Dates

EventsDates
Online Registration begins.

22nd July 2019 (10.00 hours)
Online Registration concludes21st August 2019 (23.59 hrs)
Application Fee payment from22nd July - 21st August

BECIL Vacancies 2019

Post CodeName of PostNo. of vacancy
NE-01Station Controller /Train Operator09
NE-02Customer Relations Assistant16
NE-03Junior Engineer( Electrical)12
NE-04Junior Engineer( Mechanical)04
NE-05Junior Engineer (Electronics)15
NE-06Junior Engineer (Civil)04
NE-07Maintainer /Fitter09
NE-08Maintainer/ Electrician29
NE-09Maintainer/Electronic & Mechanic90
NE-10Maintainer / Ref & AC Mechanic07
NE11Accounts Assistant03
NE12Office Assistant01

Eligibility Criteria

Educational Qualification & Age Limit

Post CodeName of PostMinimum Education QualificationAge as on 01/01/19
NE-01Station Controller /Train Operator
  • Graduate in Science OR
  • Three years of diploma in Electrical /Electronics/Electronics and Telecommunications /Civil/Mechanical /IT/Com Science OR
  • B.Tech in Electrical /Electronics/Electronics and Telecommunications /Civil/Mechanical /IT/Com Science
Min: 18 years
Max: 32 years
NE-02Customer Relations Assistant
  • 03 years Graduate in any discipline
Min: 18 years
Max: 32 years
NE-03Junior Engineer( Electrical)
  • Three years Diploma in Electrical Engineering
    /equivalent trade OR
  • B.Tech in Electrical Engineering/Equivalent
Min: 18 years
Max: 32 years
NE-04Junior Engineer( Mechanical)
  • Three years diploma in Mechanical Engineering /Equivalent OR
  • B.Tech in Mechanical Engineering
Min: 18 years
Max: 32 years
NE-05Junior Engineer (Electronics)
  • Three years diploma in Electronics/ Electronics & Communications Engineering OR
  • B.Tech in Electronics/ Electronics and Communication Engineering
Min: 18 years
Max: 32 years
NE-06Junior Engineer (Civil)
  • Three years diploma in Civil Engineering /equivalent OR
  • B.Tech in Civil Engineering
Min: 18 years
Max: 32 years
NE-07Maintainer /Fitter
  • ITI/Fitter ( NCVT/SCVT)
Min: 18 years
Max: 32 years
NE-08Maintainer/ Electrician
  • ITI / Electrician ( NCVT/SCVT)
Min: 18 years
Max: 32 years
NE-09Maintainer/Electronic & Mechanic
  • ITI / Electro Mechanic / IT /TV and Radio (NCVT/SCVT)
Min: 18 years
Max: 32 years
NE-10Maintainer / Ref & AC Mechanic
  • ITI /Ref & AC ( NCVT/SCVT)
Min: 18 years
Max: 32 years
NE11Accounts Assistant
  • B.Com or CA(Inter) /ICWA
Min: 18 years
Max: 32 years
NE12Office Assistant
  • BBA/BCA
Min: 18 years
Max: 32 years

Note

  • Candidates who are appearing at the qualifying exam, passing of which would make them educationally eligible for these posts, have not yet been informed of the results, will also be eligible for applying for these posts.
  • Candidates can apply for more than one post subject to eligibility but application money for each is required to be deposited separately
  • Candidates with higher qualificationin respective disciplines can also apply except maintainers posts for which ITI (NCVT/SCVT) in specific trades is essential )
  • Candidates can apply for more than one post as per their qualifications.
  • However, they must note that the written test for postcodes NE03, NE04, NE05, NE 06 may be held on one single day depending upon the volume of applications received.

Application Fee

  • UR & OBC (including Ex-servicemen) candidates:₹ 675
  • SC/ST/PWD candidates: ₹ 500

Selection Procedure

Station Controller/Train Operator & Customer Relation Assistant

  • Written Test (two papers), Psycho Test (qualifying) & Personal Interaction

Junior Engineer (Electrical/Electronics/Civil/Mechanical)

  • Written Test (two papers), Skill/trade test followed by Medical examination

Accounts Assistant, Office Assistant

  • Written test (two papers), followed by personal interaction /Medical examination

Maintainer

  • Written Test (one paper), followed by a skill test/ trade test and Medical examination
  • There will be no General English paper for this category and no interview

Written Test

  • The written test will consist of two papers (Paper-I and Paper-II).

Paper-I

  • There will be a total of 90 questions to be answered in 90 minutes, where each carrying equal marks.
  • There will be negative marking, for every wrong answer, 1/3 marks will be deducted.
  • Paper-I will consist of multiple choice objective type questions, bilingual (Hindi/English) including
    • General Awareness
    • Logical Ability,
    • Quantitative Aptitude and
    • Knowledge of Discipline.

Paper-II (Not for Maintainer Post, NE 07 to NE 10)

  • It will consist of objective type questions on General English to judge the knowledge of the English language.
  • There will be a total of 30 questions to be answered in 30 minutes, where each carrying equal marks.
  • There will be negative marking, for every wrong answer, 1/3 marks will be deducted.
  • Candidates will have to qualify in Paper-II (GENERAL ENGLISH) to be evaluated for Paper I ( except for maintainers).
  • Candidates who stand in the merit of Written Test of Paper I will be called for personal interaction /psycho test/ Skill test/Medical Examination in Noida/ Delhi/NCR.
  • Written Test can be conducted either through online mode.

Medical Examination

  • All candidates shall have to undergo the medical fitness test(s) and meet the medical standards prescribed by NMRC for various posts.
  • Expenses for the first time medical examination of the candidate will be borne by NMRC.
  • However, in case a candidate seeks an extension for joining or re-examination, subject to extant rules, then for the second time medical examination/re-examination, if need so arises,, the expenditure for the medical test/s will be borne by the candidate himself/herself.
  • Candidates having undergone Lasik surgery are not suitable for any post except Accounts Assistant, Office Assistant.
  • Click here to apply

Friday, 12 July 2019

RESULT OF THE CIVIL SERVICES (PRELIMINARY) EXAMINATION, 2019

On the basis of the result of the Civil Services (Preliminary)
Examination, 2019 held on 02/06/2019, the candidates with the
following Roll Numbers have qualified for admission to the Civil
Services (Main) Examination, 2019.


The candidature of these candidates is provisional. In accordance
with the Rules of the Examination, all these candidates have to apply
again in the Detailed Application Form-I (DAF-I) for the Civil Services
(Main) Examination, 2019, which will be available on the website of the
Union Public Service Commission (https://upsconline.nic.in) during
the period from 01/08/2019 (Thursday) to 16/08/2019 (Friday) till
6:00 P.M. All the qualified candidates are advised to fill up the DAF-I
ONLINE and submit the same ONLINE for admission to the Civil
Services (Main) Examination, 2019 to be held from Friday, the
20/09/2019. Important instructions for filling up of the DAF-I and its
submission will also be available on the website. The candidates who
have been declared successful have to first get themselves registered
on the relevant page of the above website before filling up the ONLINE
DAF-I. The qualified candidates are further advised to refer to the
Rules of the Civil Services Examination, 2019 published in the Gazette
of India (Extraordinary) of Department of Personnel and Training
Notification dated 19.02.2019.
It may be noted that mere submission of DAF-I does not, ipso
facto, confer upon the candidates any right for admission to the Civil
Services (Main) Examination, 2019. The e-Admit Card along with the
Time Table of the said Examination will be uploaded on the
Commission’s Website for the eligible candidates around 3-4 weeks
before the commencement of the Examination. Changes, if any, in the
postal address or email address or mobile number after submission of
the DAF-I may be communicated to the Commission at once.


Click here to download Result UPSC Civil services

Expected Cut off marks for RRB JE CBT1

We are going to give you a expected cut off of RRB JE CBT-1 2019 which will be required for getting a Government Job in Indian railways.












So, let’s have a look at the Expected Cut-Off Marks for RRB JE CBT-1 2019 Exam held from 22
nd May 2019 till 2nd June 2019.





























  1. Expected Cut-Off for RRB JE 1st Stage CBT 2019 Exam



    Category



    Expected Cut-Off (Out of 100 Marks)



    General



    55 to 65 Marks



    OBC



    50 to 55 Marks



    SC



    45 to 50 Marks



    ST



    40 to 45 Marks















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Thursday, 11 July 2019

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Monday, 8 July 2019

Electrical Important Question & Answer What is Ferranti effect?




ANS  The Ferranti effect is an increase in voltage occurring at the receiving end of a long transmission line, above the voltage at the sending end. This occurs when the line is energized, but there is a very light load or the load is disconnected


   How corona loss can be reduced?


ANS Corona loss can be reduced by increasing the diameter of the conductor and increasing the distance between two conductors

    How skin effect can be reduced?
 ANS  To reduce skin effect the shape of the wire should be less for stranded conductor than that of the solid conductor.



   What is circuit breaker?



    • ANS

    • Circuit Breaker is a device which opens when the fault occurs, the contact of the circuit breaker opens and separate the circuit from overcurrent

    • It consists of  contact lever with one end connected to electromagnet and other end connected to fixed contact if the current flow exceeds the safe operating range electromagnet gets energized and the contact opens from the fixed end thus separating the circuit





  • Again if the current reduces the electromagnet get de-energized so contact will join  to include the circuit again



 What is Relay?


ANS


  • Relay is a electromagnetic device which opens in the occurrence of fault. It is based on the principle of electromagnetic attraction and electromagnetic induction

  • When the fault occurs supply to the relay will be lost hence the contact opens to prevent the line from fault




  What is difference between relay and circuit breaker?


ANS

Relay prevents phase lines from fault where as circuit breaker prevents circuts from fault

Why star delta starter is preferred with induction motor?


Star delta starter is preferred with induction motor due to following reasons:
• Starting current is reduced 3-4 times of the direct current due to which voltage drops and hence it causes less losses.
• Star delta starter circuit comes in circuit first during starting of motor, which reduces voltage 3 times, that is why current also reduces up to 3 times and hence less motor burning is caused.
• In addition, starting torque is increased and it prevents the damage of motor winding.

State the difference between generator and alternator


Generator and alternator are two devices, which converts mechanical energy into electrical energy. Both have the same principle of electromagnetic induction, the only difference is that their construction. Generator persists stationary magnetic field and rotating conductor which rolls on the armature with slip rings and brushes riding against each other, hence it converts the induced emf into dc current for external load whereas an alternator has a stationary armature and rotating magnetic field for high voltages but for low voltage output rotating armature and stationary magnetic field is used.

Why AC systems are preferred over DC systems?


Due to following reasons, AC systems are preferred over DC systems:
a. It is easy to maintain and change the voltage of AC electricity for transmission and distribution.
b. Plant cost for AC transmission (circuit breakers, transformers etc) is much lower than the equivalent DC transmission
c. From power stations, AC is produced so it is better to use AC then DC instead of converting it.
d. When a large fault occurs in a network, it is easier to interrupt in an AC system, as the sine wave current will naturally tend to zero at some point making the current easier to interrupt.

How can you relate power engineering with electrical engineering?


Power engineering is a sub division of electrical engineering. It deals with generation, transmission and distribution of energy in electrical form. Design of all power equipments also comes under power engineering. Power engineers may work on the design and maintenance of the power grid i.e. called on grid systems and they might work on off grid systems that are not connected to the system.

What are the various kind of cables used for transmission?


Cables, which are used for transmitting power, can be categorized in three forms:
• Low-tension cables, which can transmit voltage upto 1000 volts.
• High-tension cables can transmit voltage upto 23000 volts.
• Super tension cables can transmit voltage 66 kV to 132 kV.

Why back emf used for a dc motor? highlight its significance.


The induced emf developed when the rotating conductors of the armature between the poles of magnet, in a DC motor, cut the magnetic flux, opposes the current flowing through the conductor, when the armature rotates, is called back emf. Its value depends upon the speed of rotation of the armature conductors. In starting, the value of back emf is zero.

What is slip in an induction motor?


Slip can be defined as the difference between the flux speed (Ns) and the rotor speed (N). Speed of the rotor of an induction motor is always less than its synchronous speed. It is usually expressed as a percentage of synchronous speed (Ns) and represented by the symbol ‘S’.

Explain the application of storage batteries.


Storage batteries are used for various purposes, some of the applications are mentioned below:

• For the operation of protective devices and for emergency lighting at generating stations and substations.
• For starting, ignition and lighting of automobiles, aircrafts etc.
• For lighting on steam and diesel railways trains.
• As a supply power source in telephone exchange, laboratories and broad casting stations.
• For emergency lighting at hospitals, banks, rural areas where electricity supplies are not possible.




Explain advantages of storage batteries


Few advantages of storage batteries are mentioned below:
• Most efficient form of storing energy portably.
• Stored energy is available immediately because there is no lag of time for delivering the stored energy.
• Reliable source for supply of energy.
• The energy can be drawn at a fairly constant rate.

. What are the different methods for the starting of a synchronous motor.


Starting methods: Synchronous motor can be started by the following two methods:

• By means of an auxiliary motor: The rotor of a synchronous motor is rotated by auxiliary motor. Then rotor poles are excited due to which the rotor field is locked with the stator-revolving field and continuous rotation is obtained.
• By providing damper winding: Here, bar conductors are embedded in the outer periphery of the rotor poles and are short-circuited with the short-circuiting rings at both sides. The machine is started as a squirrel cage induction motor first. When it picks up speed, excitation is given to the rotor and the rotor starts rotating continuously as the rotor field is locked with stator revolving field.

  Name the types of motors used in vacuum cleaners, phonographic appliances, vending machines, refrigerators, rolling mills, lathes, power factor improvement and cranes.


Following motors are used: -
• Vacuum cleaners- Universal motor.
• Phonographic appliances – Hysteresis motor.
• Vending machines – Shaded pole motor.
• Refrigerators – Capacitor split phase motors.
• Rolling mills – Cumulative motors.
• Lathes – DC shunt motors.
• Power factor improvement – Synchronous motors.

. State Thevenin’s Theorem:


According to thevenin’s theorem, the current flowing through a load resistance
Connected across any two terminals of a linear active bilateral network is the ratio open circuit voltage (i.e. the voltage across the two terminals when RL is removed) and sum of load resistance and internal resistance of the network. It is given by Voc / (Ri + RL).

. State Norton’s Theorem


The Norton’s theorem explains the fact that there are two terminals and they are as follows:
• One is a terminal active network containing voltage sources
• Another is the resistance that is viewed from the output terminals. The output terminals are equivalent to the constant source of current and it allows giving the parallel resistance.

The Norton’s theorem also explains the constant current that is equal to the current of the short circuit placed across the terminals. The parallel resistance of the network can be viewed from the open circuit terminals when all the voltage and current sources are removed and replaced by the internal resistance.

. State Maximum power transfer theorem


The Maximum power transfer theorem explains about the load that a resistance will extract from the network. This includes the maximum power from the network and in this case, the load resistance is being equal to the resistance of the network and it also allows the resistance to be equal to the resistance of the network. This resistance can be viewed by the output terminals and the energy sources can be removed by leaving the internal resistance behind.


Explain different losses in a transformer.


There are two types of losses occurring in the transformer:
• Constant losses or Iron losses: The losses that occur in the core are known as core losses or iron losses. Two types of iron losses are:
o eddy current loss
o Hysteresis loss.
These losses depend upon the supply voltage, frequency, core material and its construction. As long as supply voltage and frequency is constant, these losses remain the same whether the transformer is loaded or not. These are also known as constant losses.
• Variable losses or copper losses: when the transformer is loaded, current flows in primary and secondary windings, there is loss of electrical energy due to the resistance of the primary winding, and secondary winding and they are called variable losses. These losses depend upon the loading conditions of the transformers. Therefore, these losses are also called as variable losses.

 Explain different types of D.C motors? Give their applications


Different type of DC motors and their applications are as follows:-
• Shunt motors: It has a constant speed though its starting torque is not very high. Therefore, it is suitable for constant speed drive, where high starting torque is not required such as pumps, blowers, fan, lathe machines, tools, belt or chain conveyor etc.
• Service motors: It has high starting torque & its speed is inversely proportional to the loading conditions i.e. when lightly loaded, the speed is high and when heavily loaded, it is low. Therefore, motor is used in lifts, cranes, traction work, coal loader and coal cutter in coalmines etc.
• Compound motors: It also has high starting torque and variable speed. Its advantage is, it can run at NIL loads without any danger. This motor will therefore find its application in loads having high inertia load or requiring high intermittent torque such as elevators, conveyor, rolling mill, planes, presses, shears and punches, coal cutter and winding machines etc.

. Explain the process of commutation in a dc machine. Explain what are inter-poles and why they are required in a dc machine.


Commutation: It is phenomenon when an armature coil moves under the influence of one pole- pair; it carries constant current in one direction. As the coil moves into the influence of the next pole- pair, the current in it must reverse. This reversal of current in a coil is called commutation. Several coils undergo commutation simultaneously. The reversal of current is opposed by the static coil emf and therefore must be aided in some fashion for smooth current reversal, which otherwise would result in sparking at the brushes. The aiding emf is dynamically induced into the coils undergoing commutation by means of compoles or interpoles, which are series excited by the armature current. These are located in the interpolar region of the main poles and therefore influence the armature coils only when these undergo commutation.

. Comment on the working principle of operation of a single-phase transformer.


Working principle of operation of a single-phase transformer can be explained as
An AC supply passes through the primary winding, a current will start flowing in the primary winding. As a result, the flux is set. This flux is linked with primary and secondary windings. Hence, voltage is induced in both the windings. Now, when the load is connected to the secondary side, the current will start flowing in the load in the secondary winding, resulting in the flow of additional current in the secondary winding. Hence, according to Faraday’s laws of electromagnetic induction, emf will be induced in both the windings. The voltage induced in the primary winding is due to its self inductance and known as self induced emf and according to Lenze’s law it will oppose the cause i.e. supply voltage hence called as back emf. The voltage induced in secondary coil is known as mutually induced voltage. Hence, transformer works on the principle of electromagnetic induction.

. Define the following terms:-


• Reliability,
• Maximum demand,
• Reserve-generating capacity,
• Availability (operational).


Reliability: It is the capacity of the power system to serve all power demands without failure over long periods.
Maximum Demand: It is maximum load demand required in a power station during a given period.
Reserve generating capacity: Extra generation capacity installed to meet the need of scheduled downtimes for preventive maintenance is called reserve-generating capacity.
Availability: As the percentage of the time a unit is available to produce power whether needed by the system or not.

. Mention the disadvantages of low power factor? How can it be improved?


Disadvantages of low power factor:
• Line losses are 1.57 times unity power factor.
• Larger generators and transformers are required.
• Low lagging power factor causes a large voltage drop, hence extra regulation equipment is required to keep voltage drop within prescribed limits.
• Greater conductor size: To transmit or distribute a fixed amount of power at fixed voltage, the conductors will have to carry more current at low power factor. This requires a large conductor size.



 State the methods of improving power factor?


Methods of improving power factor:
• By connecting static capacitors in parallel with the load operating at lagging power factor.
• A synchronous motor takes a leading current when over excited and therefore behaves like a capacitor.
• By using phase advancers to improve the power factor of induction motors. It provides exciting ampere turns to the rotor circuit of the motor. By providing more ampere-turns than required, the induction motor can be made to operate on leading power factor like an overexcited synchronous motor.

. State the factors, for the choice of electrical system for an aero turbine.


The choice of electrical system for an aero turbine is guided by three factors:
• Type of electrical output: dc, variable- frequency ac, and constant- frequency ac.

• Aero turbine rotational speed: constant speed with variable blade pitch, nearly constant speed with simpler pitch- changing mechanism or variable speed with fixed pitch blades.

• Utilization of electrical energy output: in conjunction with battery or other form of storage, or interconnection with power grid.

 What are the advantages of VSCF wind electrical system?


Advantages of VSCF wind electrical system are:
• No complex pitch changing mechanism is needed.
• Aero turbine always operates at maximum efficiency point.
• Extra energy in the high wind speed region of the speed – duration curve can be extracted
• Significant reduction in aerodynamic stresses, which are associated with constant – speed operation.

. Explain the terms real power, apparent power and reactive power for ac circuits and also the units used.


• Real Power: It is the product of voltage, current and power factor i.e. P = V I cos j and basic unit of real power is watt. i.e. Expressed as W or kW.
• Apparent power: It is the product of voltage and current. Apparent power = V I and basic unit of apparent power is volt- ampere. Expressed as VA or KVA.
• Reactive Power: It is the product of voltage, current and sine of angle between the voltage and current i.e. Reactive power = voltage X current X sinj or Reactive power = V I sin j and has no other unit but expressed in VAR or KVAR.

 Define the following: Average demand, Maximum demand, Demand factor, Load factor.


• Average Demand: the average power requirement during some specified period of time of considerable duration is called the average demand of installation.
• Maximum Demand: The maximum demand of an installation is defined as the greatest of all the demand, which have occurred during a given period. It is measured accordingly to specifications, over a prescribed time interval during a certain period.
• Demand Factor: It is defined as the ratio of actual maximum demand made by the load to the rating of the connected load.
• Load Factor: It is defined as the ratio of the average power to the maximum demand.



 Explain forward resistance, static resistance and dynamic resistance of a pn junction diode.


• Forward Resistance: Resistance offered in a diode circuit, when it is forward biased, is called forward-resistance.
• DC or Static Resistance: DC resistance can be explained as the ratio of the dc-voltage across the diode to the direct current flowing through it.
• AC or Dynamic Resistance: It can be defined as the reciprocal of the slope of the forward characteristic of the diode. It is the resistance offered by a diode to the changing forward current.

 How does Zener phenomenon differ from Avalanche breakdown?


The phenomenon when the depletion region expands and the potential barrier increases leading to a very high electric field across the junction, due to which suddenly the reverse current increases under a very high reverse voltage is called Zener effect. Zener-breakdown or Avalanche breakdown may occur independently or both of these may occur simultaneously. Diode junctions that breakdown below 5v are caused by Zener Effect. Junctions that experience breakdown above 5v are caused by avalanche-effect. The Zener-breakdown occurs in heavily doped junctions, which produce narrow depletion layers. The avalanche breakdown occurs in lightly doped junctions, which produce wide depletion layers.

. Compare JFET’s and MOSFET’s.


Comparison of JFET’s and MOSFET’s:
• JFET’s can only be operated in the depletion mode whereas MOSFET’s can be operated in either depletion or in enhancement mode. In a JFET, if the gate is forward-biased, excess-carrier injunction occurs and the gate-current is substantial.
• MOSFET’s have input impedance much higher than that of JFET’s. Thus is due to negligible small leakage current.
• JFET’s have characteristic curves more flat than that of MOSFET is indicating a higher drain resistance.
• When JFET is operated with a reverse-bias on the junction, the gate-current IG is larger than it would be in a comparable MOSFET.

 Explain thin film resistors and wire-wound resistors


a. Thin film resistors- It is constructed as a thin film of resistive material is deposited on an insulating substrate. Desired results are obtained by either trimming the layer thickness or by cutting helical grooves of suitable pitch along its length. During this process, the value of the resistance is monitored closely and cutting of grooves is stopped as soon as the desired value of resistance is obtained.
b. Wire wound resistors – length of wire wound around an insulating cylindrical core are known as wire wound resistors. These wires are made of materials such as Constantan and Manganin because of their high resistivity, and low temperature coefficients. The complete wire wound resistor is coated with an insulating material such as baked enamel

 What is a differential amplifier? Also, explain CMRR.


Differential Amplifier: The amplifier, which is used to amplify the voltage difference between two input-lines neither of which is grounded, is called differential amplifier. This reduces the amount of noise injected into the amplifier, because any noise appearing simultaneously on both the input-terminals as the amplifying circuitry rejects it being a common mode signal.
CMRR: It can be defined as the ratio of differential voltage-gain to common made voltage gain. If a differential amplifier is perfect, CMRR would be infinite because in that case common mode voltage gain would be zero.






Only question ( answer this question in below comment section)

Q1)  Is two wattmeter method is applicable for balance as well as an unbalanced load?

Q2) what are the methods of transformer protection?

Q3) why silicon is used in the core of transformer?

Q4) why secondary of current transformer never open circuit?

Q5) what is phase error in a potential transformer?

Q6) what is phantom loading?





 

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