Search
Home / Technical Articles / 10 Common Questions You Can Expect In Electrical Engineering Examination Test

EE Examination Test

This examination will focus on testing the fundamentals of electrical engineering from low voltage to high voltage. Of course, you can use huge EEP’s resources for help in calculations and finding right answers.

10 Common Questions You Can Expect In Electrical Engineering Examination Test
10 Common Questions You Can Expect In Electrical Engineering Examination Test

Ok, let’s start with questions. You can either take the pencil/paper or MS Word (if you ask me – paper and pen) and give yourself a try. Answers and calculations are placed below questions, but hey – try to give your best before peeking :)


Question #1

A 3-phase transformer bank consisting of a three one-phase transformer is used to step-down the voltage of a 3-phase, 6600V transmission line. If the primary line current is 10A, calculate the secondary line voltage, secondary line current and output kVA for the following connections:

  1. Y / ∆
  2. ∆ / Y. The turns ratio is 12. Neglect losses.

Answer on this question ↓


Question #2

The magnetization characteristic of 4-pole DC series motor may be taken as proportional to current over a part of the working range, on this basis the flux per pole is 4.5 mWb/A. The load requires a gross torque proportional to the square of the speed equal to 30 Nm at 1000 rev/min. The armature is wave-wound and has 492 conductors.

Determine the speed at the which the motor will run and current it will draw when connected to a 220V supply, the total resistance of the motor being 2.0 Ω.

Answer on this question ↓


Question #3

What are the advantages of high voltage transmission? Give its limitations also.

Answer on this question ↓


Question #4

What do you mean by transmission line efficiency and regulation? Can it be negative also?

Answer on this question 


Question #5

With the help of neat sketches explain different types of distribution systems.

Answer on this question 


Question #6

Bring-out the advantageous features of the following:

  1. Gas turbines.
  2. Energy Conservation.

Answer on this question 


Question #7

Why are different levels of voltages used for generation, transmission and distribution of electric power? What are the essential differences between high voltage (H.V.) and low voltage (L.V.) switchgears?

Answer on this question 


Question #8

Where are synchronous motors used?

Answer on this question 


Question #9

Write technical explanation on any TWO of the following:

  1. Parallel operation of transformers
  2. Speed control of induction motors
  3. Applications of induction machines

Answer on this question 


Question #10

Explain the functions and basic requirements of a protective relay. Compare the merits and demerits of overhead lines with an underground distribution system.

Answer on this question 


Answers On Questions:

Below you can find calculations and answers on all above questions. You shouldn’t peek before trying to solve questions by yourself :)


Answer #1

a) Y/∆

  • Given turn ratio = 12
  • V1 = 6600V, I1 = 10A,
  • V2 = ? , I2 = ?
  • o/ρ = kVA
  • IPP = Phase current in primary winding
    ILP = Line current in primary winding
  • For Y: Vp = VL/√3 , Ip = IL
    For ∆: Vp = VL , Ip = IL/√3
  • VPP/VPS = N1 / N2
    VLP/VLS = √3 × VPP/VPS = √3 × N1/N2
    VLS = VLP/12√3 = 6600/ 12√3
    VLS = 315.33 Volts
  • IPP/IPS = N2/N1
    ILP/ILS = IPP/√3 × 1/IPS = (1/√3) × (N2/N1)
    ILS = (1/√3) × (N1 /N2) × ILP = 12√3 × 10Amp. = 120√ 3 Amp.
  • o/ρ kVA = VLS × ILS
    = (6600 × 12√3 × 10) / 12 √3
    o/ρ = 66kVA

b) ∆/Y

  • For Y: Vp = VL/√3 , Ip =IL
    For ∆: Vp = VL , Ip = IL/√3
  • VPP/VPS = N1 /N2
    VLP/VLS = VPP/√3 × VPS = 1/√3 × N1/N2 = 12 /√3
    VLS =√3 × VLP/12 = √3 × 6600/12
    VLS = 550√3 Volts
  • IPP/IPS = N2/N1
    ILP/ILS = √3 × IPP/IPS = √3N2/N1 = √3/12
    ILS = 12/√3 × 10 = 40√3Amp.
  • o/ρ kVA = VLS x ILS
    = (√3 × 6600 × 12 × 10) /  12 × √3
    o/ρ = 66kVA

Go back to question ↑


Answer #2

  • Ea = (φ N Z / 60) × (P/A) = ((4.5 × 10-3 × Ia) × N × 492 (4/2)) / 60
    Ea = 0.0738 N Ia
  • The torque developed :
    T =φ Ia Z (P/A) = (1/2π) (4.5 × 10-3 Ia) × 492 (4/2)
    T = 0.705 Ia2 (equation 1)
  • Further: Ea =V-Ia (Ra +RSe ) = 220 – 2Ia (equation 2)
  • Substituting equations 1 in 2:
    0.0738 N Ia = 220 – 2Ia
    Ia = 220/2+0.0738 N (equation 4)
  • Substituting the expression for Ia in equation (1)
    T = 0.705 (220/2 + 0.0738 N)2
  • Given, TL =KL N2
  • From the given data KL can be calculated as:
    KL = 30/10002 = 3 × 10-5 Nm/rpm
  • Under steady operation condition TL =T
    Or 3 × 10-5 N2 = 0.705 × (200 / 2+0.0738 N)2
    N = 662.6 rpm
  • Substituting for N in equation (4)
    Ia = 220 / (2 + 0.0738 × 663.2)
    Ia = 4.32 Amp.

Go back to question ↑


Answer #3

High voltage transmission is subdivided into HVAC and HVDC transmission systems.

1. HVAC transmission:

Advantages of HVAC transmission are as follows:

As the voltage is increased, the current carried by the conductors decreases. The i2R losses correspondingly get reduced. However the cost of transmission towers, transformers, switches and circuit breakers rapidly increases with increase in voltage, in the upper ranges of a.c. transmission voltages.


2. HVDC transmission:

Advantages

They (HVDC lines) are economical for bulk power transmission. The voltage regulation problem is much less in DC since only IR drop is involved. There is easy reversibality and controllability of power flow through a DC link. Also there is considerable insulation economy.

Limitations:

The systems are costly since installation of complicated converters and DC switchgear is expensive. The converters require considerable reactive power. Lack of HVDC circuit breakers hampers network operation.

Moreover there is nothing like DC transformer. Voltage transformation has to be provided on the a.c. sides of the system.

➡ Resource for further learning

Go back to question ↑


Answer #4

Definition – The ratio of receiving end power to the sending end power of a transmission line is known as the transmission efficiency of the line.

% Transmission Efficiency = (Receiving end power / Sending end power) x 100;
% Transmission Efficiency = (VR IR cosφR/ VS IS cosφS) × 100

where:

  • VR, IR and cosφR are the receiving end voltage, current and power factor while
  • VS, IS cosφS are the corresponding values at sending end
The difference in the voltage at the receiving end of a transmission line between the conditions of no load and full load is called voltage regulation and is expressed as a percentage of the receiving end voltage.

Mathematically, % voltage regulation = (VS – VR) / VR × 100;

%Voltage regulation is also given by = (I R cos φR ± I X sin φR) / ER × 100;

+ for lagging pf
– for leading pf

When the pf is leading, and the term IX sin φR is more than the I R cosφR, then the voltage regulation becomes –ve.

Go back to question ↑


Answer #5

Different types of distribution systems:

1. Radial system

In this system separate feeders radiate from a single sub station and feed the distributors at one end only. Figure shows a single line diagram of a radial system for a DC distribution where a feeder OC supplies a distributor AB at a point A. Obviously the distributor is fed at one end only i.e. point A in this case.

The radial system is employed only when power generated at low voltage and substation is located at the center of the load.

Radial system
Radial system

2. Ring main system

In this system the primaries of distribution transformers form a loop. The loop circuit starts from the substation bus bars, makes a loop through the area to be served and returns to the substation.

Ring main system
Ring main system

The sub-station supplies to the closed feeder LMOPQRS. The distributors are tapped from different points M, O, and Q of the feeder through distribution transformer. This system has the advantage of continuity of supply and less voltage fluctuations.


3. Inter-connected system

When the feeder ring is energized by two or more than two generating stations or substations it is called inter connected system.

Inter-connected system
Inter-connected system

In the above diagram of the inter-connected system the closed feeder ring ABCD is supplied by two substations S1 and S2 at D and C. Inter connected system has the advantages of increased service reliability and increasing efficiency of the system by reducing reserve power capacity.

➡ Resource for further learning

Go back to question ↑


Answer #6

a) Advantageous features of Gas Turbines:

  • Very high power-to-weight ratio.
  • Smaller in size.
  • Moves in one direction only, with less vibrations.
  • Less moving parts.
  • Low operating pressures.
  • High operation speeds.
  • Low lubricating oil cost and consumption.

b) Energy conservation advantageous features

  • It saves, the need of fuel imports for many power plants.
  • It minimises CO2 and environmental pollution.
  • It helps in global warming problem solution.
  • Less air pollution.
  • Fewer power plants and liquid natural gas ports are needed.
  • Less need to secure oil and natural gas overseas.

Go back to question ↑


Answer #7

Answer on first question:

Different Levels Of Voltages For Generation, Transmission And Distribution:

  • The generation voltage is upto 30 KV AC rms (line to line).
  • The long distance high power transmission is by EHVAC lines 220 kV, 400 kV & 760 kV Ac.
    In special cases, HVDC line is preffered. The rated voltages of HVDC lines are ±250 kV, ± 400 kV, ± 500 kV & ± 600 kV.
  • The backbone transmission system is done by EHV AC transmission lines (400 kV AC & 200 kV AC).
  • Distribution is at lower AC voltage between 132 kV AC and 3.3 kV AC.
  • Utilization is at the low voltages up to 1kV and medium voltage up to 11 kV.
  • The industrial substations receive power at distribution voltage such as 3.3 kV and step down it to 440 V AC.
    Larger industries receive power at 132 kV and internal distribution at 3.3 kV to 440 volt AC.

➡ Resource for further learning


Answer on second question:

High voltage (HV) switchgears:
  • Circuit breaker: Switching during normal and abnormal conditions, interrupt the fault currents.
  • Isolators: It is disconnecting switch to disconnect the system from line parts under no load condition.
  • Earth switch: This is used to discharge the voltage on the lines to earth after disconnecting them.
  • Surge arrester: This is used to divert the high voltage surge to earth and maintaining continuity during normal voltage.
  • Current Transformer: To step down the current for measurement, protection and control purposes.
  • Potential Transformer: To step down the voltage for the purpose of protection, measurement and control.

Low voltage (LV) switchgears:
  • MCBs (Miniature Circuit Breakers): Switching OFF during abnormal conditions to interrupt the fault current.
  • Fuses: A short length wire having low melting point, connected in series with circuit. In the event of fault, the circuit current rises abruptly and fuse wire melts to interrupt the circuit.
  • Switches: These are used to ON/OFF the power of a circuit. These are used in power / control circuits. The switches are specified as per voltage rating, current rating, number of poles, duty cycle and fault interruption capacity.

Go back to question ↑


Answer #8

Synchronous motor applications:

  1. Power factor correction
  2. Constant speed, constant load drives
  3. Voltage regulation

Over excited synchronous motors having leading power factor are widely used for improving power factor of those systems which employ a large number of induction motors and other devices having lagging power factor such as welders and fluorescent lights etc.

In constant speed applications, synchronous motors are well suited due to their high efficiency and high speed, such as centrifugal pumps, belt driven reciprocating compressors, blowers, line shafts, rubber and paper mill etc.

Voltage in long transmission lines varies greatly at the end when large inductive load are present and when inductive load disconnected suddenly, the voltage tends to rise considerably above its normal value because of line capacitance.

Therefore, by installing a synchronous motor with field regulator, the voltage rise can be controlled.

Go back to question ↑


Answer #9

1. Parallel operation of transformers

Certain conditions must be satisfied in order to avoid any local circulating currents and to ensure that the transformers share the common load in proportion to their kVA ratings.

These conditions are:

  1. The Primary winding of the transformers should be suitable for the supply system voltage and frequency.
  2. The transformer should be properly connected with regard to polarity.
  3. The voltage rating of both primaries and secondaries should be identical (same turn ratio)
  4. Percentage impedances should be equal in magnitude and have the same X/R ratio in order to avoid circulating currents and operation at different Power factors.
  5. With transformers having different kVA ratings, the equivalents impedances should be inversely proportional to the individual kVA ratings, if circulating currents are to be avoided

In case of 3 phase transformers’ parallel operation, the following conditions are also added with the above conditions.

  1. The voltage ratio must refer to terminal voltage of primary and secondary.
  2. The phase displacement between primary and secondary voltages must be the same for all transformers, which are to be connected for parallel operation.
  3. The Phase sequence must be the same.
  4. All 3-phase transformers must have same construction either core or shell.

➡ Resource for further learning


2. Speed control of induction motors

The speed control of induction motor is obtained by various methods:

  1. Frequency variation.
  2. Pole changing.
  3. Changing motor circuit resistance.
A slip ring (wound) motor is an adjustable speed motor. The rotor winding terminals are connected to slip rings. A 3-phase star connected rheostat is provided separately with motor. By changing the resistance in rotor circuit, the torque speed characteristics of machine can be changed.

The Speed and Torque can be changed as:

T × α × r2 /s and Slip‘s’ α × r2

By increasing rotor resistance, The torque increases and slip ‘s’ increases, Therefore, speed decrease as:

T × α × 1/N

The speed reduces in slip ring induction motor through rheostat control. And this can be obtained only if the motor is loaded under no load condition. The no-load speed changes are very little with the variation in the load resistance.


3. Applications of Induction Motors

The squirrel cage induction motor with single or double pole changing windings are available as follows:

ClassApplications
Variable torque, power output α N2Fans, centrifugal pumps
Constant –torque power output α NConveyors, stokers, reciprocating compressors, printing presses
Inverse torque power output rating constantMachine Tools, lathes, boring mills, drill, planers

The Multi-speed motors are of slip ring type used for hoist, conveyor and elevator.

Go back to question ↑


Answer #10

Answer on first question:

Functions of a protective relay:

The protective relaying senses the abnormal conditions as a part of the power system and gives an alarm or isolate that part from the healthy system. The relays are compact, self-contained devices, which respond to abnormal conditions.

Whenever abnormal conditions occur, the relay close its contacts. Thereby trip circuit of the circuit breaker is closed and circuit breaker will open. So that, the faulty part will be disconnected from the supply.

Basic requirements of a protective relay:

Selectivity, discrimination – The protective relaying should select the faulty part of the system and should isolate, as far as possible only faulty part from the remaining healthy system.

Speed, Time – It is the time between fault instant and closing of relay. A rapid contact fault cleaning i.e. 0.07 second with 60 kA rms value of current, has no damage to the system but if it is 7 sec, the bus bar will destroy complete. Therefore, relay time must be minimum as much as possible (i.e. in millisecond.)

Fault clearing time – Fault clearing time = Relay time + Breaker time

Sensitivity, power consumption – It refer to the smallest value of actuating quantity at which protection starts operating in relation with the minimum value of fault current in the protection zone

Sensitivity factor ‘KS = IS /IO = Min. short-circuit current / Min. operating current in protection

Stability – A quality of protective system by the virtue of which , the protective system remains operating and stable under certain specified conditions such as system disturbances, through faults, transient etc.

Reliability – The protective relaying should not fail to operate in the event of faults in protected zone. The reliability of protective systems depends on diverse aspects such as protective gear manufactures, Electricity Boards & Associates.

Adequateness – The adequateness of protection is judged by considering following aspects:

  • Rating of protecting machines.
  • Location of protecting machines.
  • Probability of abnormal condition due to internal and external causes.
  • Cost of machine, importance.
  • Continuity of supply as affected by failure of machine.

➡ Resource for further learning


Answer on second question:

Merits
Overhead power transmission linesUnder ground Power cable
1. Easy maintenance and repair1. Less disturbances for other system
2. Low cost of installation2. System looks neat and beautiful
3. Mostly used in transmission system due to effective voltage up to 400 kV3. Mostly used in distribution system (Medium and low voltages up to 11kW)
4. Less skilled staff is required4. Less lightening thunder effect

Demerits
Overhead power transmission linesUnder ground Power cable
1. Chances of frequent failure due disturbance from other system1. High initial cost
2. Less safe2. High maintenance and repair cost
3. Lightning protection is required3. Problem of charging current in HV transmission
4. Chances of tapping or theft more4. Highly skilled staff is required

Go back to question ↑


Reference // Typical questions and answers; Subject: ELECTRICAL ENGINEERING

Premium Membership

Premium membership gives you an access to specialized technical articles and extra premium content (electrical guides and software).
Get Premium Now ⚡

About Author

author-pic

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on

3 Comments


  1. Claudio Biccai
    Dec 29, 2017

    Dear Sirs,

    Probably I did not understand the question for exercise 1.
    However, my 3 phase power calculation is 114.3 kVA.
    Could you verify my doubt?
    Thanks in advance and best regards.
    Claudio.


    • constance
      Jan 06, 2018

      same here. But im thinking 66kvA is for one of the single phase trnx.


  2. Amit Kumar Prusty
    Dec 27, 2017

    Thank you so much Sir. I am the regular follower of EEP. I have a request for an article on Siemens AC Drives. Total installation, types of AC Drives with uses and commissioning of AC Drives .

Leave a Comment

Tell us what you're thinking... we care about your opinion!

Premium Membership

Get access to premium electrical guides, technical articles and much more!