BEL Trainee Engineer Electrical Mock Test - 3 - Electrical Engineering (EE) MCQ

Concept:

Inertia constant (H) is given by,

Calculation:

Given that, inertia constant (H) = 4 sec

MVA rating (S) = 100 MVA

KE stored in rotor = H × S = 4 × 100 = 400 MJ

Unit impulse function:

It is defined as

Properties:

1.

2.

3. x(t) δ(t – t0) = x(t0)

4.

5.

6.

The discrete time version of the unit impulse is defined by

Properties:

1. x[n] δ[n] = x[0] δ[n]

2.

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4.

5. δ[an] = δ[n]

Concept:

The sign convention for power calculation is given below.

When current entering the positive terminal, the power absorbed is positive and the power delivered is negative.

Power absorbed (P) = VI

Power delivered or supplied (P) = -VI

When current entering the negative terminal, the power delivered is positive and the power absorbed is negative.

Power delivered or supplied (P) = VI

Power absorbed (P) = -VI

In any element power absorb is equals to negative times of power delivered or supplying to element.

P_absorb= - P_supp

Calculation:

In the given circuit, the current enters the negative terminal. So, the power absorbed is negative.

Power absorbed, P = - (10) (8.5) = -85 Watts

Concept:

Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane. In particular, zeros will not lie on the right side of the s-plane.

For a minimum phase system,

Where P & Z are finite no. of poles and zeros of G(s)H(s)

Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane. In particular, zeros lie on the right side of the s-plane.

Stable system: A system is said to be stable if all the poles lie on the left side of the s-plane.

Application:

As one zero lies in the right side of the s-plane, it is a non-minimum phase transfer function.

As there no poles on the right side of the s-plane, it is a stable system.

Concept:

In a energy meter,

Meter constant = (Number of revolutions) / (Energy consumed (E) in kWh)

E (in kWh) = voltage x current x cos ϕ × time x 10^-3

Calculation:

Given that,

Voltage = 230 V

Current = 4 A

2300 revolutions in 5 hours

Number of revolutions in one hour = 2300/5 = 460

Now energy (E) consumed by the resistive load in one hour is

E = 230 x 4 x 1 x 10^-3 = 0.920 kWh

Hence meter constant of domestic energy meter is

Meter constant = (460) / (0.92)

Meter Constant = 500

• Temperature can influence the magnetic characteristics of materials.
• The rise in the temperature of solid increases the magnitude of the thermal vibrations of atoms.
• The atomic magnetic moments are free to rotate; hence, with rising temperature, the increased thermal motion of the atoms tends to randomize the directions of any moments that may be aligned.

Concept:

KP = position error constant =

Kv = velocity error constant =

Ka = acceleration error constant =

Steady state error for different inputs is given by

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and steady-state error for parabolic-input.

Calculation:

Velocity error coefficient,

Concept:

The maximum efficiency in any machine occurs, when the Copper losses (W_cu) is equal to the iron losses/constant losses (W_cu).

Copper losses (W_cu) = iron losses (W_i) at the maximum efficiency

Then, Total losses = 2 × iron losses (W_i)

Copper losses are proportional to the square load current (i.e. Square of X% full load)

∴ x² (full load Cu losses) = iron losses

Condition for the maximum efficiency at x of full load.

x of full load

Calculation:

Given that the transformer has maximum efficiency at ¾ t of full load

Concept:

In master-slave flip flop, the minimum clock time is equal to propagation delay

Also

Where f_m = Maximum clock frequency in Hz,

T_c = Minimum clock time in seconds,

Calculation:

⇒ f_m = 1 M Hz

• A power MOSFET has three terminals called a drain, source and gate. It is a voltage-controlled device.
• As its operation depends upon the flow of majority carriers only, it is a unipolar device. Therefore, time delays caused by the removal or recombination of minority carriers are eliminated. Thus, it can work at high switching frequencies.
• These are so constructed as to avoid punch through
• The channel length is relatively small and the channel width is relatively large.
• These can be put in parallel to handle large currents

Punch through: It is observed when the depletion region on the source side of the body-drift p-n junction reaches the source region at drain voltages below the rated avalanche voltage of the device. This provides a current path between source and drain and causes a soft breakdown characteristic as shown.

Power electronic circuits can be classified as follows.

1. Diode rectifiers:

• A diode rectifier circuit converts ac input voltage into a fixed dc voltage.
• The input voltage may be single phase or three phase.
• They find use in electric traction, battery charging, electroplating, electrochemical processing, power supplies, welding and UPS systems.

2. AC to DC converters (Phase controlled rectifiers):

• These convert ac voltage to variable dc output voltage.
• They may be fed from single phase or three phase.
• These are used in dc drives, metallurgical and chemical industries, excitation systems for synchronous machines.

3. DC to DC converters (DC Choppers):

• A dc chopper converts dc input voltage to a controllable dc output voltage.
• For lower power circuits, thyristors are replaced by power transistors.
• Choppers find wide applications in dc drives, subway cars, trolley trucks, battery driven vehicles, etc.

4. DC to AC converters (Inverters):

• An inverter converts fixed dc voltage to a variable ac voltage. The output may be a variable voltage and variable frequency.
• In inverter circuits, we would like the inverter output to be sinusoidal with magnitude and frequency controllable. In order to produce a sinusoidal output voltage waveform at a desired frequency, a sinusoidal control signal at the desired frequency is compared with a triangular waveform.
• These find wide use in induction motor and synchronous motor drives, induction heating, UPS, HVDC transmission etc.

5. AC to AC converters: These convert fixed ac input voltage into variable ac output voltage. These are two types as given below.

i. AC voltage controllers:

• These converter circuits convert fixed ac voltage directly to a variable ac voltage at the same frequency.
• These are widely used for lighting control, speed control of fans, pumps, etc.

ii. Cycloconverters:

• These circuits convert input power at one frequency to output power at a different frequency through a one stage conversion.
• These are primarily used for slow speed large ac drives like rotary kiln etc.

6. Static switches:

• The power semiconductor devices can operate as static switches or contactors.
• Depending upon the input supply, the static switches are called ac static switches or dc static switches.

• Errors due to change in impedance of the coil and change in the magnitude of the eddy current have only occurred in ac measurements.
• Hysteresis errors and errors due to stray magnetic field fields have occurred in both AC & DC measurements.
• In ac measurements, the change in frequency causes an error due to the change in the magnitude of eddy currents set up in the metal portion of the instruments.
• The reactance of the operating coil changes by the change in frequency.
• At high frequencies, the meter will give a reading lower than the actual value and at low frequencies, meter give a reading higher than the actual.
• This type of error can be eliminated only if the impedance of the coil is independent of the frequency. This is achieved by having a capacitance in shunt with the swamping resistance.

• Friction and windage losses do not depend upon the load condition of the machine; They are the function of the speed only, thus for the most synchronous machine they are constant
• The core losses (hysteresis and eddy current losses) depends upon the speed and magnitude of the rotating magnetic field
• Copper losses (both stator and field) depends on the load

Important Point:

The flow of power through a synchronous motor, from stator to rotor and then to shaft output, is shown in the figure.

Ploss = Pscl + Pcore + Pfcl + Pf,w + Pstray

Where:

Pscl = stator copper losses

Pfcl = field copper losses

Pcore­ = core losses

Pf,w = friction and windage losses

Pstary = stray losses

Concept:

In AC bridges, there are two conditions to get bridge balance condition.

1. The magnitude of impedances satisfy the relationship, Z1Z4 = Z2Z3

2. The phase angles of impedances satisfy the relationship, ∠θ1 + ∠θ4 = ∠θ2 + ∠θ3

These two are independent conditions for balance and both of them must be satisfied for the bridge to be balanced.

Calculation:

From the given bridge, the condition to get balance is

Z₁ R₃ = R₂ Z₄

Inverted V – curve:

The excitation and power factor of synchronous motor can be summed up in the following graph. This is called an inverted V curve of the synchronous motor.

From the inverted V – curve,

• The synchronous motor operates at a lagging power factor when the excitation is under excited.
• The motor operates at a unity power factor at normal excitation.
• The motor operates at a leading power factor when the excitation is over-excited.

Important Points:

• Under excited alternator works at the leading power factor
• The normal excited alternator works at the unity power factor
• The overexcited alternator works at lagging power factor

Given the network is:

In the given network,

R₂, R₄, and R₅ are directional overcurrent relays. R₁, R₃ and R₆ are non-directional overcurrent relays.

For the fault on line 2 i.e. L₂, R3 and R4 must be operated. If R₄ is not operated then R₁ and R₆ will operate.

Therefore, back up for R₄ are R1 and R6

The relay R₃ is directly connected to relay R₄, so that relay R₃ can’t provide remote backup.

Output = S + QR̅

The sentence "They will have completed the work by the time we get there" in the Active Voice can be transformed into the Passive Voice as:

A. The work will have been completed by the time we get there.

This option accurately represents the original sentence in the Passive Voice, maintaining its intended meaning.

In the given context we are talking about the way in which he convinced his parents for the school trip and since he did it very quickly, it is evident that it was a very easy thing to do by his standard.

A Stalking horse: Pretence 
A slap on the wrist: A very mild punishment 
A piece of cake: Something that is very easy to achieve 
A toss-up: An uncertain result that can go either way 
Among the given options, it is clear that option C is correct in the given context as it expresses the actual meaning of the given context.
This makes option C the correct choice among the given options.

The correct answer is closed-circuit television.

• The full form of CCTV is a closed-circuit television.

Key Points

• CCTV is an acronym for closed-circuit television.
• The inventor of the CCTV camera was Walter Bruch.
• It was first used in 1942 in Germany.
• It is used for video surveillance.
• It uses video cameras to transmit a signal to a specific place, on a limited set of monitors. 

The correct option is (4) By default, it has 3 placeholders.

Key Points

• Choose Home > New Slide to add a new slide to your presentation. The Title and Content slide layout are now added. This slide arrangement, which is the default, has placeholders for a title, bulleted text, pictures, and illustrations.
• Microsoft creates a Title slide by default when you create a new presentation because it expects the first slide in your presentation will be for the title. Two text placeholders are included on a title slide.
• The font family and font size of the regular input text are automatically carried over to the placeholder text, but there may be times when you want to modify the color of the placeholder text.

Additional Information

• Create a presentation storyboard or outline using the outline view. It merely shows the text on your slides; no graphics or other visual elements are displayed.
• Only a static string of text can be used as placeholder help information, which may not always be enough to convey the intended message.

The correct answer is .

Key Points

• In MS Word, Shift- Ctrl - K is used for small-cap in MS Word.
• Other few common keyboard shortcuts used in MS Word are given below:
  • In MS Word, Shift- Ctrl - > is used to increase the font size in the document.
  • In MS Word, Shift- Ctrl - D is used to double-underline the selected content.
  • In MS Word, Shift- Ctrl-A is used to open the list bullet.

Additional Information

• The leading creators of MS Word software are Charles Simonyi and Richard Brodie, launched in 1983 under Microsoft corporation.
• Microsoft word is recognised as a word processor software which means that a user can rewrite, erase or modify a document as per the requirement.
• The documents created in MS Word are very high in quality. MS Word creates reports, stories, letters, resumes etc. 
• It allows users to edit or modify the new or existing document with colour or any other formats like bold, italic and underline. 
• In MS Word, the users can write the stories in paragraph format.
• The file extension used in Ms Word is .doc or .docx.

The correct answer is Microsoft Outlook.

Key Points:

• Microsoft SQL Server is a relational database management system developed by Microsoft. 
• MySQL Database Service is a fully managed database service to deploy cloud-native applications. 
• Microsoft Outlook is the email service provided by Microsoft.
  • Hence this is the answer.
• Microsoft Access is the only application software you need to create apps, utilize databases, and store data in SQL Server and Microsoft Azure SQL.

The correct answer is OMR.

Important Points

• Optical mark recognition
  • It is the input device used to scan a marked answer book of a multiple choice question paper.
  • An optical answer sheet or bubble sheet is a special type of form used in multiple-choice question examinations. 
  • Optical mark recognition is used to detect answers.

Additional Information

• OCR
  • ​OCR (Optical Character Recognition) recognizes all the characters from the paper document.
  • It collects and stores them into an editable document.
• ​MICR
  • ​The term magnetic ink character recognition (MICR) refers to the line of numbers that appears at the bottom of a check.
  • The MICR line is a group of three numbers, which are the check number, account number, and bank routing number.
• Card reader
  • A card reader is a data input device that reads data from a card-shaped storage medium.

The correct answer is Thumbnail.

Key Points

• A thumbnail is referred to as a graphical object which represents the miniature version of a page in MS Word 2016.
• The main job of the thumbnail is to identify a file by its contents.
• The user can click the thumbnail to open the specific file.

Additional Information

• The leading creators of MS Word software are Charles Simonyi and Richard Brodie, launched in 1983 under Microsoft corporation.
• Microsoft word is recognised as a word processor software which means that a user can rewrite, erase or modify a document as per the requirement.
• The documents created in MS Word are very high in quality. MS Word creates reports, stories, letters, resumes etc. 
• It allows users to edit or modify the new or existing document with colour or any other formats like bold, italic and underline. 
• The file extension used in Ms Word is .doc or .docx.

Sure, let's find the correct answer for the given problem.

144 : 10 :: 169 : ?

- 144 is a perfect square (12²) and its square root is 12.
- 12 is then reduced to 10 by subtracting 2 (12 - 2 = 10).
- Similarly, 169 is a perfect square (13²) and its square root is 13.
- Following the same pattern, subtract 2 from 13 (13 - 2 = 11).

Thus, the correct answer is:

B: 11

- 169 : 11

As 24:60 = (2/5)

Similarly, (120/300) = (2/5)

According to the information given in the question, we can draw the following diagram.

Thus we can see if a man has to travel to point E from point A  (through these points by the shortest distance), he has to follow a path point A → B → F → E.

Thus travelling from point A, he will pass the first point B.

Hence, the correct answer is "Point B".

Checking Statement I:

Here, as Kirti correctly remembers that her birthday is in the second half of the year, so her birthday is in July or August or September or October or November or December.

Hence data in statement I is not sufficient to answer the question.

Checking Statement II:

Here, as Kirti’s father correctly remembers that her birthday is after 31st October but before 1st December, so we can say that Kirti’s birthday is in November.

Hence data in statement II is sufficient to answer the question.

Hence, Option B is correct.