BEL Trainee Engineer Electrical Mock Test - 4 - Electrical Engineering (EE) MCQ

Number of address lines, n = 12

No. of memory locations = m = 2ⁿ = 2¹²

No. of data line = 4 = No. of bits/location

Total No. of chips given = 8

Total capacity = No. of chips given × No. of memory location (N) × No. of data line

= 8 × 2¹² × 4 bits

= 2³ × 2¹² × 2² bits

= 2¹² × 2² bytes

∴ 1 byte = 2³ = 8 bits

= 2¹⁰ × 2⁴ bytes

= 16 k bytes

∴ 2¹⁰ = 1 k bytes

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To solve Problems with independent Sources:

To find V_oc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (V_th).

To find R_th:

To calculate Thevenin’s Resistance we should replace all independent current sources with an Open circuit and Independent voltage sources with a Short circuit.

Application:

Let 2 Ω be the Load of the given circuit and open circuit to get Thevenin voltage.

A circuit diagram can be drawn as

By applying KVL,

9 - 4I - 4I - 6I = 0

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given differential equation is,

⇒ s2 Y(s) + 3s Y(s) + 2 Y(s) = U(s) + s U(s)

There is an 8-bit flag register out of which only 5 bits are used

RRC A (Rotate Right Accumulator with Carry) instruction:

• The RRC A instruction rotates the eight bits in the accumulator and the one bit in the carry flag right one-bit position.
• Bit D₀ of the accumulator is rotated into the carry flag while the original value of the carry flag is rotated into bit D₇ of the accumulator.
• Bit D₇ of the accumulator is rotated into bit D₆, bit D₆ into bit D₅, and so on.
• No other flags are affected by this instruction.
• Rotation operation with RRC A involves 9 -bits, eight bits of the accumulator, and one carry bit(CY).​

RR A (Rotate Right Accumulator):

The RR A instruction rotates the eight bits in the accumulator right one-bit position. Bit D₀ of the accumulator is rotated into bit D₇, bit D₇ into bit D₆, and so on. No flags are affected by this instruction.

RL A (Rotate Left Accumulator):

The RL A instruction rotates the eight bits in the accumulator left one-bit position. Bit D₇ of the accumulator is rotated into bit D₀, bit D₀ into bit D₁, bit D₁ into bit D₂, and so on. No flags are affected by this instruction.

Note:

SWAP A is an instruction used for swapping between upper and lower nibbles of the accumulator. It is not used for rotation.

Concept:

The back emf in a DC motor is given by

ϕ is the flux per pole

Z is the total number of conductors

N is the speed

P is the number of poles

A is the number of parallel paths

For lap wound, A = P

For wave wound, A= 2

Mechanical power developed (P) = E_b I_a

Calculation:

Pole P = 4, Z = 400, A = 4, l = 20 cm, D= 30 cm, N = 1500 rpm, I_a= 25 A and B = 0.4 T

As the motor is Lap Connected, therefore A = P i.e. a number of parallel paths is equal to the number of poles.

Flux(ϕ) = Flux density(B) × Area(A)

Area/pole = πDL/P

Area =

Therefore, Flux = 0.4 × 0.0471 = 0.01884 Wb

Latching Current: It is the minimum anode current required to maintain the thyristor in the ON state immediately after a thyristor has been turned on and the gate signal has been removed.

Holding Current: It is the minimum anode current to maintain the thyristor in the on-state.

Latching current is always greater than holding current.

Therefore, SCR can be turned off by reducing the anode current below holding current.

Concept:

Fault current for LG fault

Fault current for three phase fault =

where,

X₁ = Positive sequence reactance

X₂ = Negative sequence reactance

X₀ = Zero sequence reactance

X_n = Grounding reactance

Calculation:

Given that,

X₁ = 0.25 pu

X₂ = 0.15 pu

X₀ = 0.05 pu

Fault current for LG fault is

Fault current for three phase fault,

Given that both the fault currents are equal.

⇒ X_n = 0.1 pu

Base kV = 30 kV

Base MVA = 50 MVA

• A control system is a collection of components assembled to produce a desired response for a given input.
• Reference (command) input is a signal supplied to the control system which represents the desired value (or variation) of the controlled output.
• Disturbance input is an unwanted input that tends to adversely affect that tends to adversely affect the value of the output of a system.
• A controller is essentially the comparator which compares the given input with the reference input and generates the error signal.
• Control signal is the output of the controller that will be used to bring the output of the system as close to the desired value as possible.

Concept:

Illumination is defined as the luminous flux received by the surface per unit area.

It is usually denoted by the symbol ‘E’ and is measured in lux or lumen/m2 or meter candle or foot candle.

Illumination,

Solid angle

Illumination

Calculation:

Given that, distance (d) = 5 m

Illumination (E) = 6 lux

⇒ 6 = CP/52

⇒ CP = 150

∴ current through the operating coil = 3.125 – 2.75 = 0.375 A

Alternator:

An alternator is a generator that produces AC (Alternating Current) supply and it converts mechanical energy into electrical energy. It is also called an AC generator or synchronous generator. It runs at a constant speed of synchronous speed.

Alternators are generally designed to supply electric power because of the following reasons.

• Constant Voltage
• Constant frequency
• Capability to deliver active as well as reactive power

The reactive power supply capability of an alternator is determined by the short circuit ratio of the alternator.

Short Circuit Ratio (SCR):

• It affects the operating characteristics, physical size, and cost of the machine
• The large variation in the terminal voltage with a change in load takes place for the lower value of the short circuit ratio of a synchronous generator
• For the small value of the short circuit ratio (SCR), the synchronizing power is small
• A synchronous machine with the high value of SCR had a better voltage regulation and improved steady-state stability limit, but the short circuit fault current in the armature is high

Option 3 is the correct answer: Central Processing Unit (CPU) in a computer controls, coordinates and supervises the operations of the computer.

• Central Processing Unit (CPU) is also known as the brain of the computer.
• It is the key component that processes the data to give useful information.
• Commonly referred to as Processor it is made up of a number of micro transistors mounted on a chip.
• It consists of two basic components:
  • Arithmetic and Logical Unit (ALU): This unit of CPU performs all the arithmetic and logic operations.
  • Control unit: It directs the flow of data and information.
• Random Access Memory (RAM):
  • It is the volatile memory of the computer where information is temporarily stored.
  • The data is stored until the system is working and loses it as soon as power is switched off.
• Read-Only Memory (ROM):
  • It is a non-volatile memory in which the information is stored permanently.
  • The content does not get erased off even when the system is switched off.

Concept:

Signal flow graph

• It is a graphical representation of a set of linear algebraic equations between input and output.
• The set of linear algebraic equations represents the systems.
• The signal flow graphs are developed to avoid mathematical calculation.

Maon gain formula is used to find the ratio of any two nodes or transfer function.

T F =

Where P_k = k^th forward path gain

Δ = 1- ∑ individual loop gain + ∑ two non-touching loops gain - ∑ the gain product of three non-touching loops + ∑ gain of four non-touching loops

Shotcut: while writing Δ take the opposite sign for the odd number of non-touching loops snd the same sign for the even the number of non-touching loops.

Δ_K is obtained from Δ by removing the loops touching the K^th forward path.

Calculation:

For the given SFG two forward paths

Since all loops are touching the paths P_K1 and P_K2 so Δ_K1 = Δ_K2 = 1

We have Δ = 1- ∑ individual loops + ∑ non-touching loops gain

Loops are

As all the loops are touching each other we have

Δ = 1 – ( L1 + L2 + L3 + L4)

Δ = 1 – ( - 4 – 4s^-1 – 2s^-2 -2s^-1 )

Δ = 5 + 6s^-1 + 2s^-2

Freak of nature (Idiom): something or someone that is unusual, rare, or abnormal in some way; beyond or outside the natural world.
Funny feeling (Idiom): an intuition or premonition about something; a sense of foreknowledge about a situation, condition, or set of circumstances.
Talk shop (Idiom): discuss matters concerning one's work, especially at a social occasion when this is inappropriate.
Wear off (Idiom): lose effectiveness or intensity.
It is obvious that ‘freak of nature’ makes perfect sense in the given blank.
Hence, the correct answer is option B. 

The passive voice of imperative sentences which suggest order, suggestion or request can be made in two ways:

Active: Verb + object

Passive:

Let + object + be + past participle
You are requested/ ordered/ suggested + to + verb (Ist form) + object
Going by first way of passive voice, the passive voice of the given sentence would be: Let the door be shut.

obsolete : no longer in use

One word substitution is Linguist.

Linguist: a person skilled in foreign languages.
Grammarian: a person who studies and writes about grammar.
Polyglot: knowing or using several languages.
Bilingual: speaking two languages fluently.

The correct answer is Flash memory.

Key Points

• Flash memory is an electronic non-volatile memory used for storing and transferring data between PC and digital devices or can be used as a standalone storage device like a USB.
• It is a type of Electrically Erasable Programmable Read Only Memory or abbreviated as EEPROM and can be electrically erased and reprogrammed in large blocks. Hence option 4 is correct.
• A Flash memory is composed of NOR or NAND gates.

Additional Information

• DRAM or Dynamic Random Access Memory is a specific type of RAM that stores bits of data in transistors. DRAM is located close to the processor, so that the computer can quickly access it for all the processes.
• PROM or Programmable ROM is a memory that can be programmed once after it is created. Once it is programmed, the data written is permanent and cannot be deleted. An example of PROM is a computer BIOS.
• EPROM or Erasable Programmable ROM is a memory that can retain the data even if there is no power supply. This data can be erased and reprogrammed by using UV light. The process of programming an EPROM is called Burning and the box into which it is plugged to program it is EPROM Burner.

The correct answer is Second.

Key Points

• Second-generation computers (1955-1964) were made of transistors.
• The second generation of computers, which used transistors instead of vacuum tubes, were released in the late 1950s and early 1960s as interest in computer technology grew quickly.
• The second generation of computers, however, was entirely built using transistors rather than vacuum tubes.

Additional Information

The correct answer is ENIAC.

Key Points

• The first computer with no mechanical parts created by JW Mauchly and J Presper Eckert in 1943 was called the ENIAC.
• ENIAC stands for Electronic Numerical Integrator and Computer.
• It was the first general-purpose electronic digital computer, and it was used for a variety of purposes, including ballistics calculations, weather forecasting, and nuclear weapons research.
• The ENIAC was a huge machine, weighing 30 tons and taking up 1,800 square feet of space.
• It had 17,468 vacuum tubes, 7,200 crystal diodes, and 1,500 relays. It could perform 5,000 additions per second, which was a huge improvement over previous computers.
• The ENIAC was a major breakthrough in the development of computers, and it paved the way for the development of modern computers.

Additional Information

Here are some additional details about the ENIAC:

• It was built at the University of Pennsylvania's Moore School of Electrical Engineering.
• It was completed in 1945.
• It was decommissioned in 1955.
• It is now on display at the Smithsonian National Museum of American History in Washington, D.C.

The correct answer is a - ii, b - iii, c - iv, d - i.

Key Points

• Optical Mouse
  • It is a computer mouse that uses a light source, typically a light-emitting diode (LED), and a light detector, such as an array of photodiodes, to detect movement relative to a surface.
  • It uses special-purpose image processing chips, as the mouse takes over 1,000 images/ps below the surface level to detect movement through reflected light changes.
  • It uses camera technology and digital processing to compare and track the position of the mouse.
• Track Ball
  • ​It is an input device used to enter motion data into computers or other electronic devices.
  • It is designed with a moveable ball on the top, which can be rolled in any direction.
  • A moving ball is mounted on a stationary device that can be rotated mechanically (manually) using the fingers
• Digitizing(Graphic) Tablet
  • ​It is a computer input device that enables a user to hand-draw images, animations and graphics, with a special pen-like stylus.
  • Alternatively, it is a tool used to convert hand-drawn images into a format suitable for computer processing.
• Joystick
  • ​It is an input device that can be used for controlling the movement of the cursor or a pointer in a computer device.
  • It was first invented by C. B. Mirick at the U.S. Naval Research Laboratory, and patented in 1926.
  • It includes a base and a stick that can be move to the different direction.
  • It is used to indicate position in video game play.

From the final arrangement, we get, R doesn’t belong to that group as except for R, all the others are facing outside.
Hence, Option D is correct.
References:
Q sits second to the right of P.
Inferences:
From the above reference, we get two different cases:
In case 1, P sits at the corner or in the middle of the side of the table.

References:
R is not an immediate neighbor of P.
R faces inside.
R sits third to the left of W.
Inferences:
From the above references, we get different cases for R, in case 1(a) in which R faces P and in case 1(b) R faces Q.

References:
T sits third to the left of U.
V sits second to the left of U.
Inferences:
So, from this case 1(b) and case 2 will be eliminated as there is no possible place for V in these cases after placing U and T. Hence, we get our final arrangement in which S sits on the immediate left of P and V sits on the immediate right of R.
Final arrangement:

From Statement I: From this statement, we get the following inference:

All the vowels are coded as 3rd next letter as per English alphabetical series and the consonants are coded as 2nd previous letter as per English alphabetical series.

So, the code of ‘HEALTHY’ is ‘FHDJRFW’.

Clearly, Statement I alone is sufficient to answer the question.

From Statement II: From this statement, we get the following inference:

All the vowels are coded as 3rd next letter as per English alphabetical series and the consonants are coded as 2nd previous letter as per English alphabetical series.

So, the code of ‘HEALTHY’ is ‘FHDJRFW’.

Clearly, Statement II alone is sufficient to answer the question.

Therefore, the question can be answered using either statement I alone or statement II alone.

Hence, option C is the correct answer.

As Wick is a part of Candle similarly Wheel is a part of Bicycle.

P sits to the immediate left of S.

Hence, option A is correct.

From the final arrangement, we get S sits exactly opposite to U.
Hence, Option D is correct.
References:
Q sits second to the right of P.
Inferences:
From the above reference, we get two different cases:
In case 1, P sits at the corner or in the middle of the side of the table.

References:
R is not an immediate neighbor of P.
R faces inside.
R sits third to the left of W.
Inferences:
From the above references, we get different cases for R, in case 1(a) in which R faces P and in case 1(b) R faces Q.

References:
T sits third to the left of U.
V sits second to the left of U.
Inferences:
So, from this case 1(b) and case 2 will be eliminated as there is no possible place for V in these cases after placing U and T. Hence, we get our final arrangement in which S sits on the immediate left of P and V sits on the immediate right of R.
Final arrangement:

1st term = 1² = 1
2nd term = 2² = 4
3rd term = 3² = 9
4th term = 4² = 16
5th term = 5² = 25
Thus, Missing number = 6² = 36