Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  RRB JE Electrical (CBT II) Mock Test- 9 - Electrical Engineering (EE) MCQ

RRB JE Electrical (CBT II) Mock Test- 9 - Electrical Engineering (EE) MCQ


Test Description

30 Questions MCQ Test - RRB JE Electrical (CBT II) Mock Test- 9

RRB JE Electrical (CBT II) Mock Test- 9 for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The RRB JE Electrical (CBT II) Mock Test- 9 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 9 MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE Electrical (CBT II) Mock Test- 9 below.
Solutions of RRB JE Electrical (CBT II) Mock Test- 9 questions in English are available as part of our course for Electrical Engineering (EE) & RRB JE Electrical (CBT II) Mock Test- 9 solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt RRB JE Electrical (CBT II) Mock Test- 9 | 150 questions in 120 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
RRB JE Electrical (CBT II) Mock Test- 9 - Question 1

Radiation burn mainly due to _________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 1

Depending on the photon energy, gamma radiation can cause very deep burns, with 60Co internal burns, are common. Beta burns tend to be shallow as beta particles are not able to penetrate deep into the person; these burns can be similar to sunburn. Alpha rays don't penetrate very deeply into the skin.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 2

How many coulombs of charge flow through a circuit carrying a current of 15 A in 2 minutes?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 2

15 A current in 2 minute or 120 seconds implies 15 x 120 = 1800 coloumbs

RRB JE Electrical (CBT II) Mock Test- 9 - Question 3

The maximum power in the shown load is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 3

Then by using the following Ohm’s Law equations:

RRB JE Electrical (CBT II) Mock Test- 9 - Question 4

Quantities that measure the dispersion in the form of ratio, percentage or coefficient are called______.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 4

Quantities that measure the dispersion in the form of ratio, percentage or coefficient are called relative measures of dispersion. These quantities have no unit of measurement and dimension and are used to compare the dispersion in two or more data sets measured in different units.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 5

Match List-I (types of DC machines) with List-II (applications) and select the correct answer using the given options.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 5

Applications of shunt generator:

Used for general lighting

Used to charge battery because they can be made to give constant output voltage

Used for giving the excitation to the alternators

Used for small power supply (such as a portable generator)

Applications of series generator:

Used for supplying field excitation current in DC locomotives for regenerative breaking

Used as boosters to compensate the voltage drop in the feeder in various types of distribution systems such as railway service

Used in series arc lightening

Applications of compound generator:

Cumulative compound generators (over compounded) are generally used for lighting, power supply purpose and for heavy power services because of their constant voltage property; These are also used for driving a motor

The flat compounded generators are generally used for small distance operation, such as power supply for hotels, offices, homes and lodges

The differential compound wound generators are used for arc welding where huge voltage drop and constant current is required because of their large demagnetization armature reaction

RRB JE Electrical (CBT II) Mock Test- 9 - Question 6
The internal impedance of a source is 3 + j4 Ω, it is desired to supply maximum power to a resistive load. The load resistance should be
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 6

According to maximum power transfer theorem in the AC circuit, to draw maximum power the load impedance must be complex conjugate of source circuit internal impedance.

Here purely resistive load is chosen, hence for maximum average power transfer, the load resistance is equal to the magnitude of source internal impedance.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 7

Find the value of the resistor R1 from the given circuit.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 7

Given that, voltage across resistor R1 (VR1) = 6 V
Current flows through resistor R1 (IR1) = 4 A
Resistance (R1) = 6/4 = 1.5 Ω

RRB JE Electrical (CBT II) Mock Test- 9 - Question 8

In Fleming’s right hand rule, the middle finger indicates the direction of:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 8

Fleming's Right-hand Rule shows the direction of induced current when a conductor attached to a circuit moves in a magnetic field. It can be used to determine the direction of current in a generator's windings.

As shown on the above figure, if the thumb, the forefinger, and the middle finger of the right hand are bent at right angles to one another with the thumb pointed in the direction of motion of a conductor relative to a magnetic field and the forefinger in the direction of the field, then the middle finger will point in the direction of the induced electromotive force.

Lenz's law: When a voltage is generated by a change in magnetic flux according to Faraday's law, the polarity of the induced voltage is such that it produces a current whose magnetic field opposes the change which produces it.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 9

Quality factor is defined as

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 9

Quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

In a series RLC, Quality factor

In a parallel RLC,

It is defined as, resistance to reactance of reactive element.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 10

The force experienced by a current carrying conductor lying parallel to a magnetic field is ________

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 10

The force experienced by a current carrying conductor lying in a magnetic field is with an angle θ is
F = Bil sin θ
As the conductor is lying parallel to a magnetic field, θ = 0°
Hence the force = zero

RRB JE Electrical (CBT II) Mock Test- 9 - Question 11

In an Induction type meter, maximum torque is produced when the phase angle between two fluxes is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 11

An induction type meter consists of two electromagnets, called shunt magnet and series magnet of laminated construction.

A coil having a large number of turns of fine wire is wound on the middle limb of shunt magnet. This coil is known as voltage coil or pressure coil and it is connected across the supply mains. It has many turns and is arranged to be as highly inductive as possible. This causes the current and therefore the flux, to lag the supply voltage by nearly 90°.
An adjustable copper shading rings are provided on the central limb of the shunt magnet to make the phase angle displacement between the magnetic field set up by shunt magnet and the supply voltage is approximately 90°.
The series electromagnet is energized by a coil which is known as current coil and it is connected in series with the load so that it carries the load current. The flux produced by this magnet proportional to and in
phase with the load current.
The moving system essentially consists of a light rotating aluminium disk mounted on a vertical spindle or shaft. The fluxes produced by shunt and series magnet induce eddy currents in the aluminium disc. The interaction between these two magnetic fields and eddy currents set up a driving torque in the disc.
As in energy meter instrument, we have two fluxes and two eddy currents and therefore two torques are produced by
(i) First flux interacting with eddy current produced by second flux and
(ii) Second flux interacting with eddy currents generated by the first flux
Maximum torque is produced when the phase angle between these two fluxes is 90°

RRB JE Electrical (CBT II) Mock Test- 9 - Question 12
The energy meter installed at a residence charges the consumer for use of:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 12

In an energy meter, energy is given by

E = VI cos ϕ t

Here VI cos ϕ represents true power or real power.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 13
For large networks generally
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 13

For large networks generally, the node analysis is preferred because it involves lesser number of equations.

Kirchhoff's Current Law (KCL) or Node Analysis

Kirchhoff current law states that the algebraic sum of all currents entering a node of a circuit is always zero.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 14

The supplier shall provide and maintain for the consumer’s use a suitable earthed terminal in as accessible position at or near the point of commencement of supply. Which IE rule states that?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 14

IE 33: The supplier shall provide and maintain on the consumer’s premises for the consumer’s use a suitable earthed terminal in an accessible position at or near the point of commencement of supply

IE 67: Every earthing system belonging to either the supplier or the consumer shall be tested for its resistance to earth on a dry day during dry season not less than once a year.

IE 88: It explains about guarding of underground cables

RRB JE Electrical (CBT II) Mock Test- 9 - Question 15

In which of the following can induction heating be employed?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 15
  • Induction heating be employed in conducting magnetic material
  • Induction produces an emf (electromagnetic field) in a coil to transfer energy to a workpiece to be heated
  • When the electrical current passes along a wire, a magnetic field is produced around that wire
  • Induction heating is a method of heating conductive material by subjecting it to an alternating electromagnetic field, usually at frequencies between 100 and 500 kHz
RRB JE Electrical (CBT II) Mock Test- 9 - Question 16

In a 3-phase induction motor a balanced three phase supply is with electrical angle separation

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 16

The winding of 3-phase induction motor is connected in star or delta and these windings are displaced from each other by 120°.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 17

Which among these fuses is very fast in operation?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 17

Semiconductor fuse is very fast in operation compared to other fuses. It is the only protection for IGBT transistor in case of high fault current.

Second to semiconductor fuse is HRC (High Rupturing Capacity) fuse, where fast protection is required.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 18

If M is mutual inductance between fixed coil and moving coil and θ is deflection, then:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 18

Let, I1 = current in the fixed coils, I2 current in the moving coil
So deflecting torque Td = I1I2 . dM/dθ This shows that the deflecting torque depends in general on the product of current I1 and I2 and the rate of change of mutual inductance.
This deflecting torque deflects the moving coil to such a position where the controlling torque of the spring is equal to the deflecting torque. Suppose θ be the final steady deflection.
Therefore, controlling torque TC = kθ, where k = spring constant (N-m/rad)
At Final steady position, Td = TC
I1 2  dM/dθ = k θ
Or, the deflection θ = I1 2/k dM/dθ

RRB JE Electrical (CBT II) Mock Test- 9 - Question 19
A circuit breaker under normal conditions should be inspected once in:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 19
  • A circuit breaker is an automatically operated electrical switch designed to protect an electrical circuit from damage caused by excess current from an overload or short circuit
  • Its basic function is to interrupt current flow after a fault is detected
  • A circuit breaker under normal conditions should be inspected once in 3 or 6 months
RRB JE Electrical (CBT II) Mock Test- 9 - Question 20
The inter connectivity of computers to a common server within a small area is known as _________.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 20
  • A local area network (LAN) is a computer network that interconnects computers within a limited area such as a residence, school, laboratory, university campus or office building
  • In a server-based LAN, devices may connect directly to the server or indirectly via a router or switch
  • A wide-area network (WAN) is a computer network that extends over a large geographical distance/place
  • The World Wide Web (WWW), is an information space where documents and other web resources are identified by Uniform Resource Locators (URL)
RRB JE Electrical (CBT II) Mock Test- 9 - Question 21

At half power points of a resistance curve, the current is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 21

At half power points of a resistance curve, the current is 0.707 times the maximum current which is shown in figure.

Important point regarding half-power points are:

  • The current is I max/√2
  • Impedance is √2 R or √2 Zmin
  • P1 = P2  = P max/2
  • The circuit phase angle Ø 45 º or π/4 radian 
  • Tangent of the circuit phase angle at the off-resonance frequencies f1 and f2, Q = Tan θ = Tan 45° = 1
RRB JE Electrical (CBT II) Mock Test- 9 - Question 22

A Norton equivalent of a circuit consists of a 4 A current source in parallel with a 2 resistor. The Thevenin's equivalent of this circuit is a ________ V source in series with a 2 resistor.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 22

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.

Norton’s Theorem: Any two-terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a current source and a parallel resistor.

Now we can say that Norton’s theorem is the converse of Thevenin’s theorem.
Application:
A Norton equivalent of a circuit consists of a 4 A current source in parallel with a 2 resistor. It can be represented as follows:

By using source transformation, the above circuit becomes:

Now, it is the Thevenin's equivalent of the given circuit with 8 V source in series with a 2 Ω resistor.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 23
What was the theme of the World Brain Day 2019?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 23
  • World Brain Day is observed on July 22 every year.
  • The day aims at promoting interest of neurology. Previously, World Brain Day has covered topics like dementia, epilepsy and stroke, in cooperation with other societies.
  • World Brain Day 2019 theme is Migraine: The Painful Truth.
  • This year, the focus is on migraine, which is known to be one of world's most common brain diseases.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 24

If the rotor circuit resistance of a three-phase induction motor is increased

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 24

The torque of an induction motor is given by

At starting slip, s = 1,

At starting, R 2 << X 2


Hence with the increase in the rotor resistance, starting torque also increases
The max torque of an induction motor is given by
T max = 1/2 x 2
Hence with the increase in the rotor resistance, the maximum torque developed remains unchanged.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 25

A power station has a maximum demand of 2 MW. The annual load factor is 40% and the plant capacity factor is 15%. What is the reserve capacity of the plant?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 25

Plant capacity factor
= Peak load/Plant capacity x load factor
0.15 = 2//Plant capacity x 0.4 
∴ Plant capacity = 5.33 MW
∴ Reserve capacity
= Plant capacity – Peak load
= 5.333 – 2 = 3333 kW

RRB JE Electrical (CBT II) Mock Test- 9 - Question 26
Which one of the following four metals would be displaced from the solution of its salts by other three metals?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 26

Concept:

A metal of high reactivity displaces a metal of low reactivity from its salt solution.

The reactivity of metal is decided by its reactivity series.

Reactivity series:

K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Hg > Ag > Au.

Application:

Since Mg, Zn, Cu are more reactive than Silver Ag, they can replace silver from its salt solution.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 27
Who is the author of the book named "A Suitable Boy"?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 27
  • Vikram Seth is the author of the book named ‘A Suitable Boy’.
  • The novel revolves around the story of four families over a period of 18 months.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 28
Which of the following materials when used as the viewing surface of a CRO gives a bluish glow?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 28

The actual conversion of electrical to light energy takes place on the display screen when electrons strike a material known as a phosphor.

A phosphor is a chemical that glows when exposed to electrical energy. A commonly used phosphor is the compound zinc sulfide.

When pure zinc sulfide is struck by an electron beam, it gives off a greenish glow. The exact colour given off by a phosphor also depends on the presence of small amounts of impurities. For example, zinc sulfide with silver metal as an impurity gives off a bluish glow and with copper metal as an impurity, a greenish glow.

The phosphor known as yttrium oxide gives off a red glow when struck by electrons, and yttrium silicate gives off a purplish-blue glow.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 29

How many types of writs can be issued in Indian Constitution?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 29

As per Article 32 the Supreme Court can issue the writs of following nature

RRB JE Electrical (CBT II) Mock Test- 9 - Question 30

If a forward current of PN junction diode is 1 mA then the value of dynamic resistance is ________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 30

Concept:

  • The dynamic resistance can be defined from the I-V characteristic of a diode in forward biased
  • It is defined as the ratio of small change to voltage to a small change in current, 
  • It is the inverse of the slope of I-V characteristics curve

The dynamic resistance is given by the inverse of the slope of i-v characteristics

Calculation:
Given that, current (I) = 1 mA
We know that, voltage (VT) = 26 mV
Dynamic resistance Rd =V/1 = 26 Ω

View more questions
Information about RRB JE Electrical (CBT II) Mock Test- 9 Page
In this test you can find the Exam questions for RRB JE Electrical (CBT II) Mock Test- 9 solved & explained in the simplest way possible. Besides giving Questions and answers for RRB JE Electrical (CBT II) Mock Test- 9, EduRev gives you an ample number of Online tests for practice
Download as PDF