Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  RRB JE Mock Test Series Electrical Engineering 2025  >  RRB JE Electrical (CBT II) Mock Test- 10 - Electrical Engineering (EE) MCQ

RRB JE Electrical (CBT II) Mock Test- 10 - Electrical Engineering (EE) MCQ


Test Description

30 Questions MCQ Test RRB JE Mock Test Series Electrical Engineering 2025 - RRB JE Electrical (CBT II) Mock Test- 10

RRB JE Electrical (CBT II) Mock Test- 10 for Electrical Engineering (EE) 2024 is part of RRB JE Mock Test Series Electrical Engineering 2025 preparation. The RRB JE Electrical (CBT II) Mock Test- 10 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 10 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE Electrical (CBT II) Mock Test- 10 below.
Solutions of RRB JE Electrical (CBT II) Mock Test- 10 questions in English are available as part of our RRB JE Mock Test Series Electrical Engineering 2025 for Electrical Engineering (EE) & RRB JE Electrical (CBT II) Mock Test- 10 solutions in Hindi for RRB JE Mock Test Series Electrical Engineering 2025 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- 10 | 150 questions in 120 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study RRB JE Mock Test Series Electrical Engineering 2025 for Electrical Engineering (EE) Exam | Download free PDF with solutions
RRB JE Electrical (CBT II) Mock Test- 10 - Question 1

The iron loss in a 100 KVA transformer is 1 kW. and at full load copper loss is 2 kW. The maximum efficiency occurs at a load of-

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 1

RRB JE Electrical (CBT II) Mock Test- 10 - Question 2

The induced e.m.f. will be maximum when a conductor cuts the magnetic field at an angle of

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 2

The induced emf when a conductor cuts the magnetic field is

e = Bil sinθ

B is magnetic flux density

i is the current flowing through the conductor

l is the length of the conductor

θ is the angle

Induced emf is maximum at θ equals to 90 degrees.

1 Crore+ students have signed up on EduRev. Have you? Download the App
RRB JE Electrical (CBT II) Mock Test- 10 - Question 3

The wattage rating for a ceiling fan motor will be in the range of:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 3

The wattage rating for a ceiling fan motor is in the range of 50 to 150 W. Fans range in size from 36 inches to 56 inches use 55 to 100 watts, a typical 48 inch ceiling fan will use 75 watts.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 4

Single phase induction motor is also known as:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 4

A fractional-horsepower motor (FHP) is an electric motor with a rated output power of 1 HP (746.9 or 746 Watts) or less.

Single phase induction motors have very less rating hence these are also known as fractional horsepower motors.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 5

In the spot welding, the composition and thickness of the base metal determines:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 5

Spot welding means the joining of two metal sheets and fusing them together between copper electrode tips at suitably spaced intervals by means of heavy electric current passed through the electrodes.

In this type of welding, the composition and thickness of the base metal determines:

Holding time

The amount of weld current

The amount of squeeze pressure

RRB JE Electrical (CBT II) Mock Test- 10 - Question 6

After the starting winding is disconnected from the circuit the motor continues to run only on the-

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 6

A single phase motor cannot create a rotating magnetic field. It will just standstill and oscillate. To see this disconnect the start winding and you will see the motor will not start up. If you twist motor shaft with fingers it will run in whichever direction you twist. To self start a single phase motor one must create a rotating magnetic field on the stator. Two windings are used and the difference in impedance creates a phase delay in the magnetic fields produced by each winding allowing the motor to start.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 7

When the number of poles is equal to zero then how many branches of root locus tends towards infinity?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 7

number of branches of root locus tends towards infinity is equal to
= P - Z
P = number of poles of open loop system
Z = number of zero of open loop system.
in given problem P = Z then branches toward infinity is zero.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 8

Leakage current flows through the thyrister in__________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 8

In forward blocking mode junction J 2 and in reverse blocking mode junction J 1 and J 3 acts as capacitor hence leakage current present in both direction.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 9
As a result of greenhouse effect, the average temperature of the earth’s atmosphere is gradually increasing. This is called _____.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 9
As a result of the greenhouse effect, the average temperature of the earth’s atmosphere is gradually increasing. This is called Global warming. The exchange of incoming and outgoing radiation that warms the surface of Earth is known as the greenhouse effect. The greenhouse gases include carbon dioxide, methane, water vapors, nitrous oxide, and other gases.
RRB JE Electrical (CBT II) Mock Test- 10 - Question 10
Transistors are:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 10
  • A transistor is a device that regulates current or voltage flow and acts as a switch or gate for electronic signals
  • It is low voltage and low current device because semiconductors are sensitive to high voltages and current and can be easily destroyed
RRB JE Electrical (CBT II) Mock Test- 10 - Question 11

Consider the following statements.

Statement A: Electricity is distributed from a single point in a radial distribution system

Statement B: There are less fluctuation in supply at the consumer’s terminal

Which of the following is correct?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 11

When the distributor is connected to a substation on one end only with the help of feeder, then the system is called a radial distribution system.

The feeders, distributors and service mains are radiating away from the substation hence name given as radial system.

There are combinations of one distributor and one feeder, connecting that distributor to the substation.

In the above diagram, one end of distributors are connected to the substation with the help of feeders and the other end will be connected to the service mains.

In this distribution system, electricity is distributed from a single point in a radial distribution system. The fluctuation in the supply is less at the consumer’s terminal.

Both the given statements are true, but B is not the correct explanation of A.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 12
The disease which is not caused by water pollution is
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 12
Asthma is a condition in which a person's airways become inflamed, narrow and swell and produce extra mucus, which makes it difficult to breathe. It is caused due to air pollution.
RRB JE Electrical (CBT II) Mock Test- 10 - Question 13

If the shunt capacitance is reduced then the string efficiency of an insulator is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 13

The string efficiency is defined as the ratio of voltage across the string to the product of the number of strings and the voltage across the unit adjacent string.

Because of the shunt capacitance effect between insulators and tower, the voltage across each insulator is non-uniformly distributed.
If the shunt capacitance is reduced then the voltage across insulators will get uniformly distributed. Then string efficiency will increase.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 14

The following data is known about a resultant current wave which is made up of two components:
1. A direct current of 10 A
2. A sinusoidal alternating current of 50 Hz with a peak value of 10
A Calculate the r.m.s. value of the resultant wave.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 14

RRB JE Electrical (CBT II) Mock Test- 10 - Question 15
Magnitude of the electric shock on human body
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 15

Electric shock occurs upon contact of a (human) body part with any source of electricity that causes a sufficient magnitude of current to pass through the human body.

This magnitude of electric shock on human body is the line current.
RRB JE Electrical (CBT II) Mock Test- 10 - Question 16

A 16 mA current source has an internal resistance of 10 k ohm. How much current will flow in a 2.5 k ohm load connected across its terminals

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 16

The current through the source can be calculated using current division:
 IL = I (Rs/Rs + RL)
IL = 16 (10/10 +2.5) = 160/12.5 m A

RRB JE Electrical (CBT II) Mock Test- 10 - Question 17

In power systems, regulating transformer are used to control

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 17
  • The transformer which changes the magnitude and phase angle at a certain point in the power system is known as the regulating transformer
  • It is mainly used for controlling the magnitude of the bus voltage and for controlling the power flow, which is controlled by the phase angle of the transformer
  • Hence these are used to control the load flow
  • The regulating transformer is of two types; One is used for changing the magnitude of voltage which is called online tap changing transformer and the other is called phase shifting transformer
RRB JE Electrical (CBT II) Mock Test- 10 - Question 18

In figure, the value of R should be

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 18

The equivalent resistance in the circuit is,
 Req = 10 + (15║15║ R) 
= 10 + (7.5 ║R) 
= 10 + 7.5R/7.5 + R
Given that, current flowing though the circuit (I) = 3 A
We know that, V = IReq
⇒ 40 = 3 (10 + 7.5R/7.5 + R ) 
⇒ 10/3 = 7.5R/7.5 + R
⇒ 75 + 10R = 22.5R
⇒ 12.5R = 75 ⇒ R = 6 Ω

RRB JE Electrical (CBT II) Mock Test- 10 - Question 19
Arc heating occurs when the air between electrodes of opposite polarity becomes:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 19
  • When a high voltage is applied across an air gap, the air in the gap gets ionized under the influence of electrostatic forces and becomes conducting medium.
  • Current flows in the form of a continuous spark, called the arc.
  • It is to be noted that a very high voltage is required to establish an arc across an air gap but to maintain an arc small voltage may be sufficient.
RRB JE Electrical (CBT II) Mock Test- 10 - Question 20

Two wattmeters are used to measure the power in a three-phase balanced load. The wattmeter readings are 8.2 kW and 7.5 kW. The active power will be:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 20

Wattmeter reading in two wattmeter method are,
W1 = VLIL cos (30 - ϕ)
W2 = VLIL cos (30 + ϕ)
The active power will be = W1 + W2 = 8.2 + 7.5 = 15.7 W
Important:
tan Ø√ 3 ( W1 - W2)/( W+ W2)
Power factor = cos ϕ

RRB JE Electrical (CBT II) Mock Test- 10 - Question 21
A potential difference between two points of a wire carrying 4-ampere current is 0.5 volt. The resistance of the wire is
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 21

Potential difference = 0.5 V
I = 4 A
R = ?
R = V/I
V stands for potential difference
R = 0.5/4
R = 1/8
R = 0.125

RRB JE Electrical (CBT II) Mock Test- 10 - Question 22
Which of the following is the information retrieval service?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 22
  • The internet is a globally connected network system that transmits data via various types of media. Sometimes it is referred to as a ‘network of networks’
  • The World Wide Web, or simply web, is a way of accessing information over the medium of the internet
  • A web browser is a software application for accessing information on the World Wide Web
RRB JE Electrical (CBT II) Mock Test- 10 - Question 23

Which of the following is the most preferable management of waste?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 23

The order of waste management hierarchy, from most to least favoured is:
Prevention-Reuse-Recycle-Recovery-Disposal

Prevention: Preventing and reducing waste generation.
Reuse: Giving the products a second life before they become waste.
Recycle: Any recovery operation by which waste materials are reprocessed into products, materials or substances whether for the original or other purposes
Recovery: Some waste incineration based on a political non-scientific formula that upgrades the less inefficient incinerators.
Disposal: Different processes to dispose of the waste are landfilling, incineration, pyrolysis, gasification and other finalist solutions.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 24
If a DC series motor is operated on AC supply, it will
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 24

AC series motors are also known as the modified DC series motor as their construction is very similar to that of the DC series motor. If a DC series motor is operated on AC supply,

  • An AC supply will produce a unidirectional torque because the direction of both the currents (i.e. armature current and field current) reverses at the same time
  • Due to the presence of alternating current, eddy currents are induced in the yoke and field cores which results in excessive heating of the yoke and field cores
  • Due to the high inductance of the field and the armature circuit, the power factor would become very low
  • There is sparking at the brushes of the DC series motor
  • It has poor efficiency
RRB JE Electrical (CBT II) Mock Test- 10 - Question 25
The pH of acid rain should always be less than
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 25

When the quantity of acids in the raining water is more than the average, then such rain is called 'Acid rain'.

We are aware that normally rainwater has a pH of 5.6. When the pH of the rainwater drops below 5.6, it is called acid rain.

Acid rain refers to the ways in which acid from the atmosphere is deposited on the earth’s surface. Oxides of nitrogen and sulphur which are acidic in nature can be blown by the wind along with solid particles in the atmosphere and finally settle down either on the ground as dry deposition or in water, fog and snow as wet deposition.

The bad effects of acid rain

  • When acid rain falls and flows as groundwater to reach rivers, lakes etc, it affects plants and animal life in the aquatic ecosystem
  • Acid rain is harmful to agriculture, trees and plants as it dissolves and washes away nutrients needed for their growth
  • It causes respiratory ailments in human beings and animal
  • It may also cause corrosion in many buildings bridges, monuments, fencing etc
  • It causes irritation in the eyes and skin of human beings
  • This rain reduces the lustre of the metals too
  • Acid rain damages buildings and other structures made of stone or metal
  • The Taj Mahal in India has been affected by acid rain
RRB JE Electrical (CBT II) Mock Test- 10 - Question 26

The maximum number of electrons that can be accommodated in M shell is:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 26
  • Electron Shell (also called a Principle energy level) is the region where electrons revolve around the nucleus of an atom.
  • The general formula of filling electron in the shell is 2(n2); where n is the number of the shell.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 27

In an open device, current through it is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 27

In an open device, the current through it is zero and the voltage is unknown
Important:
Short Circuit: It is an electrical circuit that allows a current to travel along an unintended path with zero or very low electrical impedance. This results in an excessive amount of current flowing into the circuit.
Ideally, it has zero resistance, hence high current will flow and no voltage drop across the terminals.
Open Circuit: It is a circuit where no current flows. Any circuit which does not have a return path to flow current is an open circuit.
It has infinite resistance, hence no current will flow.
Short circuit and open circuit can be represented as shown in the figure.

RRB JE Electrical (CBT II) Mock Test- 10 - Question 28
The e-mail program is also known as
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 28
The e-mail program is also known as an E-mail client. The email client makes you read, organize and reply to messages as well as send new emails. It also handles attachments, which lets you exchange arbitrary computer files via email.
RRB JE Electrical (CBT II) Mock Test- 10 - Question 29
Calculate the time (in seconds) taken by the capacitor of a series RC circuit having a capacitance of 0.01 mF and resistance of 300 Ohms, to get fully charged.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 29

The time constant of series RC circuit is, τ = RC = 300 × 0.01 × 10-3 = 0.003 sec

Time taken by the capacitor to get fully charged = 5τ = 5 × 0.003 = 0.015 sec
RRB JE Electrical (CBT II) Mock Test- 10 - Question 30

A Satellite has a mass of 100 kg and is initially travelling at 2.64 × 103 m/s . To correct its orbit, a thruster is fired for 2.25 seconds which changes the speed of the satellite to 4.89 × 103 m/s.

The force generated by the thruster is:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 10 - Question 30

The force exerted is given by:


View more questions
163 tests
Information about RRB JE Electrical (CBT II) Mock Test- 10 Page
In this test you can find the Exam questions for RRB JE Electrical (CBT II) Mock Test- 10 solved & explained in the simplest way possible. Besides giving Questions and answers for RRB JE Electrical (CBT II) Mock Test- 10, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

Download as PDF

Top Courses for Electrical Engineering (EE)