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RRB JE Electrical (CBT II) Mock Test- 5 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series Electrical Engineering CYM_Marker_5 - RRB JE Electrical (CBT II) Mock Test- 5

RRB JE Electrical (CBT II) Mock Test- 5 for Electrical Engineering (EE) 2025 is part of RRB JE Mock Test Series Electrical Engineering CYM_Marker_5 preparation. The RRB JE Electrical (CBT II) Mock Test- 5 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 5 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- 5 below.
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RRB JE Electrical (CBT II) Mock Test- 5 - Question 1

Magnetic reluctance, or magnetic resistance can be expressed as

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 1

The total reluctance is equal to the ratio of the MMF in a passive magnetic circuit and the magnetic flux in this circuit. In an AC field, the reluctance is the ratio of the amplitude values for a sinusoidal MMF and magnetic flux.
R = F/Φ

RRB JE Electrical (CBT II) Mock Test- 5 - Question 2

Where was All India Muslim League formed?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 2

On December 30, 1906, Muslim league was formed under the leadership of Aga Khan, the Nawab of Dhaka.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 3

For slip ring induction motors:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 3

The torque equation of a three-phase induction motor is given by,

Torque is maximum when slip, s = R2/X2
At this condition, maximum torque is given by

The maximum torque is independent of slip.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 4

A shunt motor rotating at 1500 r/min is fed by a 120 V source. The line current is 51 A and the shunt field resistance is 120 ohms. If the armature resistance is 0.1 ohm, calculate the current in the armature.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 4

Given that, line current (IL) = 51 A
Shunt field resistance (Rsh) = 120 Ω
Voltage (V) = 120 V
Shunt current (Ish) = 120/120 = 1 A
Armature current (la) = IL - Ish = 51 - 1 = 50 A

RRB JE Electrical (CBT II) Mock Test- 5 - Question 5
Market activities in economy involve ______.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 5
Market strategies are those activities that a company engages in to boost sales, increase brand awareness, make profit, position its product or services, and determine pricing. These include marketing, advertising and sales.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 6
Which of the following is the characteristic of good conductor?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 6
Resistance and reactance opposes the flow of current through the conductor. Impedance is the vector sum of Resistance and reactance. So, If a material has high Resistance or Reactance or Impedance then it cannot be the good conductor. High conductivity is the correct answer.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 7
In Indus Valley Civilization, site Dholavira, has been located in which state of India?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 7
Dholavira is an archaeological site at Khadirbet in Bhachau Taluka of Kutch District, in the state of Gujarat in western India, and one of the two largest Harappan sites in India.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 8
Which of the following prizes is given for fiction?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 8
The Man Booker Prize for Fiction is a literary prize awarded each year for the best original novel written in the English language and published in the UK.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 9
The voltage of a circuit is measured by a voltmeter whose input impedance is low as compared to the output impedance of the circuit. The error caused will be due to
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 9

When the voltage of a circuit is measured by a voltmeter whose input impedance is low as compared to the output impedance of the circuit, the instrument gives less accurate results. This is known as the error due to the loading effect.

To reduce this effect, we need to use high impedance voltmeter compared to the load impedance.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 10

Core is insulated by ________ in a belted cable:

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

Belted Cables: The conductors (usually three) are bunched together and then bounded with an insulating paper ‘belt’. In such cables, each conductor is insulated using paper impregnated with a suitable dielectric. The gaps between the conductors and the insulating paper belt are filled with a fibrous dielectric material such as Jute or Hessian.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 11

The value of the variable resistor for maximum power transfer will be

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 11

Maximum power transfers to Rwhen R­L = Rth

To find Thevenin's equivalent resistance, we need to short circuit the voltage source and find the equivalent resistance across load resistance.

RL = Rth = 2 + 3 = 5 Ω

RRB JE Electrical (CBT II) Mock Test- 5 - Question 12
Maximum efficiency of D. C. Machines occurs when:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 12

In DC machines, maximum efficiency occurs when variable losses are equal to constant losses.

Variable losses = Constant losses

Constant losses include windage losses, brush losses, hysteresis losses and eddy current losses.

Variable losses include copper losses in armature winding and field winding.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 13
A single Phase-3 Wire AC system is made up of-
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 13

A single phase 3 wire AC system is made up of 2 conductor wire and 1 neutral wire

A single phase 2 wire AC system is made up of 1 conductor wire and 1 neutral wire

RRB JE Electrical (CBT II) Mock Test- 5 - Question 14

Most harmful environmental pollutants are

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 14

Nonbiodegradable chemicals are most harmful environmental pollutants as these cannot be broken down into simpler, harmless substances in nature. Whereas, biodegradable pollutants can be rapidly decomposed into simpler and harmless substances by natural processes.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 15

At a speed other than synchronous speed:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 15
  • At any speed other than synchronous speed, its rotor would experience an oscillating torque of zero average value as the rotating magnetic field repeatedly passes the slower moving rotor
  • Normally, a short-circuited winding similar to that of an induction machine is added to the rotor to provide starting torque
RRB JE Electrical (CBT II) Mock Test- 5 - Question 16

The power to a balanced three-phase system is given by

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 16

We know that for 3ϕ system
 Total power = √3 VL IL cos Ø
∴ power factor = cos Ø = P/√3 VL IL = 20 x 103/√3 x 400 x 100/√3 = 0.5 
PF = cos Ø = 0.5 leading 

RRB JE Electrical (CBT II) Mock Test- 5 - Question 17

In the armature-voltage control method of speed control of a DC shunt motor, as the speed is increased:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 17

In the armature-voltage control method of speed control of a DC shunt motor, as the speed is increased the torque remains constant and the power increases which is shown in above figure.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 18

Sewage pollutant can cause

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 18

RRB JE Electrical (CBT II) Mock Test- 5 - Question 19
Given J is the current density at a given location in a resistive material, E is the electric field at that location, and σ is a material-dependent parameter called the conductivity, Ohm’s law can be expressed as ____
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 19

Given J is the current density at a given location in a resistive material, E is the electric field at that location, and σ is a material-dependent parameter called the conductivity, Ohm’s law can be expressed as

J = σE

Where,

j → is the current density

E → is the electric field inside the conductor

And σ is the electrical a conductivity of the material

RRB JE Electrical (CBT II) Mock Test- 5 - Question 20

The process of reducing the magnetic flux density to zero is called ________ of the core.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 20

  • The process of reducing the magnetic flux density to zero is called demagnetization of the core
  • As the current becomes negative, H becomes negative, but B does not become negative instantaneously
  • The value of magnetic field strength required to wipe out the residual flux density is called as the coercive force HC
RRB JE Electrical (CBT II) Mock Test- 5 - Question 21

In armature control method, back emf (Eb) of DC motor is directly proportional to ________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 21

An armature control method is used in the DC shunt motor.
In a DC motor, the back emf is directly proportional to speed and flux.
Eb ∝ Nϕ
Where N is speed and ϕ is flux.|
In a DC shunt motor, flux is constant.
So, the back emf is directly proportional to speed.
Important Point:
Armature resistance control:
We know that,
N ∝ Eb
N ∝ V- I a(Ra + R)/ø
We can control the speed by increasing armature resistance. This method gives only below base speeds. This method is a constant torque and variable power drive.
Field control method:
We know that
N ∝ Eb
N ∝ V- I a(Ra + R)/ø
By varying flux, we can increase the speed more than its base speed. This method is constant power and variable torque drive.
The characteristics are shown below

RRB JE Electrical (CBT II) Mock Test- 5 - Question 22

The poorest voltage regulation of a transformer at full load is:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 22

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.
E2 = no-load secondary voltage
V2 = full load secondary voltage
Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.
Voltage regulation E2- V 2/E2
It can also be expressed as,

Regulation

+ sign is used for lagging loads and
- ve sign is used for leading loads

From the above graph, the poorest voltage regulation of a transformer at full load is at 0.8 lagging power factor.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 23
What is the SI unit for magnetic moment?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 23

The SI unit for the magnetic moment Newton-meter/Tesla

Tesla is a unit of flux

Newton-meter is unit of torque

Important:

The magnetic moment of a magnet is a quantity that determines the torque it will experience in an external magnetic field.

It is considered to be a vector having a magnitude and direction. The direction of the magnetic moment points from the South Pole to the North Pole of the magnet.

The magnetic field produced by the magnet is proportional to its magnetic moment.

RRB JE Electrical (CBT II) Mock Test- 5 - Question 24

For the circuit shown in figure the total impedance is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 24

(3 + j4) and (3 – j4) are connected in parallel and this combination is in series with 17/6 Ω resistance.
Total impedance (Z) = 17/6 + [(3 + j4) ||(3 – j4)]

RRB JE Electrical (CBT II) Mock Test- 5 - Question 25
A transformer on no load has an induced emf of 230V (rms), draws a current of 2A (rms) and having angle between them is 60°, then the value of core loss is ________.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 25

Concept:

Core loss (PC) = VI­C

IC = I cos ϕ

V is induced emf

I is current

ϕ is phase difference between induced emf and current

Calculation:

Given that, induced emf (V) = 230 V

Current (I) = 2 A

Phase difference (ϕ) = 60°

C = I cos ϕ = 2 cos 60° = 1 A

Core loss (PC) = 230 W
RRB JE Electrical (CBT II) Mock Test- 5 - Question 26
In a ceiling fan employing capacitor run motor
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 26
  • A capacitor start capacitor run motor is used in a ceiling fan
  • It essentially consists of a running winding and a starting winding
  • The capacitor is connected in series with the starting winding to make it start
  • Secondary winding surrounds the primary winding and at starting these both are connected in parallel
RRB JE Electrical (CBT II) Mock Test- 5 - Question 27
The purpose of providing dummy coils in dc machine armature is to
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 27

Dummy coils:

  • The wave winding is possible only with number of conductor or coils.
  • But sometimes the armature slots available in the armature winding do not meet the requirement of the winding. (It means that the available number of slots is more than the required number of conductors).
  • Some slots are kept without armature winding in that case and dummy coil or coils are employed in that slots.
  • The dummy coils are similar to other coils except their ends are cut, short and taped. They do not connect with the commutator bars.
  • The dummy coils are simply to provide mechanical balance for the armature.
  • As they do not connect with commutator bars, they do not affect the electrical characteristics of the winding.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 28
__________ is done to improve efficiency and reliability of dc generator.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 28

The advantages of the parallel operation of DC generators are -

Continuity of service: If a single large generator is used in the power plant, then in case of its breakdown, the whole plant will be shut down. If power is supplied from a number of small units operating in parallel, then in case of failure of one unit, the continuity of supply can be maintained by other healthy units.

Efficiency: Generators run most efficiently when loaded to their rated capacity. Electric power costs less per kWh when the generator is efficiently loaded. Therefore, when load demand on power plant decreases, one or more generators can be shut down and the remaining units can be efficiently loaded.

Increase Plant Capacity: The demand for electricity is increasing day by day. To meet the requirement of power generation, an additional new unit can run in parallel with the running units.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 29
The colour of the light given out by a sodium vapor discharge lamp is:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 29

Initially the sodium is in the form of a sloid, deposited on the walls of inner tube. When sufficient voltage is impressed across the electrodes, the discharge starts in the inert gas (neon). It operates as a low-pressure neon lamp with pink colour.

The temperature of the lamp increases gradually, and the metallic sodium vaporizes and then ionizes thereby producing the monochromatic yellow light. This lamp takes 10-15 min to give its full light output. The yellowish output of the lamp makes the object appears grey.
RRB JE Electrical (CBT II) Mock Test- 5 - Question 30

The luminous efficiency of the high-pressure mercury vapour lamps ranges from:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 5 - Question 30

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