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RRB JE Electrical (CBT II) Mock Test- 4 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - RRB JE Electrical (CBT II) Mock Test- 4

RRB JE Electrical (CBT II) Mock Test- 4 for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The RRB JE Electrical (CBT II) Mock Test- 4 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 4 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- 4 below.
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RRB JE Electrical (CBT II) Mock Test- 4 - Question 1

The Supreme Court decision which stated that the Directive Principles of State policy cannot override fundamental rights is_________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 1

In State of Madras Vs. Champakam Dorairajan case, the Supreme Court decisions stated that the Directive Principles of State policy cannot override fundamental rights.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 2

Magnetic fields don't represent a "flow" of anything, and no ...... is dissipated in reluctances.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 2

Electric currents represent the flow of particles (electrons) and carry power, part or all of which is dissipated as heat in resistances. Magnetic fields don't represent a "flow" of anything, and no power is dissipated in reluctances.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 3

Acceptable "Noise Pollution Level" in India ranged between which of the following?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 3

Noise is defined as unwanted sound. Sound which pleases the listeners is music and that which causes pain and annoyance is noise. Anything above this (40-45 dB) threshold limit is considered to be noise.

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

Breaking capacity of a circuit breaker is usually expressed in:

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

A circuit breaker should be rated with short circuit making capacity. As the rated short circuit making current of the circuit breaker is expressed in peak value, it is always more than rated short circuit breaking current of the circuit breaker.

Making capacity of a circuit breaker is equal to 2.55 times symmetrical breaking current.

Breaking capacity is always expressed in rms value.

Both breaking and making capacity are generally expressed MVA.

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

In power systems, regulating transformer are used to control

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

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- 4 - Question 6

The part of the Himalayas lying between Satluj and Kali rivers is known as ____________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 6

Kumaun Himalayas, west-central section of the Himalayas in northern India, extending 200 miles (320 km) from the Sutlej River east to the Kali River. The range, comprising part of the Siwalik Range in the south and part of the Great Himalayas in the north.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 7

Consider the following statements and select the correct option.

Statement A: When current more than the fixed current flows in the circuit, fuse wire melts.

Statement B: Fuse wire is made up of a low melting point material.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 7

A fuse is an electrical safety device that operates to provide overcurrent protection of an electrical circuit. Its essential component is a metal wire or strip that melts when too much current flows through it, thereby interrupting the current.

The fuse wire should have low melting point so that it can get melt when current more than rated flows in the circuit

RRB JE Electrical (CBT II) Mock Test- 4 - Question 8

The superposition theorem applies to________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 8

As superposition principal is applicable for linear,active, bi lateral network.
Current & voltage are linear to each other.
but power = I2 x R
Power is direct propositional to square of current , which is non linear.
There for it is applicable only for current & voltage calculation.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 9
Which river of India is called Vridha Ganga?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 9
The Godavari also was known as the Dakshin Vahini Ganga and Vridha Ganga. It originates at an elevation of 1067 meters in the Brahmagiri Hills of the Western Ghats in Nashik District of Maharashtra and continues its journey over 1465 km in the south-east to meet the Bay of Bengal at Narasapuram in West Godavari district of Andhra Pradesh.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 10
A diode is not used for?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 10
A diode is a specialized electronic component with two electrodes called the anode and the cathode. Most diodes are made with semiconductor materials such as silicon, germanium, or selenium. Some diodes are comprised of metal electrodes in a chamber evacuated or filled with a pure elemental gas at low pressure. Diodes can be used as rectifiers, signal limiters, voltage regulators, switches, signal modulators, signal mixers, signal demodulators, and oscillators.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 11

The reluctance of magnetic circuit varies as_________.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 11

Reluctance = length of wire/area x permeability
Reluctance, therefore reluctance varies as length/ Area.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 12

Which element is most electronegative among Arsenic, Nitrogen and Phosphorus?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 12
  • Nitrogen is the most electronegative among Arsenic, Nitrogen and Phosphorus.
  • As per the periodic table, Nitrogen has an atomic number of 7 and hence its electron configuration is 1s2 2s2 2p3.
  • Amongst all the others nitrogen has the highest tendency to attract a shared pair of electrons towards itself.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 13
The rotor of an induction motor cannot run at synchronous speed because:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 13

Synchronous speed is the speed of rotation of the magnetic field in a rotary machine. An induction motor always runs at a speed less than synchronous speed.

The rotating magnetic field which is produced in the stator will generate flux in the rotor which will make the rotor to rotate. But due to the lagging of rotor flux current with respect to stator flux current, the rotor will never reach to its rotating magnetic field speed i.e. the synchronous speed.

If the motor is running at synchronous speed, then the torque generated by the motor equal to zero. If rotor runs at synchronous speed as the field of the stator, this will cause the slip to become zero, then there will be no torque generated in rotor.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 14
The optical phenomenon that is primarily responsible for the observation of rainbow on a rainy day is ________.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 14
RRB JE Electrical (CBT II) Mock Test- 4 - Question 15
Alternators are constructed with:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 15

The field winding of an alternator is placed on the rotor and is connected to DC supply through two slip rings.

The 3-phase armature winding is placed on the stator. This arrangement has the following advantages:

  • It is easier to insulate stationary winding for high voltages for which the alternators are usually designed
  • The stationary 3-phase armature can be directly connected to load without going through large, unreliable slip rings and brushes
  • Only two slip rings are required for DC supply to the field winding on the rotor. Since the exciting current is small, the slip rings and brush gear required are of light construction
  • Due to the simple and robust construction of the rotor, higher speed of rotating DC field is possible. This increases the output obtainable for a machine of given size
RRB JE Electrical (CBT II) Mock Test- 4 - Question 16
The capacitor bank installed in the rectifier system of any electroplating plant is meant for:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 16
The capacitor bank installed in the rectifier system of any electroplating plant is to improve the power factor and the line regulation of the mains feeding the rectifier system.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 17
Which of the following folk/tribal dances is associated with Karnataka?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 17
Yakshagana is a popular folk-theatre form of Karnataka with a long history of nearly four hundred years. It is a unique harmony of musical tradition, eye-catching costumes, and authentic styles of dance, improvised gestures and acting with its extemporaneous dialogue appealing to a wide range of the community. It is a vibrant and vigorous living form of theatre art.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 18

Which of the following is an operating system?

I. Ubuntu

II. Linux

III. Unix

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 18

Ubuntu, Linux, Unix, Microsoft Windows and macOS all fall under the category of operating systems.

An Operating system is basically a low-level system software and an important program that runs on every computer. Its basic function is to perform basic tasks such as controlling peripherals, managing computer software and hardware resources, keeping track of directories and files in the storage drive, sending output to the display and more.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 19

The superposition theorem can be applied to solve networks when the network contains:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 19

The superposition theorem can be applied to solve networks when the network contains two or more than two sources of anyone, or both types.

The superposition theorem for electrical circuits states that for a linear system the response (voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances. This theorem is essentially based on linearity.

RRB JE Electrical (CBT II) Mock Test- 4 - Question 20

Two heaters of rating 1 kW, 250 V are connected in series across 250 V supply, the power taken by the heaters will be:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 20

Let's first find the current for each heater

P = V × I

I = P/V = 1000/250 = 4 Amp.

Now we need to find resistance of each heater.

V = I × R or R = V/I

I.e. R = 250/4 = 62.5 ohms.

∴ resistance of 2 heaters = 125 ohms.

Now we need to find the total power

P = 4 × 125 = 500 watts = ½ kW

RRB JE Electrical (CBT II) Mock Test- 4 - Question 21

A principal node is a

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 21

A principal node is a junction where branch current can combine or divide

Important:

Node:

A point or junction where two or more circuit’s elements (resistor, capacitor, inductor etc.) meet is called Node. In other words, a point of connection between two or more branches is known as a Node.

Branch:

That part or section of a circuit which locate between two junctions is called the branch. In a branch, one or more elements can be connected, and they have two terminals.

Loop:

A closed path in a circuit where more than two meshes can occur is known as Loop i.e. there may be many meshes in a loop, but a mesh does not contain on one loop.

Mesh:

A closed loop which contains no other loop within it or a path which does not contain on other paths is called Mesh

RRB JE Electrical (CBT II) Mock Test- 4 - Question 22
Induction generators deliver power at ______ power factor
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 22

Induction generator always works with leading power factor since it will take large amount of reactive power to produce sufficient amount of working flux so that armature reaction is always magnetizing hence it will work always with leading pf

Important Point

  • induction generator is basically an induction motor, which runs above the synchronous speed
  • when it acts as a generator it will supply the active power back to source, but for this supply of active power it needs reactive power as input to keep its winding excited
  • in case if the induction motor is connected to the grid, it will draw the required reactive power for the excitation of windings, but if it is standalone system (ie. not connected to grid) then a capacitor bank will be always connected, and this will provide leading reactive power to keep the winding excited for the process of mechanical to electrical energy conversion
  • since the reactive power is supplied by the capacitor, the induction generator is operating in leading power factor
RRB JE Electrical (CBT II) Mock Test- 4 - Question 23

A sinusoidal voltage V(t) = 200 sinωt is applied to a series LCR circuit with L = 10 mH, C = 100 nF and R = 20 Ω. Find the amplitude of current at resonance

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 23

Concept
When the LCR circuit is set to resonance, the resonant frequency is 
Quality factor is 
At resonance current is Maximum i.e. I0
I O = V0/R
Calculation:
At resonance current is maximum i.e. I0
I O = V0/R = 200/20 = 10A

RRB JE Electrical (CBT II) Mock Test- 4 - Question 24

If sag on a transmission is increase, the tension will:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 24

The sag is given by
S = w L 2/8T
Where w is the weight per unit length of the conductor
T is the tension of the conductor
L is the length of the span
If sag on the transmission is increased, the tension will decrease

RRB JE Electrical (CBT II) Mock Test- 4 - Question 25
If the total resistance in a series circuit doubles, current will:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 25

When the resistances are connected in series, the equivalent resistance will be the sum of all the resistances. In series circuit, the current flows through circuit is same.

We know that V = IR

For constant voltage, resistance is inversely proportional to current.

If the total resistance in series circuit doubles, the current will become half.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 26

Who of the following was the last Governor General of Bengal?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 26

By Charter Act of 1833 Governor of the state of Bengal designated as Governor-General of India.

 

RRB JE Electrical (CBT II) Mock Test- 4 - Question 27
In a Zener diode with high breakdown voltage
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 27
This type of breakdown occurs at the reverse bias voltage above 8 V and higher. It occurs for lightly doped diode with large breakdown voltage. As minority charge carriers (electrons) flow across the device, they tend to collide with the electrons in the covalent bond and cause the covalent bond to disrupt.
RRB JE Electrical (CBT II) Mock Test- 4 - Question 28
An ideal rectifier should have –
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 28

An ideal rectifier should have:

  • Efficiency = 100%
  • AC components (Vac) = 0
  • Transformer Utilization Factor = 1
  • Power Factor = 1
RRB JE Electrical (CBT II) Mock Test- 4 - Question 29
What is the voltage value for high voltage substation?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 29

According to the operating voltage, the substations may be categorized as:

  • High voltage substations – involving voltages between 11 kV and 66 kV
  • Extra high voltage substations – involving voltages between 132 kV and 400 kV
  • Ultra-high voltage – operating voltage above 400 kV

Important Points:

By nature of duties, the substations may be categorized as:

  • Step-up or primary substations
  • Primary grid substations
  • Step-down or distribution substations

By service rendered, the substations may be categorised as:

  • Transformer substations
  • Switching substations
  • Converting substations

By design, the substations may be categorised as:

  • Indoor type substations
  • Outdoor substations
  • Pole mounted substations
  • Foundation mounted substations
RRB JE Electrical (CBT II) Mock Test- 4 - Question 30

In delta connected circuit, when one resistor is open, the power

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 4 - Question 30

Power consumed in case of balanced delta connection  P = V2 ph/Req
Equivalent resistance in this case: Req 2R/3
Now, the power will be P1 = 3V ph/2R
If one resistor is removed from a 3 - phase delta connected circuit, then the circuit will be converted into single phase circuit having two resistors in series.
Equivalent resistance in this case: Req = 2R
Now, the power will be PV ph/2R
⇒ P= P1/3

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