RSMSSB JE Electrical Mock Test - 1 - Electrical Engineering (EE) MCQ

The correct answer is Madhya Pradesh.

Key Points

• Inter-state border of Rajasthan:
  • Border of Rajasthan and Punjab – (89 km)
  • Rajasthan and Haryana border - (1262 km)
  • Rajasthan and Uttar Pradesh border – (877 km)
  • Rajasthan and Madhya Pradesh – (1600 km)
  • Border of Rajasthan and Gujarat – (1022 km)
• Length of the terrestrial boundary of Rajasthan
  • The total terrestrial extent of Rajasthan – is 5,920 km.
  • Length of the inter-state border – 4,850 km.
  • Length of International Border – 1,070 km.
    • Sriganganagar - 210 km
    • Bikaner - 168 km
    • Jaisalmer - 464 km
    • Barmer - 228 km​​

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The correct answer is Alwar Bharatpur.

Key Points

• Bam Dance is performed in the Mewat region (Alwar - Bharatpur).
  • It is performed to celebrate a new crop.
  • It is performed with a Big drum(Bam) by the male and female on the occasion of Holi.
  • The Rasiya song, is also known as Bam-Rasiya.
  • Musical instruments used are Drum, Thali, Manjira, Dholak, etc.

The correct answer is Jaynarayan Vyas was one of the founding members of the Prajamandal.

Key Points

• Dholpur Prajamandal:
  • It was established in the year 1936.
  • It was founded by Krishnadutt Paliwal, Mr Mulchand, Mr Jwala Prasad Jigasu.
  • Shri Krishna Dutt Paliwal was made the president of this Praja Mandal.
• The aim of this project was to protect accountability, governance and civil rights.
• Movements were made by this Praja Mandal demanding responsible governance and civil rights.
• Om Prakash Sharma Ram Dayal Ram Prasad Bankel Lal Keshav Dev Kedarnath, etc., many activists were arrested.

The correct answer is Neither 1 nor 2.

• As per Kshatriya theory of origin Rajputs based on their lineage as Suryavanshi and Chandravanshi.
• As per Agnikunda theory of origin Rajputs clan from Agnikunda are Chauhans, Chalukyas, Parmaras and Pratiharas.
• Hence both the statements are incorrect.

Additional Information

• Kshatriya theory of the origin:
  • This theory was propounded by Gauri Shankar Ojha and says that the Rajputs are not from a foreign origin and they are descendants of the mythological Kshatriya Heroes like Rama.
  • The theory divides the Rajput based on their lineage as Suryavanshi & Chandravanshi, which they trace from Surya and Chandra.
  • They worship fire as the Aryans did and worship of fire was not the tradition of the Foreigners only.
• Agnikunda Theory:
  • This theory comes from the Prithvirajraso of Chandarbardai.
  • According to this theory, Rajputs were the result of Yagya performed by Hrishi Vashistha at “Guru Shikhar” in Mount Abu.
  • The four Rajput clans from Agnikunda are Chauhans, Chalukyas, Parmaras and Pratiharas.
  • Muhnot Nainsi & Suryamal Mishra also supports this theory.

The correct answer is Rigveda.

Key Points

• The Meena kingdom (Fish kingdom) was called the Matsya Kingdom in Sanskrit and was mentioned in the Rig Veda.
• Matsya or Meena (Sanskrit for fish) was the name of a Kshatriya tribe and the state of the Vedic civilization of India.
• It lay the south of the kingdom of Kurus and west of the Yamuna which separated it from the kingdom of Panchalas.
• It roughly corresponded to the former state of Jaipur in Rajasthan and included the whole of Alwar with portions of Bharatpur.
• The capital of Matsya was at Viratanagar (modern Bairat) which is said to have been named after its founder king Virata.

Additional Information

• Rigveda
  • ​The Rigveda or Rig Veda is an ancient Indian collection of Vedic Sanskrit hymns.
  • The Rigveda is the oldest known Vedic Sanskrit text.
• ​Samaveda
  • ​ The Sama Veda deals with melodies.
• Yajurveda
  • ​The Yajur Veda contains the rituals of the Yagna.
• ​Atharva veda
  • ​ The Atharva Veda contains magic spells.

The correct answer is Bhavai.

Key Points

• The famous theatre director Shanta Gandhi was associated with Bhavai drama styles.
• Shanta Gandhi was born on 20-12-1917 in Nashik in the state of Maharashtra, India.
• She was an Indian Theatre Director, Dancer & Playwright known for her work in Gujarati and Hindi Cinemas.
• She was also a founder member of IPTA (Indian People's Theatre Association) & Avehi Abacus Project.
• She was also a Chairperson of NSD (National School of Drama), Delhi from 1982 to 1984.
• She received many government awards, including the Padma Shri (1984).
• A famous play written in Bhavai style is Shanta Gandhi's- Jasma Oden.

The correct answer is Kota.

Key Points

• Shiv temple of Charchoma located in Kota.
• Shiv temple of Charchoma temple is not of Gurjar-Pratihar style of architecture.
• Durra temple in Kota and Shiva temple in Charchoma are the example of Gupta architecture in Rajasthan.

Additional Information

• There are notable examples of architecture from the Gurjara-Pratihara era, including sculptures and carved panels.
• Their temples, constructed in an open pavilion style.
• One of the most notable Gurjara-Pratihara styles of architecture was Khajuraho, built by their vassals, the Chandelas of Bundelkhand.
• The stone Nataraj/Natesha murti, in “chatura pose with jatamakuta and trinetra” and almost four feet tall, is a rare depiction of Lord Shiva in the Prathihara style.

The correct answer is Waist.

Key Points

• A series of silver chains formed into a belt are worn at the hips and are generally known as kandora or kardhani, while the men would wear a silver or gold belt.
• Kandora ornament is worn on Waist.
• Kandora is similar to Kardhani but its design is different.

Additional Information

• Other important ornaments are:
• Head: Borla, Shishufal, Rakhdi, Tikadi, Sankali, Taawat.
• Hand: Takma, Patt, Punchiyan, Nogari.
• Waist: Kandori, Karghani, Tagadi, Satka.
• Foot: Payal, Nupur, Nevari, Lachha, Todia.

• The balanced star connected system does not contain a neutral point and it is a three-wire system.
• But star connected system which has neutral point considered as an unbalanced system.
• While delta connected balanced and unbalanced system does not have neutral point.

• Delta connected system is a three-wire system. It has 3 phases only, it does not contain a neutral point.

• Star connected system is a four-wire system with three phases and one neutral wire.

Magnetic flux density:
Magnetic flux density is the amount of flux passing through a defined area that is perpendicular to the direction of the flux, i.e.
Mathematically, this is defined as:
B = ϕ/A
B = Magnetic flux density
ϕ = Magnetic Flux (weber, Wb)
A = Area (m2)

• SI unit of the magnetic flux density (B) is Tesla (T).
• Tesla is equivalent to one weber per meter squared or weber/meter2
• The CGS unit of B is gauss where 1 gauss = 10-4 tesla.

Important Notes:

The permanent magnets are made from hard ferromagnetic materials (steel, cobalt steel, carbon steel etc). Since these materials have high retentivity, the magnet is quite strong. Due to their high coercivity, they are unlikely to be demagnetized by stray magnetic fields.

Permanent Magnets:

Permanent magnets are magnets with magnetic fields that do not dissipate under normal circumstances. They are made from hard ferromagnetic materials, which are resistant to becoming demagnetized.

Permanent magnets are made from a material that will inherit the properties of a strong magnetic field when exposed to it.

Properties:

Residual induction:

The residual induction is any magnetic induction that remains in a magnetic material after removal of an applied saturating magnetic field, measured in gauss or tesla. Residual induction is also known as magnetic remanence.

Coercivity:

• Coercivity (or the coercive field) is the property of a material to resist demagnetization due to the intensity of the material's magnetic field
• Coercivity is measured by the extent to which a demagnetizing field must be applied to reduce the material's magnetism to zero
• Permanent magnets are composed of materials with a high coercivity which retains their inherited magnetic fields under most conditions, unless intentionally demagnetized

Hysteresis loop:

• Wider hysteresis loops have high retentivity, coercivity, and saturation due to their larger hysteresis loop area
• These loops are typically found in hard magnetic materials
• Due to the size, these hysteresis loops have low initial permeability which leads to higher energy dissipation
• For these reasons, they are utilized in permanent magnets that have high resistance to demagnetization
• Demagnetization is more difficult to achieve in these wider hysteresis loops because there is a larger area to cover when reversing the hysteresis loop direction back to its original paramagnetic state

Note:

Diamagnetic materials

• Weak, negative susceptibility to magnetic fields
• Diamagnetic materials are slightly repelled by a magnetic field
• All the electrons are paired so there is no permanent net magnetic moment per atom
• Most elements in the periodic table, including copper, silver, and gold, are diamagnetic

Paramagnetic materials

• Small, positive susceptibility to magnetic fields
• These materials are slightly attracted by a magnetic field
• Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field
• Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum

Ferromagnetic materials

• Have a large, positive susceptibility to an external magnetic field
• They exhibit a strong attraction to magnetic fields and can retain their magnetic properties after the external field has been removed
• Ferromagnetic materials have some unpaired electrons, so their atoms have a net magnetic moment
• Iron, nickel, and cobalt are examples of ferromagnetic materials.

In the given question load consists of pure inductance L = 0.1 H, then the voltage equation for given SCR circuit is

Take integration on both sides,

Where E = Source voltage
i = Latching current
t = pulse width
Now we can calculate minimum pulse width of gate pulse required to properly turn – ON SCR,

• When the current carrying contacts of the circuit breaker are moved apart, an arc is formed, which insist for a short period after the separation of contacts
• This arc is dangerous on account of the energy generated in it in the form of heat which may result in explosive force
• The circuit breaker should be capable of extinguishing the arc without causing any disturbances to the equipment
• This phenomenon takes place at zero current
• There are two methods of arc extinction in circuit breakers; They are high resistance method and low resistance or zero current interruption method

Concept:

To eliminate the n^th harmonic by use of short pitch coil,

Where α is a chording angle

Application:

Given that, n = 5

∴ α = 36°

Characteristics of Common Emitter Amplifier:

1) The voltage gain and current gain are high

2) The power gain is high

3) There is a phase relationship of 180 degrees in input and output

4) The input and output resistors are medium

load commutated chopper circuit has consists of 4 thyristors and one capacitor. 4 SCRs forming 2 pairs of 2 SCRs conduct alternately.

Concept:

Delta - Star Transformer:

In this type of connection in a transformer, primary side is connected in delta fashion while the secondary side is connected in star.

The main use of this connection is to step up the voltage i.e. at the beginning of high tension transmission system.

Also it is clear that there is a phase shift of 30° between primary line voltage and secondary line voltage.

Calculation:

Given:
11000/415 V delta-star transformer
Power rating (S) = 40 kVA
We know

I_L = Line current,
V_L =Line voltage
For primary side:


Now,
Voltage transformation ratio for transformer is

Short Trick:

As we know transformer transfers energy from primary to secondary. It cannot step up or step down energy because it will violate the law of transformer remains same.
Therefore, We can directly calculated the line current on the secondary side.
i.e.


I_L = 55.648 A

Medium lines - Nominal T method:

In this method, the whole line to neutral capacitance is assumed to be concentrated at the middle point of the line and half the line resistance and reactance are lumped on either side as shown in the figure.

V_S = V₁ + I_SZ/2
V₁ = V_R + I_RZ/2
I_C = I_S – I_R = V₁Y






From the above equations of V_S and I_S,

Concept:

The time-domain equation that relates the terminal voltage to the terminal current for an inductor as shown is:

Taking the Laplace transform, we get:

---(1)
The above equation is the sum of voltages sLI(s) and –Li(0^-). This is represented as:

Equation (1) can also be written as:

The circuit representing the above Equation is as shown:

Application:

Replacing the inductor by its s-domain equivalent, we get:

The parallel equivalent of 0.4s and 40 Ω resistor will be:

Applying current division rule, we get:

Concept:

The generated emf in AC rotating machine is given by:

where, E = EMF
ϕ = Flux
- sign is due to Lenz law

Calculation:
Let us assume ϕ = sin(ωt)

From the above phasor diagram, E lags ϕ by 90°
Note: Anticlockwise rotation is considered to be leading and clockwise to be lagging.

Differential relay:

• Differential relay operation depends on the phase difference of two or more electrical quantities.
• It works on the principle of comparison between the phase angle and the magnitude of the same electrical quantities.
• The differential relay is used for the protection of the feeder, large busbars, etc.

Differential protection for feeders consist of
1) Voltage balance differential relay
2) Current balance differential relay

Note:
Generally Translay relay is used for feeder protection

• The translay relay is a differential relay
• The arrangement is similar to overcurrent relay, but the secondary winding is not closed on itself
• These types of relays are used in the feeder protection and the scheme is called Translay scheme
• In this scheme, two such relays are employed at the two ends of feeder as shown in the Fig

• The secondaries of the two relays are connected to each other using pilot wires
• The connection is such that the voltages induced in the two secondaries oppose each other
• The copper coils are used to compensate the effect of pilot wire capacitance currents and unbalance between two currents transformers
• Under normal operating conditions, the current at the two ends of the feeder is same
• The primaries of the two relays carry the same currents inducing the same voltage in the secondaries
• As these two voltages are in opposition, no current flows through the two secondaries circuits and no torque is exerted on the discs of both the relays
• When the fault occurs, the currents at the two ends of the feeder are different, hence unequal voltages are induced in the secondaries
• Hence the circulating current flows in the secondary circuit causing torque to be exerted on the disc of each relay
• But as the secondaries are in opposition, hence torque in one relay operates to close the trip circuit while in other relay the torque just holds the movement in unoperated position
• The care is taken that at least one relay operates under the fault condition

Concept:
The load impedance is given by:
Z = R + j X (for lagging load)
Z = R - j X (for leading load)

The power factor is given by:
cosϕ R/Z

Calculation:

Given, Z = 20 - j20

cosϕ = 0.707 leading

Concept:

• Ideally, a transformer draws the magnetizing current and lags the primary applied voltage by 90°.
• But the transformer also has a core loss current component which will be in phase with applied voltage.
• No-load current is nothing but the vector summation of these two currents.
• Hence, the no-load current will not lag behind applied voltage by exactly 90° but it lags somewhat less than 90°.
• It is in practice generally about 75°.

Additional Information

Circuit diagram for a transformer on no-load:

Where,
V1 is the applied primary voltage
Iw is the working component of current through Ro (Magnetizing resistance)
Iμ is the magnetizing component of current through Xo (magnetizing reactance)
N1 and N2 are primary and secondary turns ratio

• In the case of no-load, the second terminal of the transformer is open.
• There is no path available for the current to flow on the secondary side.
• Hence, the transformer does not draw current from the source.
• A small ampere of current flows through the primary transformer (no-load current I ) called excitation current (used for excitation of the core).
• No-load current (Io ) is further divided into Iw and Iμ.

Phasor diagram when transformer on no-load:

Concept:
AM: In amplitude modulation, the amplitude of the carrier signal varies in accordance with the instantaneous amplitude of the modulating signal.
The time-domain representation of an amplitude-modulated signal is given as:
s(t) = Ac [1 + ka m(t)] cos 2πfct
s(t) = Ac cos 2πfct + Ac ka m(t) cos 2πfct
Where the carrier Ac cos ωct is modulated in amplitude and ka is the amplitude sensitivity of the modulator.
The frequency-domain representation of an amplitude-modulated signal is given as:

∴ We can see that the bandwidth of an AM signal is twice that of the maximum frequency present in the message signal.
Important Point

Concept:

Per unit quantity:

Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units

⇒ Per unit impedance Zpu = Zactual / Zbase

⇒ Zpu = ZΩ × MVAb / (kVb)2

Conversion of one per unit impedance into another per unit impedance is given by

3 phase circuit fault has maximum short-circuit current.

Three-phase bolted fault has more maximum short-circuit current than 3 phase to ground fault.

Important Points

The different type of faults in power systems are:

• Single line to ground fault (LG)
• Line-to-line fault (LL)
• Double line to ground fault (LLG)
• Three-phase faults (LLL or LLLG)

Frequency of occurrence:

• Among the given faults, LG or line to ground fault is most common and occurs frequently.
• The order of frequency of occurrence is given below.

LG > LL > LLG > LLL

Severity of faults:

• Among the given faults, LLLG or 3 phase faults are most severe. LG or line to ground fault is least severe.
• Line to line fault is more severe than line to ground fault while double line to ground fault is one level severe than LL.
• The order of severity of faults is given below.

LLL > LLG > LL > LG

Additional Information

The decreasing order of fault currents at terminals of a synchronous generator is given below
I_f(LG) > I_f(LLG) > I_f(3ϕ) > I_f(LL)

The fault current is maximum for line to ground fault.

Concept:

The resonant frequency of a series RLC circuit is given by:

where, ω_o = Resonant frequency

L = Inductance

C = Capacitance

Calculation:

Given, L = 2 H

C = 20 μF = 20 × 10^-6 F

ω0 = 0.158 × 103 rad/sec

I₁ = 150 ± 1A

I2 = 250 ± 2A

I = I₁ +I₂

and, σ₁ = 1 A

and, σ₂ = 2 A

The standard deviation of I is,

or, σ =

Hence,

Resultant Current = (400 ± 2.24) A

Resistance wire strain gauge:

The electrical resistance strain gauge is a resistance element that changes resistance when subject to strain or temperature change.

Gauge factor is defined as the ratio of per unit change in resistance to per unit change in length.

The relationship between the Gauge factor and Poisson's ratio is:

where, G_f = Gauge factor

v = Poisson's ratio

Resistance changes due to piezoresistive effects

Calculation:

Given, = 4.2

Neglecting resistance changes due to piezoresistive effects

4.2 = 1 + 2v

v = 1.6

Definition: Travelling wave is a temporary wave that creates a disturbance and moves along the transmission line at a constant speed. Such type of wave occurs for a short duration (for a few microseconds) but causes a much disturbance in the line.

The transient wave is set up in the transmission line mainly due to switching, faults and lightning.

Crest: It is the maximum aptitude of the wave, and it is expressed in kV or kA.

Front / Rise: It is the portion of the wave before the crest and is expressed in time from the beginning of the wave to the crest value in milliseconds or µs.

Tail / Fall: The tail of the wave is the portion beyond the crest. It is expressed in time from the beginning of the wave to the point where the wave has reduced to 50% of its value at its crest.
Polarity: Polarity of the crest voltage and value. A positive wave of 400 kV crest 1 µs front/rise and 50 µs tail/fall will be presented as +400/1.0/50.0 or 400/1/50.