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HPCL Electrical Engineer Mock Test - 3 - PSSSB Clerk MCQ


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30 Questions MCQ Test - HPCL Electrical Engineer Mock Test - 3

HPCL Electrical Engineer Mock Test - 3 for PSSSB Clerk 2025 is part of PSSSB Clerk preparation. The HPCL Electrical Engineer Mock Test - 3 questions and answers have been prepared according to the PSSSB Clerk exam syllabus.The HPCL Electrical Engineer Mock Test - 3 MCQs are made for PSSSB Clerk 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for HPCL Electrical Engineer Mock Test - 3 below.
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HPCL Electrical Engineer Mock Test - 3 - Question 1

The perimeter of a triangle is 36 cm and the inradius of the triangle is 5.5 cm. Then the area of the triangle is:

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 1

Given:

The inradius of the triangle = 5.5 cm

Perimeter = 36 cm

Formula used:

Semi perimeter of triangle (s) = a/2

Area of triangle = Inradius × S

Calculation:

According to the question,

Semi perimeter of triangle = (36/2) = 18 cm

Now,

Area of triangle = (5.5 × 18) cm2

⇒ 99 cm2

∴ The required area is 99 cm2.

HPCL Electrical Engineer Mock Test - 3 - Question 2

A road roller takes 750 complete revolutions to move once over to level a road. The diameter of the road roller is 84 cm and its length is 1 m. Then, the area of the road, that is being levelled, is: [ use π = ]

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 2

Given:

Diameter = 84 cm

Length = 1 m

Formula used:

Curved surface area of cylinder = 2πrh

Calculation:

According to question,

⇒ 2πrh = 2 × 22/7 × 42 × 100

⇒ 2 × 22 × 6 × 100

Area of the road being levelled = 750 × 2 × 22 × 6 × 100

⇒ 19800000 = 1980 m2

∴ The correct answer is 1980 m2.

HPCL Electrical Engineer Mock Test - 3 - Question 3

The following series is provided and you need to answer the question accordingly:

A B C D E F G H I J K L M N O P R S T U V W X Y Z

In this series find the letter which is fifth to the left from the thirteenth letter from your right.

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 3

Given: A B C D E F G H I J K L M N O P R S T U V W X Y Z

Logic: The letter which is fifth to the left from the thirteenth letter from your right.

The 13th from the right is 'M'.

The 5th letter to the left from the thirteenth letter from right (Which is M) is 'H'.

So, H will be the answer.

Hence, the correct answer is "Option 1".

HPCL Electrical Engineer Mock Test - 3 - Question 4
If 'clock' is called 'table', 'table' is called 'paper', 'paper' is called 'pen', 'pen' is called 'grinder' and 'grinder' is called 'iron', what will show the 'time'?
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 4

Given:

Clock is called Table.

Table is called Paper.

Paper is called Pen.

Pen is called Grinder.

Grinder is called Iron.

Logic: Clock shows time but Clock is called Table.

So, the correct answer is "Table".

HPCL Electrical Engineer Mock Test - 3 - Question 5

Choose the word that is opposite in meaning to the given word.

Ignominy

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 5

The correct answer is 'Option 1' i.e. 'Esteem'.

Key Points

  • The word that is opposite in meaning to the given word "Ignominy" is: 'Esteem'
  • "Ignominy" refers to public shame or disgrace(अपमान).
  • "Esteem" is the opposite of "Ignominy". It refers to respect and admiration.(सम्मान).

Therefore, the correct answer is: 'Esteem'.

Additional Information

  • Here's the explanation for the other options:
    • "Disrepute" - It means the state of being held in low regard by the public.(बदनामी).
    • "Stigma" - It refers to a mark of disgrace associated with a particular circumstance, quality, or person.(कलंक).
    • "Infamy" - It refers to the state of being well known for some bad quality or deed.(कुख्याति).
HPCL Electrical Engineer Mock Test - 3 - Question 6

Directions: Each item in this section has a sentence with three underlined parts, labelled as (a), (b) and (c). Read each sentence to determine whether there is any error in any underlined part and indicate your response on the Answer Sheet against the corresponding letter, i.e., (a) or (b) or (c). If you find no error, your response should be indicated as (d).

He suffered (a)/ from fever when he was interviewed (b)/ for the job. (c)/ No error (d)

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 6
The correct answer is Option (1) i.e. a.
Key Points
  • The phrase "He suffered" should be in the past continuous tense to indicate an ongoing condition.
  • "Suffered" should be changed to "was suffering" to accurately reflect that the fever was ongoing at the time of the interview.
  • Present continuous tense better fits the context of the sentence describing a state during a specific past event.
  • Changing "suffered" to "was suffering" makes the sentence grammatically correct and contextually appropriate.
Therefore, the correct answer is- a.
Additional Information
  • Option b: "When he was interviewed" is correct, describing the specific time at which the suffering occurred.
  • Option c: "For the job" is correct, specifying the purpose of the interview.
  • Option d: "No error" is incorrect because there is an error in part (a).
HPCL Electrical Engineer Mock Test - 3 - Question 7

Directions: Each item in this section has a sentence with three underlined parts, labelled as (a), (b) and (c). Read each sentence to determine whether there is any error in any underlined part and indicate your response on the Answer Sheet against the corresponding letter, i.e., (a) or (b) or (c). If you find no error, your response should be indicated as (d).

The politicians (a)/ parted ways (b)/ due towards ideological differences. (c)/ No error (d)

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 7

The correct answer is 'C' i.e. this part of the sentence has an error.

Key Points

  • The phrase "due towards" is incorrect. The correct phrase should be "due to".
  • "Due to" is a prepositional phrase that is used to give a reason or cause for something.
  • In the given sentence, "due to" should be used to indicate the cause of the politicians parting ways.
  • Therefore, the correct sentence should read: "The politicians parted ways due to ideological differences."

Hence, the correct answer is option 3.

Correct sentence:

The politicians parted ways due to ideological differences.

Additional Information

  • Option 1 (a): "The politicians" - This part is correct as it correctly identifies the subject of the sentence.
  • Option 2 (b): "parted ways" - This part is also correct as it correctly describes the action taken by the politicians.
  • Option 4 (d): "No error" - This option is incorrect because there is an error in part (c) of the sentence.
HPCL Electrical Engineer Mock Test - 3 - Question 8

R, N, O, W, Q and C are sitting around a circular table facing inside the center. R is sitting immediately to the left to the W, who is facing O, O is sitting immediately to the left to the N. C is not an immediate neighbor of W. Who is sitting to the left of Q?

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 8

1. R, N, O, W, Q and C are sitting around a circular table facing the center.

2. R is sitting to the immediate left of W, who is facing O, O is sitting to the immediate left of N.

3. C is not an immediate neighbor of W.

The final arrangement will be as shown below:

W is sitting to the left of Q.

Hence, ‘W’ is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 9

The following question consists of a statement followed by two arguments I and II. You have to decide which of the arguments is a STRONG argument:

Statement: Should new big industries be started in Mumbai?

Arguments:

I. Yes. It will create job opportunities.

II. No. it will further add to the pollution of the city.

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 9

Statement: Should new big industries be started in Mumbai?

Now,

Arguments:

I. Yes. It will create job opportunities.True (starting new big industries in Mumbai, citing the potential for job creation as a positive factor. Indeed, the establishment of large industrial units can generate employment opportunities for the local population and contribute to economic growth).

II. No. it will further add to the pollution of the cityFalse (Stringent environmental regulations and effective implementation of pollution control measures can mitigate the negative impacts of industrial activities on the city's environment).

So, Only argument I is strong.

Hence, the correcet answer is "Option 3".

HPCL Electrical Engineer Mock Test - 3 - Question 10

Find the missing numbers (X and Y) in the series and find the value of Y + X:

20, 22, 24, 26, X, 32, 36, Y

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 10

The logic is:

X + Y = 28 + 40

=> X + Y = 68

Hence, ‘68’ is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 11

If and θ is an acute angle, then the value of is:

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 11

Given:

Concept used:

secθ = Hypotenuse/Base, tanθ = perpendicular/base, cosθ = base/hypotenuse

Pythgorus theorem

Hypotenuse2 = Perpendicular2 + Base2

Calculation:

Let Perpendicular = P, Base = B, Hypotenuse = H

⇒ {(H/B) + (P/B)}/{(H/B) - (P/B)} = 5/1

⇒ (H + P)/(H - P) = 5/1

⇒ H + P = 5 ----(1)

⇒ H - P = 1 ----(2)

Solve (1) and (2)

⇒ H = 3, P = 2

Hypotenuse2 = Perpendicular2 + Base2

⇒ Base = √5

⇒ {3 × (√5/3)2 + 1}/{3 × (√5/3)2 - 1}

⇒ (8/3)/(2/3)

⇒ 4

∴ The value is 4.

HPCL Electrical Engineer Mock Test - 3 - Question 12

The table given below shows the saving and expenditures of 5 companies.


The total expenditure of all the companies is how much percent less than the total savings?

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 12

Given:

Calculation:

Total savings = Sum of savings for all companies

Total savings = 400 + 900 + 700 + 400 + 500 = 2900

Total expenditure = Sum of expenditure for all companies

Total expenditure = 200 + 500 + 400 + 300 + 1100 = 2500

Percentage difference = ((Total savings - Total expenditure)/Total savings) × 100

Percentage difference = ((2900 - 2500)/2900) × 100

Percentage difference = (400/2900) × 100

Percentage difference ≈ 13.79%

Therefore, the total expenditure of all the companies is approximately 13.79% less than the total savings.

HPCL Electrical Engineer Mock Test - 3 - Question 13

Choose the word that means the same as the given word.

Edacious

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 13

The correct answer is 2) Gluttonous.

Key Points

  • The word "edacious" means having a voracious or excessively eager appetite, often associated with being gluttonous or excessively hungry.
    (तीव्र या अत्यधिक भूख होना). It describes someone or something that has a strong desire for consuming food.
    • Example: "The edacious eater devoured the entire pizza in minutes, leaving no slice behind."
    • In this example, "edacious" describes the person's insatiable appetite and their quick consumption of the entire pizza.
  • In conclusion, the word that means the same as "edacious" is 2) Gluttonous, as it conveys the idea of having an excessively eager appetite or being inclined towards overeating(अधिक खाने की ओर प्रवृत्त होना).
    • Example: Despite already having a full dinner, Tom couldn't resist indulging in a gluttonous feast of desserts.

Additional Information

  • Now let's examine the other options:
  • Solvent: (अन्‍य पदार्थ को अपने में विलीन करने वाला तरल; विलायक) This option does not have the same meaning as "edacious." "Solvent" refers to being financially stable or able to pay debts, which is unrelated to the concept of being excessively hungry.
    • Example: "Despite the economic downturn, the company remained solvent and was able to pay its employees."
  • Insolvent: (धन की कमी के कारण ऋण चुकाने में असमर्थ; दिवालिया)This option does not have the same meaning as "edacious." "Insolvent" refers to being in a state of financial distress or bankruptcy, which is unrelated to the concept of having a voracious appetite.
    • Example: "Due to mounting debts, the business became insolvent and had to file for bankruptcy."
  • Satiated: (भूख को पूरी तरह शांत करना; तृप्त करना)This option does not have the same meaning as "edacious." "Satiated" means being fully satisfied or no longer hungry or thirsty, which is the opposite of the intense hunger associated with "edacious."
    • Example: "After enjoying a delicious meal, he felt satiated and content."
HPCL Electrical Engineer Mock Test - 3 - Question 14
Select the most appropriate option to fill in the blank.
Without warning, the volcano _______ a large amount of ash, forcing the residents to _______ the area.
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 14

The correct answer is "ejected; evacuate."

Key Points

  • "Ejected" - This is the past tense of the verb "eject", which means to discharge or expel something, in this case, a large amount of ash from the volcano.
  • The phrase "Without warning" implies a sudden or unexpected event that happened in the past, and so, the past tense verb "ejected" is the most contextually appropriate form.
  • It clearly communicates that this happened as a single event in the past.
  • "Evacuate" - This verb means to move people from a dangerous place to a safe place.
  • In the sentence, it is preceded by "to," which requires the base form of a verb.
  • As "force" implies a need for an action as response to a situation, "evacuate" fits perfectly to denote the action that residents were required to take as a direct result of the ash eruption from the volcano.

Therefore, the correct answer is "Option 3."

HPCL Electrical Engineer Mock Test - 3 - Question 15
Ratio of height and radius of a cylinder is 3 : 2. If volume of cylinder is 12000π, then find radius of cylinder
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 15

Given:

Ratio of height and radius of cylinder 3 : 2

Volume of cylinder = 12000π

Formula:

Volume of cylinder = πr2h

Calculation:

Ratio of height and radius of cylinder = 3 : 2 = 3x : 2x

According to the question

π × (2x)2 × 3x = 12000π

⇒ 4x2 × 3x = 12000

⇒ x3 = 12000/12

⇒ x = ∛1000

⇒ x = 10

∴ Radius of the cylinder = 2 × 10 = 20 cm
HPCL Electrical Engineer Mock Test - 3 - Question 16

Read the question below followed by two statements. Study them and decide which statement(s) is/are sufficient to answer the question.

Question: P, Q, R, S, and T have different heights. Who among the following is tallest?

Statements:

I) Neither P nor R is the tallest. S is taller than P.

II) P is taller than R and T. S is taller than Q who is not shorter than P.
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 16

Statements:

I) Neither P nor R is the tallest. S is taller than P.

S > P

II) P is taller than R and T. S is taller than Q, who is not shorter than P. So, we have the following arrangement.

⇒ P is taller than R and T.

→ P > T > R, or

→ P > R > T ....... (i)

⇒ S is taller than Q

→ S > Q ........ (ii)

Q is not shorter than P.

→ Q > P ........ (iii)

Combining all these, we get:

S > Q > P > R/T > T/R

Therefore, S is the tallest.

Only statement II is sufficient to find the tallest person among P, Q, R, S and T.

Hence, option 3 is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 17
Two spheres of radii R1 and R2 (R2 > R1) are connected by a conducting wire. Each of the spheres has been given a charge q. Now:
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 17

Concept:

Flowing of charge does not depend upon the charge amount or size of the two bodies.

If there is an electric potential difference there will be a flow of charge, if no potential difference, there will not be a flow of charge i.e. current flow is not possible.

Thus, in all electrical circuits when the current flows, it means there is an electrical potential difference between two points.

Application:

Two spheres of radii R1 and R2 (R2 > R1) are connected by a conducting wire. Each of the spheres has been given a charge q.

Now Potential of both the spheres will be equal because of both sphere has the same charge.

HPCL Electrical Engineer Mock Test - 3 - Question 18

Match List-I with List-II:

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 18

In Maxwell’s inductance bridge resistance is connected in parallel with the standard capacitor.

Hay’s bridge:

In Hay’s bridge, a resistance is connected in series with a standard capacitor.

Hay’s bridge is used for the measurement of the inductance of coils with high-quality factors.

Wein bridge:

The circuit diagram of the Wein bridge oscillator is shown below:

The phase angle criterion for oscillation is that the total phase shift around the circuit must be 0°.

The total phase shift around circuit 0° occurs when the bridge gets balanced, which is at resonance.

∴ The frequency of oscillation is given by:

Schering Bridge:

Schering bridge is used to measure the dissipation factor and capacitance.

HPCL Electrical Engineer Mock Test - 3 - Question 19

The root locus of the open loop transfer function G(s)H(s) = is

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 19

Concept:

When the difference between the no. of poles and the zeroes is equal to 1, the root locus formed is a circle.

The angle of asymptotes in a root locus is given by:

where, q = 0,1........(P-Z-1)

θ = 180°

Calculation:

Given, G(s)H(s) =

Poles = 0, -2

Zeroes = -3

The root locus starts from the pole & terminates at zero.

HPCL Electrical Engineer Mock Test - 3 - Question 20
The pick-up value of a relay is 7.5 A and the fault current in the relay coil is 30 A. Its plug-setting multiplier is ______.
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 20

The Correct answer if option (2)

Concept:

Protective Relay:

  • Pick-up value is the minimum relay operating coil current above which the relay starts its operation.
  • If the relay operating coil current is less than the pick-up value relay does not operate.
  • If both are equal, relay is on the threshold.

The current setting is defined as the ratio of the pick-up value of the relay to the relay current rating.

The current setting is also expressed in amperes, in Amperes Current setting is equal to the pick-up value.

Plug setting Multiplier (PSM):

PSM = Primary current of CT / (CT ratio × Relay current setting (A) (or)

PSM = (secondary fault current) / (Relay current setting (A) or pick-up current)

Calculation:

Given that,

Pick up current = Relay current setting (A) = 7.5 A

Fault current = 30 A

PSM = (30 / 7.5) = 4

HPCL Electrical Engineer Mock Test - 3 - Question 21

The effective resistance of a 2200-V, 50-Hz, 440-kVA, 1-phase alternator is 0.5 Ω. On short circuit, a field current of 40 A gives a full load current of 200 A. The open circuit EMF for the same field excitation is 1200 V. Find the synchronous impedance.

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 21

Finding Synchronous Impedance (Zs) with the help of OCC and SCC:

Synchronous impedance is the ratio of Open circuit voltage per phase to the score circuit armature current at the same value of field current (Excitation).

  • Let the rated field current is IA, and we have to find the Open circuit voltage and short circuit armature current at this value by plotting the graph as shown,

From the above characteristics,

At rated field current (IA),

Open Circuit Voltage (VOC) = OB

Short Circuit Current (ISC) = OC

Synchronous Impedance (Zs) = Ω

From using the concept of, Z2 = R2 + X2, we easily find the Synchronous reactance.

And Armature resistance will find by using the Voltage Test Method or (VA or AV) method.

Application:

We have,

Field of of 40 A produced:

Open circuit voltage (Voc) of 1200 V

And short circuit current (Isc) of 200 A

Now graph can be drawn as,

Synchronous impedance is given as (Z) =

Hence, Z = 1200/200 = 6.0 Ω

HPCL Electrical Engineer Mock Test - 3 - Question 22

Match List-I with List-II and select the correct answer from the following options :

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 22

Concept:

The nature of the control system is defined on the basis of the value of the damping ratio.

HPCL Electrical Engineer Mock Test - 3 - Question 23

Identify the signal x(2t - 3) given x(t) as

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 23

Concept:

Time-shifting property: When a signal is shifted in the time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.

For example:

Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'.

If a > 1, then the signal is contracted by a factor of 'a' along the time axis.

if a < 1, then the signal is expanded by a factor of 'a' along the time axis.

For example:

Analysis:

Whenever there is a combination of operations care should be taken while following the sequence of operations.

Correct approach:

x[2(t - 1.5)]

First, time scale the signal by factor 2.

Then, time shift the signal by 1.5

Incorrect approach:

x[2(t - 1.5)]

First, time shift the signal by 1.5

Then, time scale the signal by 2.

or,

Correct approach:

x[2t - 3]

First, time shift the signal by 3.

Then, time scale the signal by 2

Incorrect approach:

x[2t - 3]

First, time scale the signal by 2

Then, time shift the signal by 3.

HPCL Electrical Engineer Mock Test - 3 - Question 24
The wattmetric relay with directional characteristic is used for the protection of generator against
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 24
  • The wattmetric relay with directional characteristic is used for the protection of generator against failure of prime mover.
  • An over speed relay is used for the protection of generator against over speed.
  • Earth fault relay is used for the protection of generator against rotor filed ground faults.
  • A protective scheme employing offset mho or directional impedance relay is used for the protection of generator against loss of excitation.
HPCL Electrical Engineer Mock Test - 3 - Question 25

A single-phase semi-converter is operated from 120 V, 50 Hz AC supply at a firing angle of α degrees. The load is such that the load current is continuous and ripple-free. The displacement factor of the converter is 0.866. Then the value of sin α is _______.

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 25

Concept:

The displacement factor (D.F.) of the single-phase semi-converter converter is given by

Where, α = firing angle

Calculation:

Given that,

Supply Voltage Vs = 120 V

D.F = 0.866

Firing angle = α

⇒ α/2 = 30°

α = 60°

⇒ sin α = sin 60° = √3/2 = 0.866

HPCL Electrical Engineer Mock Test - 3 - Question 26

Match List-I (Power device) with List-II (Property) and select the correct answer:

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 26

A. Thyristor – Slow device

B. MOSFET - Large on-state drop, no secondary breakdown

C. IGBT – Low on state drop

D. BJT - Low on state drop, secondary breakdown

Additional Information

Power MOSFET:

  • Power MOSFET has lower switching losses but its on-resistance and conduction losses are more.
  • MOSFET is a voltage-controlled device whereas BJT is a current controlled device.
  • MOSFET has positive temperature coefficient for resistance. This makes parallel operation of MOSFETs easy. If a MOSFET shares increased current initially, it heats up faster, its resistance rises, and this increased resistance causes this current to shift to other devices in parallel.
  • In MOSFET, secondary breakdown does not occur, because it has positive temperature coefficient.

BJT:

  • A BJT has higher switching losses but lower conduction losses.
  • A BJT has negative temperature coefficient, so current-sharing resistors are necessary during parallel operation of BJTs.
  • As BJT has negative temperature coefficient, secondary breakdown occurs. With decrease in resistance, the current increases. This increased current over the same area results in hot spots and breakdown of the BJT.

Insulated Gate Bipolar Transistor (IGBT): It is a three-terminal power semiconductor device primarily used as an electronic switch. It is a 4 layer PNPN device that combines an insulated gate N-channel MOSFET input with a PNP BJT output in a type of Darlington configuration.

​Insulated gate Bipolar Transistor is also known as Conductivity-Modulated Field Effect Transistor.

Advantages:

  • The insulated gate bipolar transistor (IGBT) is easy to turn ON and OFF.
  • The switching frequency is higher than that of power BJT.
  • It has a low on state power dissipation.
  • It has simpler driver circuit.

Disadvantages:

  • The switching frequency of insulated gate bipolar transistor (IGBT) is not as high as that of a power MOSFET.
  • High turn-off time
  • It cannot block high reverse voltages.
HPCL Electrical Engineer Mock Test - 3 - Question 27
Two spheres of radii R1 and R2 (R2 > R1) are connected by a conducting wire. Each of the spheres has been given a charge q. Now:
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 27

Concept:

Flowing of charge does not depend upon the charge amount or size of the two bodies.

If there is an electric potential difference there will be a flow of charge, if no potential difference, there will not be a flow of charge i.e. current flow is not possible.

Thus, in all electrical circuits when the current flows, it means there is an electrical potential difference between two points.

Application:

Two spheres of radii R1 and R2 (R2 > R1) are connected by a conducting wire. Each of the spheres has been given a charge q.

Now Potential of both the spheres will be equal because of both sphere has the same charge.

HPCL Electrical Engineer Mock Test - 3 - Question 28
In an AC circuit, and . The active and reactive power of the circuit are respectively
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 28

Concept:

Apparent power (S) = P + jQ = V I*

Calculation:

Given V = 200 + j40

Current (I) = 30 - j10

Conjugate current (I*) = 30 + j10

Apparent power (S) = Voltage × Conjugate current (I*)

Apparent power (S) = V I*

= 5600 + j 3200

∴ Active power (P) = 5600 watt

Reactive power (Q) = 3200 VAR

As the reactive power is positive, the load is inductive.

HPCL Electrical Engineer Mock Test - 3 - Question 29
A buck DC to DC converter is applied with a voltage of 100 V and supplies a resistive load of 50 Ω. During ON state, it has a voltage drop of 2 V. The chopping frequency is 1 kHz and the duty ratio is 50%. Average and rms output voltages respectively are:
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 29

Concept:

In a buck DC to DC converter

Vavg = δ Vd

Where δ = duty ratio

Vi = input voltage

Vd = Vi – Voltage drop

Calculation:

Duty ratio (δ) = 50% = 0.5

Vd = Vi – Voltage drop = 100 – 2 = 98 V

Vavg = δ Vd = 0.5 × 98 = 49 V

HPCL Electrical Engineer Mock Test - 3 - Question 30
A solenoid has 4000 turns over it with inductance of 0.126 H. The energy stored in the magnetic field, when a current of 2 A flows in the solenoid will be
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 30

Concept:

The inductance of a solenoid is given by:

N = Number of turns

A = Area of the solenoid

l = length of the solenoid

Also, the energy stored by the inductor is given by:

Calculation:

Given that, inductance (L) = 0.126 H

Current (I) = 2 A

= 0.252 J

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