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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 2024 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 2024 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

A man lends out an amount of Rs. 25,000 into two parts. The first part of the money is lent out at 8% simple interest and the second part of the money is lent out at 8.5% simple interest. If total annual income of the man is Rs. 2031.25, then find how much money is lent out at the rate of 8.5%?

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

Calculation:

Total principal = 25000 and total interest receive = 2031.25

So, all-over interest rate = 2031.25/25000 × 100 = 8.125%

So, the 1st part and 2nd part of the principal were lent out in the ratio 3 : 1

Now, the 2nd part of the money is = 25000 × = Rs. 6250

∴ The correct answer is Rs. 6,250

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.

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HPCL Electrical Engineer Mock Test - 3 - Question 3

A, B and C run on a circular track. Time taken by A, B, and C to complete one round are 6 min, 8 min and 3 min respectively. If they start together at 5 p.m., then at what time will they meet again at the starting point?

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

Given

Time taken by A, B, and C to complete one round: 6 min, 8 min, 3 min respectively

Concept:

The time they will meet again at the starting point will be the least common multiple (LCM) of their individual times.

Solution:

LCM of 6, 8, and 3 minutes = 24 minutes

They started at 5:00 p.m. so they will meet again at 5:00 p.m. + 24 minutes ⇒ 5:24 p.m.

Therefore, they will meet again at the starting point at 5:24 p.m.

HPCL Electrical Engineer Mock Test - 3 - Question 4

In this question, two statements are followed by two conclusions, numbered I and II. Find out which conclusion (s) is / are definitely true, based on the given statements.

Statement : D > B ≥ W, Y > W > M

Conclusions :

I. B ≥ M

II. M < D

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 4

The given statement is: D > B ≥ W, Y > W > M

Combine the statement: D > B ≥ W, Y > W > M

Conclusions:

I. B ≥ M False (as D > B ≥ W, Y > W > M gives; as B ≥ W > M; so B > M).

II. M < D True (as D > B ≥ W, Y > W > M gives; as D > B ≥ W > M; so M < D).

Hence, the correct answer is Only conclusion II is true.

HPCL Electrical Engineer Mock Test - 3 - Question 5

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 5

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 6

Arrange the words given below in a meaningful sequence:

1. Milky-way

2. Sun

3. Moon

4. Earth

5. Stars

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 6

The words given are different planetary bodies arranged in increasing order of their size.

3. Moon (Smallest among all).

4. Earth (Bigger than moon).

2. Sun (Bigger than Earth).

5. Stars (Bigger than Sun).

1. Milky-way (Biggest among all).

Hence, The correct answer is "Option 1".

HPCL Electrical Engineer Mock Test - 3 - Question 7

Fill in the blank with most appropriate option.

Tigers won't attack _______ they are hungry.

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

Key Points

  • The phrase "unless they are hungry" correctly completes the sentence by providing a condition that must not be met for the action (attack) to occur.
  • The word 'unless' introduces a conditional clause, indicating that the action mentioned (tigers attacking) will not happen unless the condition (being hungry) is fulfilled.
  • The use of 'unless' implies a specific condition that contradicts the main clause, making it the most appropriate choice for the blank.
  • Other options like 'because', 'if', and 'although' do not correctly convey the intended meaning of the sentence, which is about a condition preventing an action unless it is met.

Hence, the correct answer is option 3.

Correct sentence: Tigers won't attack unless they are hungry.

Additional Information

  • Because is used to give a reason for something, which does not fit the conditional nature of the sentence.
  • If introduces a condition but lacks the exclusive context that 'unless' provides, suggesting that the action might happen under other conditions as well.
  • Although introduces a contrast or exception, which does not align with the conditional requirement of the sentence.
HPCL Electrical Engineer Mock Test - 3 - Question 8

Direction: In the following question assuming the given statements to be true, find which of the conclusion among given some conclusion is/are definitely true and then give your answers accordingly:

Statements:

M ≥ Z ≥ B > R < T ≤ V

Conclusions:

I. M > B

II. R < V

III. M = B

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

Given statements:

M ≥ Z ≥ B > R < T ≤ V

Conclusions:

I. M > B → False (no definite relation is given between M and V)

II. R < V → True (no definite relation is given between R and V)

III. M = B → False (it is possible but not definite)

M ≥ B is True,

Hence, Only II follow and Either I or III follow is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 9

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 9

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 10

Gd is the only sister of Md. Md is married to Nd. Nd is mother of Bd. Gd is married to Kd. Jd is son of Kd. How is Jd related to Bd?

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

Preparing the family tree using the following symbols:

Possible tree diagram will be

Hence, Jd is Cousin of Bd.

HPCL Electrical Engineer Mock Test - 3 - Question 11

The value of 60 ÷ 16 × 2 - [76 ÷ 5 of 19 - 15 ÷ 8 × (21 - 25) - 3] ÷ 53 is:

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

We would use the concept of BODMAS-

Calculation:

60 ÷ 16 × 2 - [76 ÷ 5 of 19 - 15 ÷ 8 × (21 - 25) - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [76 ÷ 5 of 19 - 15 ÷ 8 × -4 - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [76 ÷ 95 - 15 ÷ 8 × -4 - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [ -  × -4 - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [ +  - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [ - 3] ÷ 53

⇒ 60 ÷ 16 × 2 - [] ÷ 53

⇒ 60 ÷ 16 × 2 - [] ÷ 53

⇒  × 2 - 

⇒  - 

⇒  = 

∴ The answer is .

HPCL Electrical Engineer Mock Test - 3 - Question 12
The breadth of a rectangular hall is 4/5 of its length. If the area of the floor is 500 m2, then what is the difference (in metres) between the length and breadth of the hall?
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 12

Given,

Area of the floor of rectangular hall = 500 m2

Formula:

Area of the rectangle = length × breadth

Calculation:

Let length of the rectangular hall be 5x

Breadth of the rectangular hall = 5x × (4/5) = 4x

According to the question

5x × 4x = 500

⇒ 20x2 = 500

⇒ x2 = 500/20

⇒ x2 = 25

⇒ x = 5

Difference between length and breadth = 5x – 4x = x = 5
HPCL Electrical Engineer Mock Test - 3 - Question 13
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 13

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 14
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 14

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 15
A and B can do a piece of work in 15 days, while B and C can do the same work in 12 days, and C and A in 10 days. They all work together for 5 days, and then B and C leave. How many days more will A take to finish the work?
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 15

Given:

A and B can do a piece of work in 15 days, while B and C can do the same work in 12 days, and C and A in 10 days.

They all work together for 5 days, and then B and C leave.

Concept used:

Total Work = Work done each day × Total time taken (in days)

Calculation:

LCM (15, 12, 10) = 60

Let the total work be the LCM of the time taken by them.

Hence, total work = 60 units

Thus,

A & B together do each day = 60 ÷ 15 = 4 units

B & C together do each day = 60 ÷ 12 = 5 units

C & A together do each day = 60 ÷ 10 = 6 units

Hence,

A + B = 4 ....(1)

B + C = 5 ....(2)

C + A = 6 ....(3)

Now, 2(A + B + C) = 15 ....(4)

Solving these equations we get,

A = 2.5; B = 1.5; C = 3.5.

Now, in 5 days work done = 5 × (A + B + C) = 5 × 7.5 = 37.5 units

Remaining work = 60 - 37.5 = 22.5 work

Time taken by A to complete the work = 22.5 ÷ 2.5 = 9 days

∴ Time taken by A to complete the work is 9 days.

HPCL Electrical Engineer Mock Test - 3 - Question 16

In a row of 26 persons facing North, Person A is 13th from the right side. After shifting him third to the left from the existing position, what is his position from the left side now?

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 16

Inital position of A is 13th from the right side but after shifting him third to the left A is at 16th position from the right.

Position from left = total person - Position from the right + 1

Position from left = 26 -16 + 1 = 11th

Hence, Option (3) is correct.

HPCL Electrical Engineer Mock Test - 3 - Question 17

Six friends are sitting in a bench facing east. Falak is third to the right of Charu. There are only two people sitting between Sargun and Karan. Charu is at the North end of the bench. Arav is sitting to the immediate left of Falak. Vaibhav and Karan are sitting immediately next to each other. Who is sitting second from the south end?

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 17

Given:

Six Friends: Falak, Charu, Sargun, Karan, Arav and Vaibhav

1.Falak is third to the right of Charu.

2. Charu is at the North end of the bench.

3.Arav is sitting to the immediate left of Falak.

4.Vaibhav and Karan are sitting immediately next to each other.

5.There are only two people sitting between Sargun and Karan.

so case 1.1, 2.1 and 2.2 gets cancelled.

Hence. Karan is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 18

Direction: In the following question assuming the given statements to be true, find which of the conclusion among given some conclusion is/are definitely true and then give your answers accordingly:

Statements:

M ≥ Z ≥ B > R < T ≤ V

Conclusions:

I. M > R

II R < V

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

Given statement:

M ≥ Z ≥ B > R < T ≤ V

Conclusions:

I. M > R → True (M ≥ Z ≥ B > R)

II. R < V → True (R < T ≤ V)

Hence, Both I and II follow is the correct answer.

HPCL Electrical Engineer Mock Test - 3 - Question 19

A voltmeter connected across 15 kΩ resistor reads 10 V in the circuit given. Voltmeter is rated at 500 Ω / volt and has a full- scale reading of 20 V. The supply voltage is:

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

Concept:

The voltmeter sensitivity (Sv) is determined by dividing the sum of the resistance of the meter (Rm)

Mathematically, sensitivity is expressed as:

E = Rated Voltage or Full scale Reading

Voltmeter sensitivity is expressed in Ohm/Volt

Rm = Sv × E

Calculation:

Sensitivity of voltmeter = 500 Ω/V

Internal resistance of voltmeter = 500 × 20 = 10 kΩ

Effective resistance across A – B is

= 10 || 15 = 6 kΩ

The voltmeter reading is 10 V

⇒ VAB = 10 V

By voltage division,

Vs = 60 V

HPCL Electrical Engineer Mock Test - 3 - Question 20

A Charge of 2 Coulombs is placed near a ground conducting plate at a distance of 1 m. The force acting between the charge of 2 C and ground conducting plate in Newton is

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 20

Concept:

Induction and Method of Images:

  • When a charge Q is brought near a conductor surface, the radially outward field has tangential components at the conductor. This displaces the free electrons that are accumulated as an induced surface charge.
  • These displaced electrons create a counterforce and hence the final tangential component of the Electric field is zero.
  • The resultant field has only normal components and appears to be a dipole’s field with an image charge below the conductor surface at the same distance as shown below:

Application:

The force between two point charges is given by:

Here Q+ = Q- = 2 C and d = 1 m

∴ r = 2d = 2 m

Note: The direction will be towards the plane.

HPCL Electrical Engineer Mock Test - 3 - Question 21

For the two-port network shown in the figure, the transmission parameter C is:

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

Concept:

Transmission or ABCD parameters

General two-port network and Transmission parameters two-port network is shown below.

Transmission 2 port network

The direction of I2 is opposite to the standard form so it is taken as negative.

Transmission parameters are defined as:

V1 and I1 are dependent and V2 and I2 are independent.

Calculation:

From the ABCD parameters matrix, we get the following equations.

V1 = A V2 – B I2 ---(1)

I1 = C V2 - D I2 ---(2)

To calculate C, we open circuit the output terminal as shown:

Using Equation (1), we get:

I1 = C V2

Since I2 = 0, the current across Zb will be I1, i.e.

V2 = I1 × Zb

HPCL Electrical Engineer Mock Test - 3 - Question 22

For following Fig. if C(s) is Laplace Transform of output and R(s) is Laplace transform of input, the equivalent transfer function T(s) will be

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

Concept:

According to Mason’s gain formula, the transfer function is given by

Where, n = No of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Given signal flow graph is

Forward paths

M1 = s × s × 1/s = s

M2 = 1/s × 1/s = 1/s2

Loops:

L1 = s × s (-1) = - s2

L2 = -1/s

L3 = s × s × 1/s × (-s) = -s2

L4= 1/s × 1/s × (-s) = - 1/s

Δ = 1 - ( - s2 - s2 - 1/s - 1/s) = 1 + 2s2 + 2/s

HPCL Electrical Engineer Mock Test - 3 - Question 23
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 23

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 24
The relationship between impedance (Z) and admittance (Y) is
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 24

The relationship between the impedance and admittance is given by:

where Z = Impedance

Y = Admittance

The impedance is analogous to admittance in the following ways:

HPCL Electrical Engineer Mock Test - 3 - Question 25
Which method of determination of voltage regulation gives more accurate results because it is based on the separation of armature leakage reactance effect?
Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 25

ZPF or Potier triangle method:

  • This method depends on the separation of the leakage reactance of armature and its effects.
  • It is used to obtain the leakage reactance and field current equivalent of armature reaction.
  • It is also called a general method of voltage regulation.
  • It is the most accurate method of voltage regulation.
  • For calculating the regulation, it requires open circuit characteristics and zero power factor characteristics.

The following assumptions are made in this method.

  • The armature resistance is neglected.
  • The O.C.C taken on no-load accurately represents the relation between MMF and voltage on load.
  • The leakage reactance voltage is independent of excitation.
  • The armature reaction MMF is constant.
HPCL Electrical Engineer Mock Test - 3 - Question 26

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 26

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 27

A Two-Port Network is said to be symmetrical when the following equalities hold good

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 27

A two-port network is said to be symmetrical if the input and output ports can be interchanged without altering the port voltages and currents.

A network is said to be reciprocal if the ratio of the response to the excitation is invariant to an interchange of the positions of the excitation and response of the network.

Conditions of reciprocity and symmetry in terms of different two-port parameters are:

HPCL Electrical Engineer Mock Test - 3 - Question 28

A uniform volume charge density ρv C/m3 is distributed inside the region defined by a cylindrical surface of cross-sectional radius a. The electric field intensity at a distance r (< a) from the cylindrical axis is proportional to:

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 28

Concept:

Gauss Law states that the net electric flux through a closed surface enclosing a volume equals the charge enclosed by it.

Since D = ϵ0 E, we can write:

is the electric flux density.

Analysis:

We construct a Gaussian surface at ρ = r as shown in the figure:

So, according to Gauss law, the total outward flux through the surface ρ = r will be equal to the charge enclosed by it.

Assuming the height of the cylinder to be ‘h’, we can write:

D(2πrh) = ρv (πr2h)

Therefore the electric field intensity at a distance r from the cylindrical axis will be:

Thus E ∝ r

HPCL Electrical Engineer Mock Test - 3 - Question 29

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

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 29

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 30

The y-parameters for the network shown in the figure can be represented by

Detailed Solution for HPCL Electrical Engineer Mock Test - 3 - Question 30

Y parameter:

I1 = V1 Y11 + V2 Y12

I2 = V1 Y21 + V2 Y22

 ...1)

 ...2)

Calculation:

For the given question Z = 5 Ω

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