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HPCL Electrical Engineer Mock Test - 2 - PSSSB Clerk MCQ


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30 Questions MCQ Test - HPCL Electrical Engineer Mock Test - 2

HPCL Electrical Engineer Mock Test - 2 for PSSSB Clerk 2025 is part of PSSSB Clerk preparation. The HPCL Electrical Engineer Mock Test - 2 questions and answers have been prepared according to the PSSSB Clerk exam syllabus.The HPCL Electrical Engineer Mock Test - 2 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 - 2 below.
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HPCL Electrical Engineer Mock Test - 2 - Question 1

The sum of all the three digit numbers that can be formed using the digits a, b and c without repetition is a multiple of:

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 1

Calculation:

Let's assume a = 1, b = 2, c = 3, then all possible 3-digit numbers formed by a, b, c without repetition are,

123, 132, 213, 231, 312, and 321, so the sum of these numbers = 1332

Here, 1332 is a multiple of 37

∴ The correct answer is 37

HPCL Electrical Engineer Mock Test - 2 - Question 2

Find the single discount equivalent to three successive discounts of 40%, 20% and 10%.

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

Given:

Three successive discounts of 40%, 20% and 10%.

Formula used;

SP = MP × × ×

Calculation:

Let the MP be Rs 100.

SP = MP × × ×

⇒ 100 × × ×

100 × × × = Rs. 43.2

So, single equivalent discount

⇒ 100 - 43.2 = 56.8

Discount % = × 100% = 56.8%

HPCL Electrical Engineer Mock Test - 2 - Question 3

The profit percent earned by selling a centre table for Rs.13440 is equal to the loss percent made by selling the same centre table for Rs.10560. At what price should the article be sold to make 20% profit?

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

Given:

Selling prices = Rs. 13440 and Rs. 10560

Concept:

The cost price is midway between the price sold at a loss and the price sold at a profit.

Solution:

⇒ Cost Price (C.P.) = (13440 + 10560)/2 = Rs. 12000

⇒ Selling Price for 20% profit = 120% of C.P. = 1.2 × 12000 = Rs. 14400

Therefore, the centre table should be sold for Rs. 14400 to make a 20% profit.

HPCL Electrical Engineer Mock Test - 2 - Question 4

If + means, -, × means /, - means + and / means ×, then find out the answer of the following question :

(48 × 6)/4

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 4

Decoding the information,

Given:

(48 × 6)/4

After replacing the signs from left to right and using the BODMAS rule,

= (48 / 6) × 4

= 8 × 4

= 32

Hence, “32” is the correct answer.

HPCL Electrical Engineer Mock Test - 2 - Question 5

R and S are a married couple. A and V are brothers. S is the only sister of V. V is married to U and X is the only son of V. How is S related to X?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 5

To understand the relationship between S and X based on the given information, let's break down the relationships step by step:

R and S are married, making them husband and wife.

A and V are brothers, indicating they share the same parents.

S is the only sister of V, meaning V is S's brother, and S is also A's sister (since A and V are brothers).

V is married to U, making them a married couple.

X is the only son of V, which means X is V's and U's son.

Since S is V's sister and V is X's father, that makes S the Paternal aunt of X.

HPCL Electrical Engineer Mock Test - 2 - Question 6

Fill in the blank with the appropriate form of the Verb that agrees with the Subject.

Each one of the workers _______ a raise in salary from the coming month.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 6

The correct answer is Option (2) i.e. "wants".

Key Points

  • Here "Each one" is the main subject which is singular.
  • "Each one of the workers" implies each worker individually which requires a singular verb.
  • "Wants" correctly agrees with the singular subject phrase.

Therefore, the correct answer is- "Each one of the workers wants a raise in salary from the coming month."

Additional Information

  • Option (1) "are wanting" does not agree with the singular subject phrase.
  • Option (3) "have wanted" does not correctly fit the singular subject in present tense.
  • Option (4) "want" is plural and does not agree with the singular subject phrase.
HPCL Electrical Engineer Mock Test - 2 - Question 7
Select the correctly spelt word.
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 7

Here the correct answer is Imprudence.

Key Points

  • The correctly spelt word is "Imprudence".
  • Imprudence means the quality of being unwise or indiscreet.
    • Example: His imprudence led to the loss of the entire investment.
  • Thus, the correct answer is Option 2.
HPCL Electrical Engineer Mock Test - 2 - Question 8

Select the most appropriate option that can substitute the underlined words in the given sentence.

The new restaurant is more better than the old one.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 8
The correct answer is: Better.

Key Points
  • The use of "more" before "better" is incorrect as "better" is already a comparative form of the adjective "good".
  • The correct comparative form of "good" is "better", and the superlative form is "the best".
  • Therefore, the correct sentence should be "The new restaurant is better than the old one."

Hence, option 2 is the correct answer.

Important Points

  • The comparative and superlative forms of adjectives are used to compare and describe different degrees of a particular quality.
  • Comparative Form:

    • Usage: The comparative form is used when you are comparing two things or two groups.

    • Formation: Most adjectives form the comparative by adding "-er" (e.g., faster, taller) or by using "more" before the adjective (e.g., more beautiful, more interesting).

    • Example: The cat is faster than the dog.

  • Superlative Form:

    • Usage: The superlative form is used when you are comparing more than two things or groups, indicating the highest degree of quality.

    • Formation: Most adjectives form the superlative by adding "-est" (e.g., fastest, tallest) or by using "most" before the adjective (e.g., most beautiful, most interesting).

    • Example: The cheetah is the fastest land animal.

HPCL Electrical Engineer Mock Test - 2 - Question 9

Choose the correct phrase from the given option and make a conditional sentence.

If she _________ her homework, she would have gotten a better grade.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 9

The correct answer is: had done

Key Points

  • The third conditional is used when the condition is impossible.
  • The condition is "if she had done her homework".
  • The result is "she would have gotten a better grade".
  • The verb in the if-clause is in the past perfect tense.
  • The verb in the main clause is in the conditional perfect tense. Therefore, the correct answer is- had done

Additional Information

  • Did is incorrect because it is in the past tense, not the past perfect tense.
  • Would do is incorrect because it is in the conditional tense, but the condition is not possible.
  • Does is incorrect because it is in the present tense, not the past perfect tense.
HPCL Electrical Engineer Mock Test - 2 - Question 10

In a certain code language, 'no more food' is written as 'ta ka da' and 'more than that' is written as 'sa pa ka'. How is 'that' written in that code language?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 10

According to the given information,

‘That’ is written as either ‘sa or pa’.

Hence. “sa or pa” is the correct answer.

HPCL Electrical Engineer Mock Test - 2 - Question 11
How many 200 mm lengths can be cut from 10 m of ribbon?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 11

Given:

200 mm lengths can be cut from 10 m of ribbon

Concept:

1m = 1000 mm

10 m = 10000 mm

Calculation;

We have,

Number of 200 mm lengths that can be cut from 10 m of ribbon

⇒ 50

∴ 50 lengths can be cut from 10 m of ribbons.

HPCL Electrical Engineer Mock Test - 2 - Question 12

Study the given pie chart and answer the following question.

Total income for 4 years is Rs. 75,00,000.

What is the total income from year 2 to year 3?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 12

% of income in year 2 = 26

% of income in year 3 = 30

So, 56% in these two years

7500000 × 56/100

⇒ 4200000

∴ The total income from year 2 to year 3 is Rs. 4200000

HPCL Electrical Engineer Mock Test - 2 - Question 13
If the 5-digit number 9x34y is divisible by 24, then what is the maximum value of (x + y)?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 13

Given:

Five digit number is given = 9x34y, which is divisible by 24.

Concept used:

Divisibility rule of 8 and 3 which is factor of 24.

Divisibility rule of 8 is last three digit of number is divisible by 8.

Divisibility rule of 3 is sum of all digits of given number is divisible by 3.

Calculation:

The number is = 9x34y

First we check divisibility rule of 8,

According to this rule, last three digits is divisible by 8,

⇒ 34y is divisible by 8,

34y / 8

When we divide 34y by 8 ,for complete divisibisible by 8 y should be 4 344 / 8 (Let y = 4)

Check the divisibility rule of 3,

According to this rule, sum of all digits is divisible by 3.

9 + x + 3 + 4 + y = 16 + x + y is divisible by 3.

Put y = 4,

16 + x + 4 = 20 + x

Which is completely divisible by 3 when x = 1 and 7

We take maximum value because question ask maximum value, x = 7

20 + 7 = 27 is divisible by 3.

X = 7 and y = 4

x + y = 7 + 4 = 11

∴Option 1 is correct.

HPCL Electrical Engineer Mock Test - 2 - Question 14

In this question, three statements are given, followed by two conclusions numbered I and II. Assuming the statements to be true, even if they seem to be at variance with commonly known facts, decide which of the conclusion(s) logically follows/follow from the statements.

Statements:

All cricketers are wealthy.

Some wealthy are Indians.

All Indians are honest.

Conclusions:

I. All Indians are cricketers.

II. Some cricketers are honest.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 14

Statements:

All cricketers are wealthy.

Some wealthy are Indians.

All Indians are honest.

The Venn diagram as per the given statements:

Conclusions:

I. All Indians are cricketers. → Does not follow (Because there is no definite relation given between Indians and cricketers, so it can be possible only but not definitely true)

II. Some cricketers are honest. → Does not follow (Because there is no definite relation given between cricketers and honest, so it can be possible only but not definitely true)

Hence, the correct answer is "Neither conclusions I nor II follow".

HPCL Electrical Engineer Mock Test - 2 - Question 15

Study the given information carefully and answer the question that follows.

Five People A, B, C, D and E are sitting in a row facing south but not necessarily in the same order. A and B are sitting together. C is sitting immediate left of A. D sits second to the right of A. C sits in the extreme left end of the row.

Which of the following sequence is true?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 15

Given:

Five People A, B, C, D and E are sitting in a row facing south.

1) A and B are sitting together. C is sitting immediate left of A.

2) C sits in the extreme left end of the row.

3) D sits second to the right of A.

4) Remaining Person will be placed in left space

​​​​​​​​​​​​​​

Hence, "Option 3" is the correct answer.

HPCL Electrical Engineer Mock Test - 2 - Question 16

Read the given statements and conclusions carefully. Decide which of the given conclusions logically follows from the statements.

Statement: “It is difficult to form true selfless friendship with women after their marriage because they always give priority to their family over their friends.” – A woman complains to her husband.

Conclusion:

I. Families should be given priority by the elders to maintain good relations within the family.

II. Friendship with a married person is not possible.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 16

Given:

Statements: “It is difficult to form true selfless friendship with women after their marriage because they always give priority to their family over their friends.” – A woman complains to her husband.

Conclusion:

I. Families should be given priority by the elders to maintain good relations within the family → False (Because it is given in the statement that women always give priority after their marriage to their family over their friends and there is no such information mentioned regarding elders.)

II. Friendship with a married person is not possible → False (Because no such information is given regarding married person.)

Hence, the correct answer is "Option 4".

HPCL Electrical Engineer Mock Test - 2 - Question 17
The output of a 12 V SMPS is monitored in an oscilloscope with AC coupling enabled and with the appropriate setting of time base and amplitude. What will be the waveform observed in the oscilloscope?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 17

Concept:

  • The electronic power supply integrated with the switching regulator for converting the electrical power efficiently from one form to another form with desired characteristics is called a Switch-mode power supply.
  • It is used to obtain a regulated DC output voltage from unregulated AC or DC input voltage.
  • Ideally, the output voltage is expected to be pure DC, but practically some ripples of the alternating input are always present at the output.

Two common coupling modes of an oscilloscope are:

DC Coupling: DC Coupling shows the whole input signal (DC + AC).

AC Coupling: AC Coupling blocks the DC component of a signal.

Application:

Given input to the oscilloscope is 12 V SMPS supply.

Since the oscilloscope is AC coupled, it will block the DC part and will show ripples alone since some ripples of the alternating input are always present at the output of SMPS.

Note: If the oscilloscope was DC coupled, the output would be 12 V DC along with ripple.

HPCL Electrical Engineer Mock Test - 2 - Question 18

The following figure shows zero sequence equivalent circuits of

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 18

Concept:

Zero sequence networks for different 3-phase loads are shown below.

The zero sequence equivalent circuits of 3-phase transformers can be drawn by using the following general circuit.

Z is the zero sequence impedance of the windings of the transformer. These are two series and two shunt switches. One series and one shunt switches are for both sides separately.

The series switch of a particular side is closed if it is star grounded and the shunt switch is closed if that side is delta connected, otherwise they are left open.

Application:

In the given circuit diagram, both the shunt switches are closed. Therefore, both the primary and secondary are delta-connected.

HPCL Electrical Engineer Mock Test - 2 - Question 19
A single-phase energy meter is operating on 200 V, 50 Hz supply with a load of 10 A for two hours at 0.8 p.f. The meter takes 1800 revolutions in that period. The meter constant is:
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 19

Energy meter:

An energy meter or watt-hour meter is used to measure the energy consumed in kWh.

It is an integrating type instrument.

Recording mechanism:

It is required for the energy meter to record the no. of revolutions made by the aluminum disc which is proportional to the energy consumed in kWh.

The meter constant can be calculated as

k = (No. of revolutions / Energy consumption in kWh)

Units of meter constant are revolutions per kWh

1 unit means 1 kWh.

Calculation:

Given:

Voltage V= 200 V

Current I = 10 A

Time = 2 hour

cosϕ = 0.8

Energy consumed = V × I × cosϕ × time = 200 × 10 × 0.8 × 2 = 3.2 kWh

Meter constant = No. of revolution by meter/Energy consumed = 1800/3.2 = 562.5 rev/kWh.

Points to remember:

Braking torque obtained by a permanent magnet inside the meter used to control the speed of the disc.

HPCL Electrical Engineer Mock Test - 2 - Question 20

With reference to TRIAC which of the following statement is WRONG?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 20
  • A TRIAC is defined as a three terminal, four layer, bi-directional semiconductor device that controls AC power, which is different from the other silicon controlled rectifiers in the sense that TRIAC can conduct in both the directions that is whether the applied gate signal is positive or negative, it will conduct. Thus, this device can be used for AC systems as a switch.
  • TRIAC gives a two-quadrant operation, but Triacs can be triggered in any of the four quadrants, and two of the four possible quadrants are needed to trigger conduction during the two (positive and negative) half cycles of the AC wave.
  • Quadrants I, and III or quadrants II, and III are the favored methods of triggering, as quadrant IV is much less sensitive to triggering because of the way the DIAC is constructed.
  • So if quadrant IV is used with any of the other three quadrants, the positive and negative half cycles would need different values of trigger current, creating an unnecessary complication.

Characteristics of a TRIAC: The Triac characteristics is similar to SCR but it is applicable to both positive and negative Triac voltages. The operation can be summarized as follows:

​​​​​​​

  • First Quadrant Operation of Triac: The voltage at terminal MT2 is positive with respect to terminal MT1 and the gate voltage is also positive with respect to the first terminal.
  • Second Quadrant Operation of Triac
  • The voltage at terminal 2 is positive with respect to terminal 1 and the gate voltage is negative with respect to terminal 1.
  • Third Quadrant Operation of Triac: The voltage of terminal 1 is positive with respect to terminal 2 and the gate voltage is negative.
  • Fourth Quadrant Operation of Triac: The voltage of terminal 2 is negative with respect to terminal 1 and the gate voltage is positive.
HPCL Electrical Engineer Mock Test - 2 - Question 21
An analog ammeter has a linear scale of 50 divisions. Its full-scale reading is 10 A and half a scale division can be read. What is the resolution of the instrument?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 21

Concept:

where, SDR = Scale division reading

FSR = Full scale reading

D = No. of divisions on scale

Resolution =

Calculation:

Given, FSR = 10 A

D = 50 divisions

Resolution =

Resolution = 0.1 A

HPCL Electrical Engineer Mock Test - 2 - Question 22

The given circuit is:

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 22

Single-phase controlled rectifier with SCR:

  • The bridge rectifier circuit is made of four SCRs T1, T2, T3, T4, and a load.
  • The input signed is applied across terminals A and B and the output DC signal is obtained across the load.
  • The four SCRs are arranged in such a way that only two SCRs conduct electricity during each half cycle.
  • SCRs T1 and T2 are pairs that conduct electric current during the positive half-cycle.
  • Similarly, SCRs T3 and T4 conduct electric current during a negative half-cycle only if T3 would be normal, but to some issue, it damaged and gets open-circuited.
  • Now due to the open-circuit, there will be no conduction in the negative half.
  • Due to the inductive nature of the load, a negative voltage spike will come.
  • As the free-wheeling diode is connected, no negative voltage spike will appear in the final waveform.
  • Finally, this circuit will only conduct in the positive half cycle.
  • The formula of average output voltage (Vavg or e0avg) for half-wave rectifier is

Where Vm = Maximum Voltage

α = Firing angle of the thyristor / SCR

HPCL Electrical Engineer Mock Test - 2 - Question 23

In a balanced Y – Y circuit, the Van = Vp∠0, Vbn = Vp∠-120, Vcn = Vp∠+120. The line voltages are

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 23

Concept:

Drawing the balanced 3-phase star connected circuit

Where, Van = phase voltage and Vab = line voltage.

Now drawing the Phasor Diagram to find the line voltage:

From the above phasor diagram,

Line voltage Vab leads the phase voltage by 30 degrees.

So, we can establish the relation between line and phase voltage of a star-connected balanced system.

Calculation:

In ABC sequence:

Phase B lags Phase A by 120o and phase C leads phase A by ,120o.

, ,

HPCL Electrical Engineer Mock Test - 2 - Question 24

The resonant frequency of the series RLC circuit shown below is:

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 24

The equivalent inductance of series opposing inductors:

In the RLC network given in the question, we can observe that inductors are connected in series opposing connection,

If two inductors are connected in series opposing connection with self inductances L1 and L2 and mutual inductance between them is M.

Then the equivalent inductance of such connection is given by

Leq = L1 + L2 - 2M

Resonance in series RLC circuit:

Seris resonance is the state of the network at which the impedance value of the network reaches a minimum and hence accepts the maximum current.

The frequency at which this state happens is called the resonance frequency.

The resonance frequency of the series RLC circuit is given by

rad/sec.

Where

L is the inductance of the coil, C is the capacitance of the capacitor

Calculation:

Given that,

C = 4 F

L1 = L2 = 4 H

M = 2 H

⇒ Leq = 4 + 4 - 2 × 2 = 4 H

And resonance frequency

= 1 / √(4×4) = 1/4 = 0.25 rad/sec

HPCL Electrical Engineer Mock Test - 2 - Question 25
The output of a 12 V SMPS is monitored in an oscilloscope with AC coupling enabled and with the appropriate setting of time base and amplitude. What will be the waveform observed in the oscilloscope?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 25

Concept:

  • The electronic power supply integrated with the switching regulator for converting the electrical power efficiently from one form to another form with desired characteristics is called a Switch-mode power supply.
  • It is used to obtain a regulated DC output voltage from unregulated AC or DC input voltage.
  • Ideally, the output voltage is expected to be pure DC, but practically some ripples of the alternating input are always present at the output.

Two common coupling modes of an oscilloscope are:

DC Coupling: DC Coupling shows the whole input signal (DC + AC).

AC Coupling: AC Coupling blocks the DC component of a signal.

Application:

Given input to the oscilloscope is 12 V SMPS supply.

Since the oscilloscope is AC coupled, it will block the DC part and will show ripples alone since some ripples of the alternating input are always present at the output of SMPS.

Note: If the oscilloscope was DC coupled, the output would be 12 V DC along with ripple.

HPCL Electrical Engineer Mock Test - 2 - Question 26

If the current through a moving iron instrument is increased by 20%, what is the percentage increase in the deflection torque?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 26

Concept:

In moving iron instruments, the deflecting torque is unidirectional (acts in the same direction) whatever may be the polarity of the current.

The deflecting torque is given by

Where,

I is operating current

L is self-inductance

θ is deflection

So that the deflection torque of the moving iron instrument is proportional to the square of the RMS value of the operating current and change in self-inductance.

Calculation:

When the current through a moving iron instrument is increased by 20%, then deflection torque is

∴ T'd = 1.44 Td

Hence deflection torque is increased by 44%

HPCL Electrical Engineer Mock Test - 2 - Question 27
Total harmonic distortion is defined as:
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 27

Total harmonic distortion or THD is a common measurement of the level of harmonic distortion present in power systems.

THD can be related to either current harmonics or voltage harmonics.

It is defined as the ratio of rms value of all the harmonic components to the rms value of the fundamental component.

Mathematically, it can be represented as

HPCL Electrical Engineer Mock Test - 2 - Question 28
A circular loop has its radius increasing at a rate of 1 m/s. The loop placed perpendicular to a constant magnetic field of 0.8 wb/m2. When the radius of the loop is 2 m, the emf induced in the will be
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 28

Given that,

Rate of radius (dr/dt) = 1 m/s

Magnetic field (B) = 0.8 wb/m2

Radius of the loop (r) = 2 m

Emf induced

= 2π × 2 × 0.8 × 1 = 3.2π V

HPCL Electrical Engineer Mock Test - 2 - Question 29
A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 29

Concept:

Voltage regulation:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.

E2 = no-load secondary voltage

V2 = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation

It can also be expressed as,

Regulation

+ sign is used for lagging loads and

- sign is used for leading loads

Explanation:

% voltage regulation

Given that, no load voltage = 100 V

full load voltage = 95 V

HPCL Electrical Engineer Mock Test - 2 - Question 30

In the circuit shown in the figure, the switch is on position 1 long enough to establish steady-state and switched to position 2 at t = 0. The current I for t ≥ 0 is

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 30

In the given circuit, the switch is on position 1 till the steady-state is reached, in the steady-state the inductor acts like a short circuit,

The circuit at t = 0-

I(0-) = IL(0-) = 50 / 10 = 5 A

As inductor does not allow a sudden change in the current so IL (0+) = 5 A

Circuit at t = 0+

Now the circuit is a simple RL source free circuit, Inductor will start discharging and when it reaches the steady stage it will completely discharge i.e the energy stored by the inductor is zero.

So at the steady-state, the current passing through the inductor is in source free RL circuit is zero

IL (0+) = 5 A = initial current.

IL (∞) = 0 A = final value

The expression for IL(t) for t ≥ 0 of source free RL circuit is given by

IL(t) = IL(∞) + [IL(0+) - IL(∞)] e-t/τ

Where

τ = time constant = L / R

τ = 0.2 / 10 = 0.02

I(t) = IL(t) = 0 + ( 5 - 0) e-t / 0.02

= 5 e-50t for t ≥ 0

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