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Interview MCQ Set 1

1. Categories for a transformer earthing of a system neutral are ___
a) 2
b) 3
c) 4
d) Many

View Answer

Answer: b [Reason:] Fortunately for transformer designers, earthing of a system neutral can only fall into one of three categories. These are: Neutral solidly earthed, Neutral earthed via an impedance, Neutral isolated.

2. Which of the following neutral earthing method is disadvantageous?
a) Neutral solidly earthed
b) Neutral earthed via an impedance
c) Neutral isolated
d) Neutrally

View Answer

Answer: c [Reason:] Due to the problems and disadvantages of the third alternative that is neutral isolated, it is unlikely that it will be encountered in practice so that it is only necessary to be able to design for the first two.

3. Minimum voltage above which electrical system needs to be earthed is _____
a) 50 V
b) 5 V
c) 10 V
d) 20 V

View Answer

Answer: d [Reason:] According to regulations and acts made in earlier years Every electrical system rated at greater than 50 V shall be connected to earth. Here minimum voltage is defined at the level of 50 V. Various other rules are defined according to the voltage ranges.

4. What is the low voltage range according to the earthing procedures?
a) 5 V – 100 V
b) 0 V – 100 V
c) 50 V – 1000 V
d) 0 V – 1000 V

View Answer

Answer: c [Reason:] Low voltage is defined as exceeding 50 V but not exceeding 1000 V and is mainly referring to 415 V distribution networks. For low voltage systems the regulations say that ‘no impedance shall be inserted in any connection with earth.

5. Low voltage systems should be solidly earthed.
a) True
b) False

View Answer

Answer: a [Reason:] Low-voltage systems must be solidly earthed. The system of protective multiple earthing, which can be advantageous on 415 V distribution networks in some situations, is permitted on low-voltage systems subject to certain other conditions but this still requires that the neutral should be solidly earthed ‘at or as near as is reasonably practicable to the source of voltage.

6. Earthed neutral allows rapid operation of protection to ___________
a) Line to line faults
b) Earthed faults
c) Other faults
d) For all faults

View Answer

Answer: b [Reason:] An earthed neutral allows rapid operation of protection immediately an earth fault occurs on the system. The earthed neutral in conjunction with sensitive earth fault protection results in the faulty section being isolated at an early stage of the fault.

7. Which method will reduce the cost of insulation between earth and cables?
a) Neutral solidly earthed
b) Neutral earthed via an impedance
c) Neutral isolated
d) Neutrally

View Answer

Answer: a [Reason:] If the neutral is solidly earthed, the voltage of any live conductor cannot exceed the voltage from line to neutral. As under these conditions the neutral point will be at zero potential, it is possible to effect appreciable reductions in the insulation to earth of cables and overhead lines, which produces a corresponding saving in cost.

8. For an un-earthed line voltage of line to earth can be ______________
a) More than line to neutral of earthed system
b) Less than line to neutral of earthed system
c) Equal to line to neutral of earthed system
d) Can’t specify

View Answer

Answer: a [Reason:] On an unearthed system the voltage to earth of any line conductor may have any value up to the breakdown value of the insulation to earth, even though the normal voltage between lines and from line to neutral is maintained.

9. High voltage line is only disadvantageous in ______________
a) Small overhead lines
b) Small underground lines
c) Long overhead lines
d) Long underground lines

View Answer

Answer: c [Reason:] The only disadvantage of connecting a high-voltage system to earth is that this introduces the first earth from the outset and it thus increases the susceptibility to earth faults. This can be inconvenient in the case of a long overhead line, particularly in areas of high lightning incidence.

10. Earth contact material resistance should be ____________
a) Less than 2 ohms
b) More than 2 ohms
c) Depends on the circuit or system
d) Very high

View Answer

Answer: a [Reason:] It is not always appreciated that it is very difficult to obtain resistance values of less than about 2 ohms from a single earth plate, and often it is still more difficult to maintain the value after the earthing system has been installed for some time.

11. If parallel arrangement is done the minimum distance between two earthing conductors should be _______
a) 10 meters
b) 5 meters
c) More than or equal to 10 meters
d) 2 meters

View Answer

Answer: c [Reason:] Where a parallel arrangement is employed, each plate, rod, etc., all such things should be installed outside the resistance area of any other. Strictly, this requires a separation of the order of 10 meters.

Interview MCQ Set 2

1. Calculate the overall voltage gain Gv of a common source amplifier for which gm = 2mA/V, RD = 10 kΩ, R0 = 10 kΩ and RG = 10 MΩ. The amplifier is fed from a signal source of Thevenin resistance of 0.5MΩ and the amplifier is coupled with a load of 10 kΩ.
a) -11.2 V/V
b) -22.4 V/V
c) -33.6 V/V
d) -44.8 V/V

View Answer

Answer: a [Reason:] advanced-electronic-devices-circuits-questions-answers-q1

(Q.2-Q.6) The MOSFEt circuit below has Vt = 1V, knW/L = 0.8 mA/V2 and VA = 40V
advanced-electronic-devices-circuits-questions-answers-q2

2. Find the value of RG so that iD = 0.1 mA, the largest possible value of RD is used while the maximum signal swing at the drain is of 1V and the input resistance at the gate is 10 MΩ.
a) 1 MΩ
b) 10 MΩ
c) 0.1 MΩ
d) 0.01 MΩ

View Answer

Answer: b [Reason:] advanced-electronic-devices-circuits-questions-answers-q2a

3. Find the value of gm at the bias point
a) 0.1 mA/V
b) 0.2 mA/V
c) .0.3 mA/V
d) 0.4 mA/V

View Answer

Answer: d [Reason:] advanced-electronic-devices-circuits-questions-answers-q3

4. If terminal Z is grounded, X is connected to a signal source having a resistance of 1 MΩ and terminal Y is connected to a load resistance of 40 kΩ, find the voltage gain from the signal source to the load.
a) 3.25 v/V
b) 6.5 V/V
c) 9.75 V/V
d) 13 V/V

View Answer

Answer: b [Reason:] advanced-electronic-devices-circuits-questions-answers-q4

5. If terminal Y is grounded find the voltage gain from X to Z with Z open-circuit.
a) 0.33 V/V
b) 0.66 V/V
c) 0.99 V/V
d) None of the mentioned

View Answer

Answer: c [Reason:] advanced-electronic-devices-circuits-questions-answers-q5

6. If terminal X is grounded and terminal Z is connected to a current source delivering a current of 10 µA and having a resistance of 100 kΩ, find the voltage signal that can be measured at Y neglecting the effect of V0 .
a) 0.34V
b) 0.68 V
c) 3.4 V
d) 6.8 V

View Answer

Answer: a [Reason:] advanced-electronic-devices-circuits-questions-answers-q6

(Q.7-Q.8) The NMOS transistor in source follower circuit shown has gm = 5mA/V and a large r0 .
advanced-electronic-devices-circuits-questions-answers-q7
7. Find the output resistance.
a) 0.1 kΩ
b) 0.2 kΩ
c) 0.3 kΩ
d) 0.4 kΩ

View Answer

Answer: b [Reason:] advanced-electronic-devices-circuits-questions-answers-q7a

8. Find the open-Circuit voltage gain.
a) 1 V/V
b) 2 V/V
c) 3 V/V
d) 4 V/V

View Answer

Answer: a [Reason:](Q.7-Q.8): advanced-electronic-devices-circuits-questions-answers-q7a

advanced-electronic-devices-circuits-questions-answers-q8

9. The NMOS transistor in the common gate amplifier as shown in the circuit below has gm = 5 mA/V. Find the input resistance and the voltage gain.
advanced-electronic-devices-circuits-questions-answers-q9
a) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V
b) Input resistance: 0.1 kΩ and Voltage gain: 3.05 V/V
c) Input resistance: 0.2 kΩ and Voltage gain: 3.05 V/V
d) Input resistance: 0.2 kΩ and Voltage gain: 7.1 V/V

View Answer

Answer: d [Reason:] advanced-electronic-devices-circuits-questions-answers-q9a

10. If the output of the source follower in (I) is connected to the input of the common gate amplifier of (II). Determine the overall voltage gain (V0 /Vi ).
advanced-electronic-devices-circuits-questions-answers-q10
a) 1.55 V/V
b) 3.55 V/V
c) 5.55 V/V
d) 7.55 V/V

View Answer

Answer: b [Reason:] advanced-electronic-devices-circuits-questions-answers-q10a

Interview MCQ Set 3

1. The maximum voltage across capacitor would be
electronic-devices-circuits-basic-questions-answers-q1
a) 3200 V
b) 3 V
c) 3 V
d) 1600 V

View Answer

Answer: a [Reason:] electronic-devices-circuits-basic-questions-answers-q1a

2. Find the resonnant frequency for the circuit.
electronic-devices-circuits-basic-questions-answers-q2
a) 346 kHz
b) 55 kHz
c) 196 kHz
d) 286 kHz

View Answer

Answer: b [Reason:]electronic-devices-circuits-basic-questions-answers-q2a

3. Determine the resonant frequency of the circuit.
electronic-devices-circuits-basic-questions-answers-q3
a) 12 9 . kHz
b) 12.9 MHz
c) 2.05 MHz
d) 2.05 kHz

View Answer

Answer: c [Reason:]electronic-devices-circuits-basic-questions-answers-q3a

4. The value of C and A for the given network function is
electronic-devices-circuits-basic-questions-answers-q4
a) 10 µF, 6
b) 5 µF, 10
c) 5 µF, 6
d) 10 µF, 10

View Answer

Answer: c [Reason:]electronic-devices-circuits-basic-questions-answers-q4a

5. H(w) = Vo/Vi = ?
electronic-devices-circuits-basic-questions-answers-q5
a) 0.6 / jw(1 + 0.2jw)
b) 0.6 / jw(5 + jw)
c) 3 / jw(1 + jw)
d) 3 / jw(20 + 4jw)

View Answer

Answer: a [Reason:] electronic-devices-circuits-basic-questions-answers-q5a

6. H(w) = Vo/Vi = ?
electronic-devices-circuits-basic-questions-answers-q6
a) 1 / jw(5 + j20w)
b) 1 / jw(5 + j4w)
c) 1 / jw(5 + j30w)
d) 5 / jw(5 + j6w)

View Answer

Answer: a [Reason:] electronic-devices-circuits-basic-questions-answers-q6a

7. The value of input frequency is required to cause a gain equal to 1.5. The value is
electronic-devices-circuits-basic-questions-answers-q7
a) 20 rad/s
b) 20 Hz
c) 10 rad/s
d) No such value exists

View Answer

Answer: d [Reason:] electronic-devices-circuits-basic-questions-answers-q7a

8. In the circuit shown phase shift equal to 45 degrees and is required at frequency w = 20 rad/s. The value of R (in kilo-ohm) is
electronic-devices-circuits-basic-questions-answers-q8
a) 200
b) 150
c) 100
d) 50

View Answer

Answer: d [Reason:] electronic-devices-circuits-basic-questions-answers-q8a

9. For the circuit shown the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is
electronic-devices-circuits-basic-questions-answers-q9
a) 20 rad/s
b) 20 Hz
c) 40 rad/s
d) 40 Hz

View Answer

Answer: a [Reason:]electronic-devices-circuits-basic-questions-answers-q9a

10. The bode diagram for the Vo/Vs for the given circuit is
electronic-devices-circuits-basic-questions-answers-q10
a) electronic-devices-circuits-basic-questions-answers-q10a
b) electronic-devices-circuits-basic-questions-answers-q10b
c) electronic-devices-circuits-basic-questions-answers-q10c
d) electronic-devices-circuits-basic-questions-answers-q10d

View Answer

Answer: d [Reason:] electronic-devices-circuits-basic-questions-answers-q10e

Interview MCQ Set 4

1. The transistor in the circuit shown below has kn = 0.4 mA/V2, Vt = 0.5 V and λ = 0. Operation at the edge of saturation is obtained when
tough-electronic-devices-circuits-questions-answers-q1
a) (W/L)RD = 0.5 kΩ
b) (W/L)RD = 1.0 kΩ
c) (W/L)RD = 1.5 kΩ
d) (W/L)RD = 2.0 kΩ

View Answer

Answer: c [Reason:] Use the standard formula for edge saturation.

2. The PMOS transistor in the circuit shown has Vt = −0.7 V, μpCox = 60 μA/V2, L = 0.8 μm, and λ = 0. Find the value of R in order to establish a drain current of 0.115 mA and a voltage VD of 3.5 V.
tough-electronic-devices-circuits-questions-answers-q2
a) 12.5 KΩ
b) 25 kΩ
c) 37.5 kΩ
d) 50 kΩ

View Answer

Answer: a [Reason:] tough-electronic-devices-circuits-questions-answers-q2a

3. The NMOS transistors in the circuit shown have Vt = 1 V, μnCOX = 120 μA/V2, λ = 0, and L1 = L2 = L3 = 1μm. Then which of the following is not the value of the width of these MOSFETs shown
a) 2 µm
b) 8µm
c) All of the mentioned
d) None of the mentioned

View Answer

Answer: d [Reason:] tough-electronic-devices-circuits-questions-answers-q3

4. The MOSFET shown has Vt = 1V, kn = 100µA/V2 and λ = 0. Find the required values of W/L and of R so that when vI = VDD = +5 V, rDS = 50 Ω, and VO = 50 mV.
tough-electronic-devices-circuits-questions-answers-q4
a) W/L = 25 and R = 4.95 kΩ
b) W/L = 25 and R = 9.90 kΩ
c) W/L = 50 and R = 4.95 kΩ
d) W/L = 50 and R = 9.90 kΩ

View Answer

Answer: c [Reason:] tough-electronic-devices-circuits-questions-answers-q4a

(Q.5-Q.7) For each of the circuits shown find the labeled voltages. For all transistors, kn(W/L) = 1 mA/V2, Vt = 2V, and λ = 0
5. Find V3
tough-electronic-devices-circuits-questions-answers-q5
a) 2.41V
b) 3.41V
c) 4.41V
d) 1.41V

View Answer

Answer: b [Reason:] V3 = 10- 4 * 2 + 1.4 = 3.4v.

6. Find V4 and V5
tough-electronic-devices-circuits-questions-answers-q6
a) 4V and -5V respectively
b) -4V and 5V respectively
c) 4V and 5V respectively
d) -4V and -5V respectively

View Answer

Answer: a [Reason:] tough-electronic-devices-circuits-questions-answers-q6a

7. Find V1 and V2
a) 2V and -4V
b) -2V and 4V
c) 2V and 4V
d) -2V and -4V

View Answer

Answer: c [Reason:] ID = 1 = 12 * 1 * (VGS – 2)2 => VGS = 3.41v. V3 = 3.41v.

(Q.8 & Q.9) For each of the circuits shown find the labeled node voltages. The NMOS transistors have Vt = 1 V and kn( W/L ) = 2 mA/V2 and λ = 0

8. Find V1 and V2
tough-electronic-devices-circuits-questions-answers-q8
a) 2.44 and -1.28 V
b) 2.44 and -2.56 V
c) 1.22 and -2.56 V
d) 1.22 and -1.28 V

View Answer

Answer: b [Reason:] tough-electronic-devices-circuits-questions-answers-q8a

9. Find V3 and V4
tough-electronic-devices-circuits-questions-answers-q9
a) 3.775V and 5V
b) 3.775V and 2.55V
c) 7.55V and 2.55V
d) 7.555V and 5V

View Answer

Answer: d [Reason:] tough-electronic-devices-circuits-questions-answers-q9a

10. For the PMOS transistor in the circuit shown kn= 8 µA/V2, W/L = 25,|Vtp| = 1V and I = 100μA. For what value of R is VSD = VSG?
tough-electronic-devices-circuits-questions-answers-q10
a) 0 Ω
b) 12.45 kΩ
c) 25.9 kΩ
d) 38.35 kΩ

View Answer

Answer: a [Reason:] VSG will be equal to VSD only when the resistance shown is zero or in other words there should not be any resistance.

Interview MCQ Set 5

1. A good liquid lubricant must possess the property of ___________
a) Low viscosity
b) High boiling point
c) High freezing point
d) Low oiliness

View Answer

Answer: b [Reason:] A good liquid lubricant must possess the properties like adequate viscosity, high boiling point, low freezing point, good oiliness. It also contains stability towards the oxidation and heat.

2. Which of the following is not the liquid lubricant?
a) Olive oil
b) Palm oil
c) Castor oil
d) Grease

View Answer

Answer: d [Reason:] Grease is the semi-solid lubricant which has high viscosity but the liquid lubricant must have an adequate amount of viscosity. It should not be more or less. Olive oil, palm oil, castor oil are liquid lubricants.

3. Castor oil is used for _______
a) Plastic industry
b) For guns
c) For sewing machines
d) Light machinery

View Answer

Answer: a [Reason:] The castor oil is mainly used for machinery at high speed and low pressure. It is used at plastic industry and medical purposes. For guns and sewing machine the neats foot oil is used. For light machinery, whale oil is used.

4. Petroleum oils are also called as _________
a) Petrol oils
b) Hydrocarbon oils
c) Fatty oils
d) Whale oil

View Answer

Answer: b [Reason:] The petroleum oils are also called as the hydrocarbon oils. They contain C12 to C15. These oils are obtained from crude petroleum. They are cheap and have very wide applications.

5. Petrol captured the market up to the percentage of _____
a) 100%
b) 90%
c) 98%
d) 75%

View Answer

Answer: c [Reason:] Petrol captured 98% of the market. Naphthenes and paraffin compounds are mainly found in mineral oils and they are more efficient in preventing the metals from corrosion.

6. The lubricating oils must be refined to remove ___________
a) Wax
b) Hydrocarbons
c) Nitrogen
d) Oxygen

View Answer

Answer: a [Reason:] The lubricating oils must be refined to remove the compounds like wax, aromatic compounds and asphaltic compound. If these compounds are not removed then they will be crystallize and stops the flow of lubricating oil.

7. Easily oxidised impurities cause ___________
a) Moisture formation
b) Nitrogen formation
c) Sludge formation
d) Ash formation

View Answer

Answer: c [Reason:] The easily oxidisable impurities cause the sludge formation. During the operating conditions, the asphaltic. Napthatic compounds will allows to form the sludge. As, they decompose at high temperatures to cause the formation of carbon also.

8. To improve the refined petroleum oils ______ are added.
a) Additives
b) Inhibitors
c) Catalysts
d) Carbon

View Answer

Answer: a [Reason:] Additives are added to the refined petroleum oils. They are refined by the processes like dewaxing etc. They are added to improve the performance of the lubricants. They are called as blended oils as additives are added.

9. A dispersion system consisting of two immiscible liquids is called ________
a) Lubricants
b) Emulsions
c) Semi solids
d) Solids

View Answer

Answer: b [Reason:] A dispersion system consisting of two immiscible liquids is called emulsions. A substance is added to the emulsions called emulsifying agents for stabilization. The emulsifying agents show polar or non polar natures accordingly.

10. In oil in water emulsions, how much water is present _______
a) 3-20%
b) 40-50%
c) 3-4%
d) 5-8%

View Answer

Answer: a [Reason:] Oil in water emulsion is obtained by adding the oil to the sufficient amount water that is about 3 to 20%. The emulsifying agents must be water soluble like sodium soaps or sodium or potassium salts or sulphides.

11. The emulsifiers in water in oil type emulsions are ________
a) Sodium salts
b) Potassium salts
c) Oxides
d) Alkaline earth soaps

View Answer

Answer: d [Reason:] The emulsifiers in water in oil type are alkaline earth metals. Sulphides, sodium and potassium salts are used in oil in water type emulsion. To prepare water in oil type emulsions, 1-10% of water and emulsifiers are added to the oil.

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