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Multiple choice question for engineering

Set 1

Circuit for Q.1-Q.2
electronic-devices-circuits-questions-answers-network-theorms-q1
1. Vth = ?
a) 1 V
b) 2 V
c) 3 V
d) 4 V

View Answer

Answer: d [Reason:] Vth = (6)(6)/(6 + 3) or 4 V.

2. Rth = ? (in ohm)
a) 2
b) 3
c) 4
d) 5

View Answer

Answer: c [Reason:] Rth = (3||6) + 2 or 4 ohm.

3. A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 ohm-1, the power dissipated by the bulb is
a) 80 W
b) 1800 W
c) 112.5 W
d) 228 W

View Answer

Answer: c [Reason:] r = Voc/Isc = 1.2 ohm Power used by bulb = (24 x 24) x 2/(1.2 + 2) x (1.2 + 2) or 11.5 Watt.

Circuit for Q.4-Q-5
electronic-devices-circuits-questions-answers-network-theorms-q4

4. In = ?
a) 1.5 A
b) 3 A
c) 6 A
d) 10 A

View Answer

Answer: b [Reason:] V1 = (15/2)/(1/4 + 1/2 + 1/2) In = Isc = V1/2 = 3 A.

5. Rn = ? (in ohm)
a) 10/3
b) 4
c) 6
d) 10

View Answer

Answer: a [Reason:] Rn = (2||4) + 2.

Circuit for (Q.6-Q.7)
electronic-devices-circuits-questions-answers-network-theorms-q6

6. Vth = ?
a) -2 V
b) -1 V
c) 1 v
d) 2 V

View Answer

Answer: c [Reason:] Vth = (2)(3)(1)/3+3 = 1V.

7. Rth = ? (in ohm)
a) 5/6
b) 6/5
c) 5/3
d) 3/5

View Answer

Answer: a [Reason:] Rth = 1||5 or 5/6 ohm.

8. The equivalent to the given circuit is
electronic-devices-circuits-questions-answers-network-theorms-q8
electronic-devices-circuits-questions-answers-network-theorms-q8a
electronic-devices-circuits-questions-answers-network-theorms-q8b

View Answer

Answer: b [Reason:] After killing all source equivalent resistance is R. Open circuit voltage is v1.

9. V1 = ?
electronic-devices-circuits-questions-answers-network-theorms-q9
a) 6 V
b) 7 V
c) 8 V
d) 10 V

View Answer

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

10. i1 = ?
electronic-devices-circuits-questions-answers-network-theorms-q10
a) 3 A
b) 0.75 mA
c) 2 mA
d) 1.75 mA

View Answer

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

Set 2

1. In a receiver the input signal is 100 V, while the internal noise at the input is 10 V. With amplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of receiver is
a) 2
b) 0.5
c) 0.2
d) None of the mentioned

View Answer

Answer: a [Reason:] NF = (100/10)/(2/0.4) or 2.

2. A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is
a) 1.8 µV
b) 8.4 µV
c) 4.3 µV
d) 12.6 µV

View Answer

Answer: a [Reason:] v2 = 4kBTR where symbols have their usual meanings, hence v = 1.8 µV.

3. A resistor R 1 k is maintained at 17C. The rms noise voltage generated in a bandwidth of 10 kHz is
a) 16 x 10(-14) V
b) 0.4 µV
c) 4 µV
d) 16 x 10(-18) V

View Answer

Answer: b [Reason:] v2 = 4kBTR where symbols have their usual meanings, hence v = 0.4 µV.

4. A mixer stage has a noise figure of 20 dB. This mixer stage is preceded by an amplifier which has a noise figure of 9 dB and an available power gain of 15 dB. The overall noise figure referred to the input is
a) 11.07
b) 18.23
c) 56.43
d) 97.38

View Answer

Answer: a [Reason:] The required answer is 7.94 + (100-1)/31.62 = 11.07.

5. A system has three stage cascaded amplifier each stage having a power gain of 10 dB and noise figure of 6 dB. the overall noise figure is
a) 1.38
b) 6.8
c) 4.33
d) 10.43

View Answer

Answer: c [Reason:] The gain at each stage is 10db, hence the required answer is 4 + (4-1)/10 + (4-1)/100 or 4.33.

(Q.6-Q.8) An amplifier when used with a source of average noise temperature 60 K, has an average operating noise figure of 5.

6. The Te is
a) 70 K
b) 110 K
c) 149 K
d) 240 K

View Answer

Answer: d [Reason:] The required answer is 60 (5-1) or 240K.

7. If the amplifier is sold to engineering public, the noise figure that would be quoted in a catalog is
a) 0.46
b) 0.94
c) 1.83
d) 2.93

View Answer

Answer: c [Reason:] The required answer is 1 + (240/290).

8. What average operating noise figure results when the amplifier is used with an antenna of temperature 30 K?
a) 9.54 db
b) 10.96 db
c) 11.23 db
d) 12.96 db

View Answer

Answer: a [Reason:] The required answer is 1 + 240/30 = 9 or 9.54 db.

9. What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7?
a) 203 K
b) 215 K
c) 235 K
d) 255 K

View Answer

Answer: a [Reason:] The required answer is 290 x (1.7-1) or 203K.

10. If a matched attenuator with a loss of 3.2 dB is placed between the source and the amplifier’s input, what is the operating spot noise figure of the attenuator amplifier cascade if the attenuator’s physical temperature is 290 K?
a) 9 db
b) 11.3 db
c) 10.4 db
d) 13.3 db

View Answer

Answer: d [Reason:] Here Te is 290[(20.89-1) + (2.089)(7.98-1)] or 4544.4K. Hence the required answer is 1 + 4544.4/290 = 212 or 13.3 db.

Set 3

1. In the non-inverting configuration of operational amplifier
a) The positive terminal is connected to the ground directly
b) The negative terminal is connected to the ground directly
c) The positive terminal is connected to the power source
d) The negative terminal is connected to the power source

View Answer

Answer: c [Reason:] Non inverting configuration requires a power source connected to the power source.

2. For ideal non-inverting operational amplifier
a) Input and output resistances are infinite
b) Input resistance is infinite and output resistance is zero
c) Input resistance is zero and output resistance is infinite
d) Input and output resistances are zero

View Answer

Answer: b [Reason:] It is an ideal characteristic of the non-inverting op amp.

3. For an ideal non-inverting operational amplifier having finite gain (A), the ratio of output voltage (v0) to input voltage (vi) is (given R2 is the feedback resistance)
a) (1+R2/R1)/(1+((1+R2/R1)/A))
b) (R2/R1)/(((1+R2/R1)/A))
c) (1+R2/R1)/(((1+R2/R1)/A))
d) (R2/R1)/(1+((1+R2/R1)/A))

View Answer

Answer: a [Reason:] It is a standard mathematical expression.

4. The gain for an ideal non-inverting operational amplifier is (given R2 is the feedback resistance)
a) R2/R1 – 1
b) R2/R1
c) -R2/R1
d) R2/R1 + 1

View Answer

Answer: d [Reason:] It is a standard mathematical expression.

5. While performing an experiment to determine the gain for an ideal operational amplifier having finite gain, a student mistakenly used the equation 1 + R2/R1 where R2 is the feedback resistance. What is the percentage error in his result? Given A is the finite voltage gain of the ideal amplifier used.
a) (R2/R1)/(A+ R2/R1) X 100%
b) (1+R2/R1)/(A+R2/R1) X 100%
c) (1+R2/R1)/(A+1+R2/R1) X 100%
d) (R2/R1)/(A+1+R2/R1) X 100%

View Answer

Answer: c [Reason:] The correct formula is (1+R2/R1)/(1+((1+R2/R1)/A)).

6. The finite voltage gain of a non-inverting operational amplifier is A and the resistance used is R1 and R2 in which R2 is the feedback resistance. Under what conditions it can one use the expression 1 + R2/R1 to determine the gain of the amplifier?
a) A ~ R2/R1
b) A >> R2/R1
c) A << R2/R1
d) None of the mentioned

View Answer

Answer: b [Reason:] The formula is valid for the ideal case in which the value of A is infinite, practically it should be very large when compared to R2/R1 .

7. Which of the following is not true for a voltage follower amplifier?
a) Input voltage is equal to output voltage
b) Input resistance is infinite and output resistance is zero
c) It has 100% negative feedback
d) None of the mentioned

View Answer

Answer: d [Reason:] All the statements are false.

8. For designing a non-inverting amplifier with a gain of 2 at the maximum output voltage of 10 V and the current in the voltage divider is to be 10 μA the resistance required are R1 and R2 where R2 is used to provide negative feedback. Then
a) R1 = 0.5 MΩ and R2 = 0.5 MΩ
b) R1 = 0.5 kΩ and R2 = 0.5 kΩ
c) R1 = 5 MΩ and R2 = 5 MΩ
d) R1 = 5 kΩ and R2 = 5 kΩ

View Answer

Answer: a [Reason:] 1 + R2/R1 = 2 and 10/(R1+R2) = 10 μA. Solve for R1 and R2.

9. It is required to connect a transducer having an open-circuit voltage of 1 V and a source resistance of 1 MΩ to a load of 1-kΩ resistance. Find the load voltage if the connection is done (a) directly and (b) through a unity-gain voltage follower.
a) 1 μV and 1 mV respectively
b) 1 mV and 1 V respectively
c) 0.1 μV and 0.1 mV respectively
d) 0.1 mV and 0.1 V respectively

View Answer

Answer: b [Reason:] When a unity gain follower is uses then input signal is equal to output signal. When connected directly, output signal is given by 1 X 1kΩ/1MΩ or 1mV.

10. Consider the figure given below. If the resistance R1 is disconnected from the ground and connected to a third power source v3, then expression for the value of
v0 is
electronic-devices-circuits-questions-answers-non-inverting-configuration-q10
a) 2v1 + 4v2 − 3v3
b) 6v1 + 8v2 − 3v3
c) 6v1 + 4v2 − 9v3
d) 3v1 + 4v2 − 3v3

View Answer

Answer: c [Reason:] When a third power source is connected to the resistance of 1kΩ, then also the potential between the two input terminals of op amps remains the same. Using this fact the expression c is obtained.

Set 4

1. Slew rate of an amplifier is defined as
a) Minimum rate of change of the output possible in a real operational amplifier
b) Maximum rate of change of the output possible in a real operational amplifier
c) Average rate of change of the output possible in a real operational amplifier
d) Ratio of the maximum and the average rate of change of the output in a real amplifier

View Answer

Answer: b [Reason:] By definition slew rate is the maximum rate of change of the output possible in a real operational amplifier.

2. Determine the slew rate of the amplifier having full power bandwidth f0 and the rated output voltage as V0. Given that the input signal is of sinusoidal nature.
a) 2πf0 V0
b) V0 / 2πf0
c) V0 / f0
d) f0 V0

View Answer

Answer: a [Reason:] v = V0sin wt dv/dt = wV0 sin wt max value of dv/dt = wV0 max value of w = w0 = 2πf0 w0 V0 = Slew Rate = 2πf0 V0.

3. The units of the full power bandwidth is
a) Watt
b) Joule
c) Seconds
d) Hertz

View Answer

Answer: d [Reason:] It has the units of frequency.

4. The full-power bandwidth, fM, is the maximum frequency at which
a) an output sinusoid with an amplitude equal to the op-amp rated output voltage (Vo max) can be produced without distortion
b) it is the range of the frequencies in which the amplitude of output signal is equal to or greater than half of the op-amp rated output voltage
c) it is the range of the frequencies in which the amplitude of output signal is equal to or less than half of the op-amp rated output voltage
d) It is the range of the frequencies in which the power gain is half or more than half of the maximum rated power gain of the op-amp

View Answer

Answer: a [Reason:] This is the only statement that satisfies the definition of the full-power bandwidth.

5. Which of the following is not limitation of the operational amplifier
a) Output voltage saturation
b) Output current limits
c) Slew rate
d) None of the mentioned

View Answer

Answer: d [Reason:] None of the mentioned are the limitations of the operational amplifier.

6. A particular op amp using ±15-V supplies operates linearly for outputs in the range −12 V to +12 V. If used in an inverting amplifier configuration of gain –100, what is the rms value of the largest possible sine wave that can be applied at the input without output clipping?
a) 120 mV
b) 60 mV
c) 84.85 mV
d) 42.42 mV

View Answer

Answer: c [Reason:] Peak value of input wave = 12/100 or 120 mV. Hence the rms value is 120/√2 or 84.85 mV.

7. For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)
a) 10 V/µs
b) 20 V/µs
c) 40 V/µs
d) 80 V/µs

View Answer

Answer: c

8. An op amp having a slew rate of 20 V/µs is to be used in the unity-gain follower configuration, with input pulses that rise from 0 to 3 V. What is the shortest pulse that can be used while ensuring full-amplitude output?
a) 0.10 µs
b) 0.15 µs
c) 0.20 µs
d) 0.30 µs

View Answer

Answer: b [Reason:] Time taken to reach 3V from 0V with slew rate of 20V/µs is 3/20 µs or 0.15 µs.

(Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt = 2 MHz, SR = 1 V/µs, and V0 max = 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.

9. If Vi = 0.5 V, what is the maximum frequency before the output distorts?
a) 31.8 kHz
b) 318 kHz
c) 3.18 kHz
d) 3.18 MHz

View Answer

Answer: a [Reason:] Vi = 0.5v, V0 = 0.5 X 10 = 5V 2πf V0 = SR or f = 31.8 kHz.

10. If f = 20 kHz, what is the maximum value of Vi before the output distorts?
a) 0.397 V
b) 0.795 V
c) 1.192 V
d) 1.590 V

View Answer

Explanation: V0 = 10Vi 2πf V0 = SR = 20πf Vi, here f is 20 kHz, SR is 1 V/µs. Hence the value of Vi is 0.795 V.

Set 5

1. What is the thickness of ‘space charge region’ or ‘transition region’ in P-N junction diode?
a) 1 micron
b) 5 micron
c) 10 micron
d) 2.876 micron

View Answer

Answer: a [Reason:] The region of the junction is depleted by mobile charges, hence it is called space charge region or depletion region or transition region which is 10-4 cm = 10-6 m= 1 micron.

2. If what of the following is doped into a semiconductor say germanium a P-N junction is formed.
a) Electrons and Protons
b) Protons and Neutrons
c) Neutrons and Electrons
d) Gallium and Phosphorus

View Answer

Answer: d [Reason:] A P-N junction is formed only when a donor impurities and acceptor impurities are added to either side of a semiconductor like silicon and germanium.

3. Which of the factors doesn’t change the diode current.
a) Temperature
b) External voltage applied to the diode
c) Boltzmann‘s constant
d) Resistance

View Answer

Answer: d [Reason:] I = Io [e(v/nVt) -1], as shown in this equation the diode current is dependent on temperature , voltage applied on the diode , Boltzmann’s constant but diode current is not dependent on resistance as it is independent of resistance.

4. The product of mobility of the charge carriers and applied Electric field intensity is known as
a) Drain velocity
b) Drift velocity
c) Push velocity
d) Pull velocity

View Answer

Answer: b [Reason:] When the semiconductors like silicon and germanium is implied by an electric field the charge carriers when get drifted by certain velocity known as drift velocity. Drift velocity is product of mobility of charge carriers and field intensity.

5. The tendency of charge carriers to move from a region of heavily concentrated charges to region of less concentrated charge is known as.
a) Depletion current
b) Drain current
c) Diffusion current
d) Saturation current

View Answer

Answer: c [Reason:] In a semiconductor the charge will always have a tendency to move from higher concentrated area to less concentrated area to maintain equilibrium this movement of charges will result in diffusion current.

6. If the drift current is 100mA and diffusion current is 1A what is the total current in the semiconductor diode.
a) 1.01 A
b) 1.1 A
c) 900m A
d) 10 A

View Answer

Answer: b [Reason:] We know that the total current in a semiconductor is equal to sum of both drift current and diffusion current. Total current = 1A + 100mA =1.1A.

7. Which of the following is reverse biased?
electronic-devices-circuits-questions-answers-online-test-q7
a) A)
b) B)
c) C)
d) D)

View Answer

Answer: c [Reason:] The P-N junction diode is forward bias when the voltage applied to p type is greater than the n type and vice versa, since the voltage applied to p type is less in C) it is the answer.

8. The drift velocity is 5V and the applied electric field intensity 20v/m what will be the mobility of charge carriers.
a) 100 m2/ (vs)
b) 4 m2/ (vs)
c) 15 m2/ (vs)
d) 0.25 m2/ (vs)

View Answer

Answer: d [Reason:] We know that mobility of charge carriers is drift velocity divide by applied electric field intensity. Mobility = drift velocity / field intensity.

9. When there is an open circuit what will be the net hole current.
a) 5A
b) 0.05A
c) 0.5A
d) 0A

View Answer

Answer: d [Reason:] If there is any current present under open circuit there will be an indefinite growth of holes at one end of the semiconductor which is practically not possible hence zero amperes.

10. Rate of change of concentration per unit length in a semiconductor is called as.
a) Concentration change
b) Concentration mixture
c) Concentration gradient
d) Concentration variant

View Answer

Answer: c [Reason:] In a semiconductor the holes as well as electrons which are the charge carriers is not equally concentrated on all regions of the semiconductor the change in their rate is referred as concentration gradient.