Anubhava Circuit Design Challenge

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Quizzes Created: 1 | Total Attempts: 363
Questions: 50 | Attempts: 363

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Anubhava Circuit Design Challenge - Quiz

INSTRUCTIONS: Every question contains negative points (+2 for correct, -1 for wrong) Please submit before time runs out. Team members can be 2-3. Once team is formed cannot be changed
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Questions and Answers
  • 1. 

    A 33 half-watt resistor and a 330 half-watt resistor are connected across a 12 V source. Which one(s) will overheat?

    • A.

      33

    • B.

      330

    • C.

      Both resistors

    • D.

      Neither resistors

    Correct Answer
    D. Neither resistors
    Explanation
    Both resistors will not overheat because the power dissipated by a resistor can be calculated using the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. In this case, both resistors have a power rating of half-watt, which means they can safely dissipate up to 0.5 watts of power. Plugging in the values, we find that the power dissipated by the 33 ohm resistor is (12^2)/(33) = 4.36 watts, and the power dissipated by the 330 ohm resistor is (12^2)/(330) = 0.44 watts. Since both values are below the power rating of the resistors, neither of them will overheat.

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  • 2. 

    When the current through the coil of an electromagnet reverses, the

    • A.

      direction of the magnetic field reverses

    • B.

      Direction of the magnetic field remains unchanged

    • C.

      Magnetic field expands

    • D.

      magnetic field collapses

    Correct Answer
    A. direction of the magnetic field reverses
    Explanation
    When the current through the coil of an electromagnet reverses, the direction of the magnetic field also reverses. This is because the magnetic field is created by the flow of electric current through the coil. When the current changes direction, the magnetic field lines also change direction, resulting in a reversal of the magnetic field.

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  • 3. 

    When a 15 V input pulse with a width equal to two time constants is applied to an RC integrator, the capacitor charges to

    • A.

      15

    • B.

      12.9

    • C.

      8.6

    • D.

      19.45

    Correct Answer
    B. 12.9
    Explanation
    When a 15 V input pulse is applied to an RC integrator, the capacitor charges up to a voltage determined by the time constant of the circuit. The time constant is the product of the resistance (R) and the capacitance (C). In this case, the width of the input pulse is equal to two time constants. Therefore, the capacitor charges up to approximately 12.9 V, which is 86% of the input voltage.

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  • 4. 

    If all the elements in a particular network are linear, then the superposition theorem would hold, when the excitation is

    • A.

      DC only

    • B.

      AC only

    • C.

      Either AC or DC

    • D.

      An Impulse

    Correct Answer
    C. Either AC or DC
    Explanation
    The superposition theorem states that in a linear network, the total response is the sum of the individual responses to each independent source acting alone. Since the question states that all the elements in the network are linear, the superposition theorem would hold for both AC and DC excitation. This means that the response of the network can be determined by considering the individual responses to each source separately and then adding them together. Therefore, the correct answer is either AC or DC excitation.

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  • 5. 

    Mesh analysis is applicable for

    • A.

      Planar networks

    • B.

      Non planar networks

    • C.

      Both planner and non-planar networks.

    • D.

      Neither planner nor non planner networks

    Correct Answer
    A. Planar networks
    Explanation
    Mesh analysis is a circuit analysis technique that is used to solve electrical circuits with multiple loops. In mesh analysis, the circuit is divided into meshes, which are closed loops that do not contain any other loops. Each mesh is then analyzed separately to determine the current flowing through it.

    Planar networks are circuits that can be drawn on a plane surface without any lines crossing over each other. Since mesh analysis requires the circuits to be divided into loops, it is applicable for planar networks. Non-planar networks, on the other hand, cannot be divided into loops without lines crossing over each other, making mesh analysis not applicable. Therefore, the correct answer is planar networks.

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  • 6. 

    Which of the following memories uses one transistor and one capacitor as basic memory unit

    • A.

      SRAM

    • B.

      DRAM

    • C.

      Both SRAM and DRAM

    • D.

      Neither of them

    Correct Answer
    B. DRAM
    Explanation
    DRAM (Dynamic Random Access Memory) uses one transistor and one capacitor as the basic memory unit. In DRAM, each memory cell consists of a transistor and a capacitor. The transistor acts as a switch to control the flow of current, and the capacitor stores the charge representing the data. This design allows for higher memory density and lower power consumption compared to SRAM (Static Random Access Memory), which uses multiple transistors per memory cell. Therefore, the correct answer is DRAM.

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  • 7. 

    A external resistance R is connected to a cell of internal resistance r, then the current is maximum when

    • A.

      R>r

    • B.

      R < r

    • C.

      R=r

    • D.

      None

    Correct Answer
    C. R=r
    Explanation
    When an external resistance R is connected to a cell of internal resistance r, the current is maximum when the external resistance R is equal to the internal resistance r. This is because the maximum power is transferred from the cell to the external resistance when the resistance of the external circuit is equal to the internal resistance of the cell. When the external resistance is larger or smaller than the internal resistance, the current decreases due to higher internal resistance or lower power transfer.

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  • 8. 

    What is referred as the average value in AC operation?

    • A.

      TriangleAverage of all values of an alternating quantity.

    • B.

      Average of all values of the phase sequences.

    • C.

      Average of all values of the (+)ve and (-)ve half.

    • D.

      Average of all values of an alternating quantity over a complete cycle.

    Correct Answer
    D. Average of all values of an alternating quantity over a complete cycle.
    Explanation
    The average value in AC operation is referred to as the average of all values of an alternating quantity over a complete cycle. This means that the average is calculated by summing up all the values of the alternating quantity during one complete cycle and dividing it by the total number of values. This gives a representative value that reflects the overall behavior of the alternating quantity over time.

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  • 9. 

    An electric field line and an equipotential surface are

    • A.

      Always 90°.

    • B.

      Always parallel.

    • C.

      Inclined at any angle.

    • D.

      None

    Correct Answer
    B. Always parallel.
    Explanation
    An electric field line and an equipotential surface are always parallel because the electric field lines indicate the direction of the electric field, which is perpendicular to the equipotential surfaces. Since the electric field lines and equipotential surfaces are perpendicular to each other, they cannot intersect and must be parallel.

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  • 10. 

    For a collpits oscillator, the capacitor values are 1micro farad and 10 micro farad. The inductor is 10 henry .Its frequency of oscillation approximately is,

    • A.

      400Hz

    • B.

      50Hz

    • C.

      1Hz

    • D.

      100Hz

    Correct Answer
    B. 50Hz
    Explanation
    The frequency of oscillation for a Colpitts oscillator is determined by the values of the capacitors and inductor used in the circuit. In this case, the capacitor values are 1 microfarad and 10 microfarad, and the inductor value is 10 henry. By analyzing the circuit equations and applying the appropriate formulas, it can be determined that the frequency of oscillation is approximately 50Hz.

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  • 11. 

    Among the following the audio frequency oscillators are 1)rc phase shift 2)weins bridge 3) Hartley 4) collpit

    • A.

      1 and 2

    • B.

      3 and 4

    • C.

      1 and 3

    • D.

      1 and 4

    Correct Answer
    A. 1 and 2
    Explanation
    The correct answer is 1 and 2. This means that among the given options, the audio frequency oscillators are RC phase shift and Wein's bridge. The other options, Hartley and Colpitt, are not audio frequency oscillators.

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  • 12. 

    The amplifiers used in radio and tv transmitters are

    • A.

      Class A

    • B.

      Class B

    • C.

      Class AB

    • D.

      Class C

    Correct Answer
    D. Class C
    Explanation
    Class C amplifiers are commonly used in radio and TV transmitters because they are highly efficient and provide high power output. These amplifiers operate in a non-linear region, where the output waveform is heavily distorted. This distortion is acceptable for radio and TV signals, as they are typically modulated onto a carrier wave. Class C amplifiers are designed to amplify only a portion of the input signal, resulting in high efficiency and reduced power consumption. The distorted waveform is then filtered and shaped to produce the desired output signal.

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  • 13. 

    In a transistor which is moderately doped

    • A.

      Collector

    • B.

      Emitter

    • C.

      Base

    • D.

      None

    Correct Answer
    A. Collector
    Explanation
    In a transistor, the collector is one of the three main regions, along with the emitter and base. The collector is typically moderately doped, meaning it has a moderate concentration of impurities. This doping level allows the collector to efficiently collect and control the majority charge carriers (electrons or holes) that flow through the transistor. The collector is responsible for collecting the current from the emitter and is often connected to an external circuit.

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  • 14. 

    Generally,the configuration used in high frequency circuits is

    • A.

      Common base

    • B.

      Common Collector

    • C.

      Common Emitter

    • D.

      None

    Correct Answer
    A. Common base
    Explanation
    In high frequency circuits, the common base configuration is often used. This configuration provides a high voltage gain and a low current gain. It is suitable for applications where the input impedance needs to be low and the output impedance needs to be high. Additionally, the common base configuration offers good high-frequency response and low noise performance, making it ideal for high-frequency circuits. The common collector and common emitter configurations are also commonly used, but they have different characteristics and are not as suitable for high frequency applications.

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  • 15. 

    Generally,the configuration used in impedence matching is,

    • A.

      Common base

    • B.

      Common Collector

    • C.

      Common Emitter

    • D.

      None

    Correct Answer
    B. Common Collector
    Explanation
    The correct answer is Common Collector. In impedance matching, the common collector configuration is commonly used. This configuration provides a high input impedance and a low output impedance, which helps in matching the impedance between the source and load. The common collector configuration also provides a voltage gain less than unity, making it suitable for impedance matching applications where voltage amplification is not required.

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  • 16. 

    In a voltage series connection the input impedence when compared to normal amplifier is

    • A.

      Decreases

    • B.

      Increases

    • C.

      No change

    • D.

      Can't determined

    Correct Answer
    B. Increases
    Explanation
    In a voltage series connection, the input impedance increases compared to a normal amplifier. This is because in a series connection, the impedance of each component adds up, resulting in a higher overall impedance. This increased impedance can affect the performance of the circuit by reducing the current flow and possibly affecting the frequency response. Therefore, the correct answer is increases.

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  • 17. 

    Commutative property of or gate is

    • A.

      A+B=B+A

    • B.

      AB=BA

    • C.

      A(B+C)=AB+AC

    • D.

      A+BC=(A+B)(A+C)

    Correct Answer
    A. A+B=B+A
    Explanation
    The commutative property states that the order of the operands does not affect the result of the operation. In the case of the OR gate, the commutative property means that the result of A OR B is the same as the result of B OR A. This is because the OR operation returns true if at least one of the inputs is true, regardless of their order. Therefore, the statement A+B=B+A is true for the commutative property of the OR gate.

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  • 18. 

    Which of the following is a digital device

    • A.

      Regulator of a fan

    • B.

      Microphone

    • C.

      Resistance of a material

    • D.

      Light switch

    Correct Answer
    D. Light switch
    Explanation
    A light switch is a digital device because it has two states: on and off. It operates using digital signals to control the flow of electricity to a light fixture. By flipping the switch, the circuit is either completed or broken, resulting in the light being turned on or off. This binary functionality makes the light switch a digital device.

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  • 19. 

    Which logic family provide minimum power dissipation

    • A.

      TTL

    • B.

      CMOS

    • C.

      ECL

    • D.

      JFET

    Correct Answer
    B. CMOS
    Explanation
    CMOS logic family provides minimum power dissipation compared to other options. CMOS stands for Complementary Metal-Oxide-Semiconductor and is known for its low power consumption characteristics. It uses both NMOS and PMOS transistors to achieve low power dissipation by consuming power only during switching transitions. This makes CMOS an energy-efficient choice for various electronic devices, such as smartphones, laptops, and integrated circuits.

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  • 20. 

    Which is the example of digital device from the given option ?

    • A.

      Sensors

    • B.

      Recorder players

    • C.

      Microprocessors

    • D.

      Thermistors

    Correct Answer
    C. Microprocessors
    Explanation
    A microprocessor is an example of a digital device because it is a small electronic chip that performs various functions by executing instructions stored in its memory. It is capable of processing and manipulating digital data, making it an essential component in many electronic devices such as computers, smartphones, and tablets. Unlike analog devices, microprocessors operate on discrete values and use binary code to perform calculations and control operations. Therefore, microprocessors are a clear example of digital devices.

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  • 21. 

    NPN transistor is not suitable for good analog switch because

    • A.

      IC −VCE characteristic curve pass directly through origin

    • B.

      The device has very high input impedance.

    • C.

      The device is asymmetrical with an offset voltage VCE off.

    • D.

      It has well defined transition frequency

    Correct Answer
    C. The device is asymmetrical with an offset voltage VCE off.
    Explanation
    The NPN transistor is not suitable for a good analog switch because it is asymmetrical with an offset voltage VCE off. This means that when the transistor is in the off state, there is still a small voltage present between the collector and emitter terminals. This offset voltage can cause inaccuracies and distortions in the analog signal being switched, making it unsuitable for precise analog applications.

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  • 22. 

    In figure v1 = 8 V and v2 = 4 V. Which diode will conduct?

    • A.

      Only D1

    • B.

      Only D2

    • C.

      Both D1 and D2

    • D.

      None

    Correct Answer
    C. Both D1 and D2
    Explanation
    Both D1 and D2 will conduct because the diode conducts current when the voltage across it is greater than the forward voltage drop. In this case, the voltage across D1 is 8V, which is greater than the forward voltage drop, and the voltage across D2 is 4V, which is also greater than the forward voltage drop. Therefore, both diodes will conduct.

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  • 23. 

    Simplification following Boolean expression gives, Expression= (A'BC')'+(AB'C)'

    • A.

      1

    • B.

      A

    • C.

      Bc

    • D.

      0

    Correct Answer
    A. 1
    Explanation
    The given Boolean expression is (A'BC')' + (AB'C)'. In this expression, (A'BC')' means the complement of A'BC', and (AB'C)' means the complement of AB'C. The '+' operator represents the OR operation. Therefore, the expression simplifies to A'BC + AB'C. The answer is 1 because the simplified expression evaluates to 1 when A=1, B=0, and C=0.

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  • 24. 

    The most commonly used amplifier in sample and hold circuit is

    • A.

      A unity gain inverting amplifier

    • B.

      A unity gain non inverting amplifier

    • C.

      An inverting amplifier with a gain of 10

    • D.

      An inverting amplifier with a gain of 100

    Correct Answer
    A. A unity gain inverting amplifier
    Explanation
    The most commonly used amplifier in a sample and hold circuit is a unity gain inverting amplifier. This type of amplifier is used because it provides a stable and accurate output voltage that is inverted from the input voltage. The unity gain ensures that there is no amplification or attenuation of the signal, while the inverting nature of the amplifier allows for the necessary polarity reversal in the sample and hold circuit.

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  • 25. 

    When the ac base voltage in a CE amplifier circuit is too high, the ac emitter current is

    • A.

      Zero

    • B.

      Constant

    • C.

      Alternating

    • D.

      Distorted

    Correct Answer
    D. Distorted
    Explanation
    When the AC base voltage in a CE amplifier circuit is too high, it can cause the amplifier to enter into saturation or cutoff regions, leading to distortion in the output signal. This distortion occurs because the transistor is unable to accurately amplify the entire AC signal, resulting in a distorted waveform. Therefore, the correct answer is distorted.

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  • 26. 

    In the circuit of figure the diode

    • A.

      Will conduct as the whole cycle

    • B.

      Will conduct in the positive half cycle

    • C.

      Will not conduct

    • D.

      Will conduct from 30° to 150° in the positive half cycle

    Correct Answer
    C. Will not conduct
    Explanation
    In the given circuit, the diode will not conduct. This is because the diode only conducts current in the forward bias condition, where the anode is at a higher potential than the cathode. However, in this circuit, the anode is connected to the negative terminal of the supply, while the cathode is connected to the positive terminal. Therefore, the diode will be in reverse bias and will not allow current to flow through it.

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  • 27. 

    Which of the following is not an essential element of a dc power supply

    • A.

      Rectifier

    • B.

      Filter

    • C.

      Voltage regulator

    • D.

      Voltage Amplifier

    Correct Answer
    D. Voltage Amplifier
    Explanation
    A DC power supply is used to convert alternating current (AC) into direct current (DC). The essential elements of a DC power supply include a rectifier, which converts AC to pulsating DC, a filter, which smoothens the pulsating DC into a steady DC, and a voltage regulator, which maintains a constant output voltage. However, a voltage amplifier is not an essential element of a DC power supply. A voltage amplifier is used to increase the amplitude of a voltage signal, which is not necessary for converting AC to DC.

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  • 28. 

    The basic purpose of filter is to

    • A.

      Minimize variations in ac input signal

    • B.

      Suppress harmonics in rectified output

    • C.

      Remove ripples from the rectified output

    • D.

      Stabilize dc output voltage

    Correct Answer
    B. Suppress harmonics in rectified output
    Explanation
    The purpose of a filter is to suppress harmonics in the rectified output. When an AC signal is rectified, it produces harmonics which are unwanted frequencies that can cause distortion and interference. A filter is used to remove or reduce these harmonics, ensuring a smoother and cleaner output signal. This is especially important in applications where a clean and stable DC voltage is required, such as in power supplies or audio amplifiers.

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  • 29. 

    Active load is used in the collector of the difference amplifier of an Op-amp:

    • A.

      To increase the output resistance.

    • B.

      To increase the differential gain.

    • C.

      To handle large signals.

    • D.

      To provide symmetry.

    Correct Answer
    B. To increase the differential gain.
    Explanation
    An active load is used in the collector of the difference amplifier of an Op-amp to increase the differential gain. The active load helps in providing a high output resistance which prevents loading effects and improves the overall performance of the amplifier. By increasing the differential gain, the amplifier becomes more sensitive to small differences in input signals, allowing for better amplification and accuracy in the output.

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  • 30. 

    How many comparators would a 12-bit flash ADC require?

    • A.

      4096

    • B.

      3096

    • C.

      4095

    • D.

      2512

    Correct Answer
    C. 4095
    Explanation
    A 12-bit flash ADC requires 2^12 - 1 comparators because each bit requires a comparator to compare the input voltage with a reference voltage. Since there are 12 bits in total, we subtract 1 from 2^12 to account for the fact that the most significant bit does not require a comparator. Therefore, the correct answer is 4095.

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  • 31. 

    The advantage of using a dual slope ADC in a digital voltmeter is that

    • A.

      Its conversion time is small

    • B.

      Its accuracy is high

    • C.

      It gives output in BCD format

    • D.

      It does not require a comparator

    Correct Answer
    B. Its accuracy is high
    Explanation
    The advantage of using a dual slope ADC in a digital voltmeter is that its accuracy is high. Dual slope ADCs are known for their ability to provide precise and accurate measurements. They achieve this by integrating the input voltage over a fixed period of time and comparing it with a known reference voltage. This method eliminates the need for a high-resolution comparator, which can introduce errors and reduce accuracy. Therefore, the high accuracy of a dual slope ADC makes it a preferred choice for applications where precise voltage measurements are required.

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  • 32. 

    A digital system is required to amplify a binary encoded audio signal. The user should be able to control the gain of the amplifier from a minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is

    • A.

      4

    • B.

      5

    • C.

      6

    • D.

      7

    Correct Answer
    D. 7
    Explanation
    To encode the gain levels in 100 increments, we need to have at least 7 bits. This is because 2^7 (128) is the smallest power of 2 that is greater than or equal to 100. With 7 bits, we can represent values from 0 to 127, which covers the range of gain levels required. Therefore, the minimum number of bits required to encode the gain levels in straight binary is 7.

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  • 33. 

    Of the values listed, the most realistic value for open-loop voltage gain of an OP-amp is ……

    • A.

      1

    • B.

      100

    • C.

      1000

    • D.

      100000

    Correct Answer
    D. 100000
    Explanation
    The open-loop voltage gain of an operational amplifier (OP-amp) is a measure of its amplification capability without any external feedback. A high open-loop voltage gain is desirable for accurate amplification of small input signals. Among the given options, 100000 is the most realistic value for the open-loop voltage gain of an OP-amp. This value indicates that the OP-amp can amplify the input signal by a factor of 100000, making it suitable for applications that require high amplification.

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  • 34. 

    Current cannot flow to ground through …….

    • A.

      Ground

    • B.

      Virtual ground

    • C.

      Mechanical ground

    • D.

      AC ground

    Correct Answer
    B. Virtual ground
    Explanation
    Current cannot flow to ground through virtual ground because virtual ground is not a physical connection to the ground. It is a concept used in electronic circuits where a voltage is artificially created to mimic a ground potential, allowing for easier analysis and design of circuits. Therefore, no actual current can flow through virtual ground to the ground.

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  • 35. 

    Calculate the output voltage if V1 = –0.2 V and V2 = 0 V.

    • A.

      6.6 V

    • B.

      4.4 V

    • C.

      2V

    • D.

      0

    Correct Answer
    C. 2V
    Explanation
    The output voltage is 2V because when V1 is -0.2V and V2 is 0V, there is no voltage drop across the resistors in the circuit. Therefore, the output voltage is equal to the sum of V1 and V2, which is 2V.

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  • 36. 

    How many op-amps are required to implement this equation?

    • A.

      1

    • B.

      2

    • C.

      3

    • D.

      4

    Correct Answer
    B. 2
    Explanation
    To implement the given equation, only 2 op-amps are required. This implies that the equation can be realized using a summing amplifier configuration with two op-amps. The summing amplifier is a commonly used circuit in which multiple input voltages are summed together and amplified. Each op-amp in the configuration can handle multiple input voltages, hence only 2 op-amps are sufficient to implement the equation.

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  • 37. 

    For N-input XOR gate,

    • A.

      Output will be 1, if number of ones in input is odd

    • B.

      Output will be 1, if number of ones in input is even

    • C.

      Output will be 1, if alternate 1’s and 0’s are present in input

    • D.

      Output will be 1, if all the bits in input are same.

    Correct Answer
    A. Output will be 1, if number of ones in input is odd
    Explanation
    The given answer is correct because an XOR gate outputs a 1 if the number of ones in the input is odd. This is because the XOR gate evaluates to true (1) when an odd number of inputs are true (1), and false (0) when an even number of inputs are true. Therefore, if the number of ones in the input is odd, the output of the XOR gate will be 1.

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  • 38. 

    What will be the output of a integrator if input is square wave ,

    • A.

      Sine wave

    • B.

      Triangular wave

    • C.

      Half rectified Sine wave

    • D.

      Square

    Correct Answer
    B. Triangular wave
    Explanation
    When an integrator receives a square wave as an input, it will produce a triangular wave as the output. This is because an integrator performs the mathematical operation of integration, which is equivalent to finding the area under the curve of the input signal. In the case of a square wave, the integrator will continuously accumulate the positive and negative areas, resulting in a triangular waveform.

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  • 39. 

    An operation in industry needs Light to blink continuously when Voltage at point A1 is greater than Voltage at point A2. Which of following combinations should be used in between Inputs and light?

    • A.

      Comparator and Ramp generator

    • B.

      Comparator and Astable Multivibrator

    • C.

      Comparator and Monostable Multivibrator

    • D.

      Only Comparator

    Correct Answer
    B. Comparator and Astable Multivibrator
    Explanation
    The correct combination to use in between the inputs and the light is a Comparator and an Astable Multivibrator. A Comparator is used to compare the voltage at point A1 with the voltage at point A2. If the voltage at A1 is greater than the voltage at A2, the Comparator will output a high signal. This high signal can then be used as an input to the Astable Multivibrator, which will generate a continuous blinking output signal to drive the light. Therefore, using a Comparator and an Astable Multivibrator will ensure that the light blinks continuously when the voltage at A1 is greater than the voltage at A2.

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  • 40. 

    How many 2:1 Multiplexers are needed to construct a 8:1 Multiplexer?

    • A.

      3

    • B.

      4

    • C.

      7

    • D.

      8

    Correct Answer
    C. 7
    Explanation
    To construct an 8:1 multiplexer, we can use a combination of 2:1 multiplexers. Each 2:1 multiplexer has 2 inputs and 1 output. By cascading these 2:1 multiplexers, we can increase the number of inputs and outputs. In this case, to construct an 8:1 multiplexer, we would need 7 2:1 multiplexers. The first 3 multiplexers would be used to combine the 8 inputs into 4 outputs, and then the remaining 4 multiplexers would be used to combine those 4 outputs into a single output.

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  • 41. 

    The number of counter states which an 8 bit stair step A/D converter has to pass through before conversion is

    • A.

      1

    • B.

      8

    • C.

      255

    • D.

      256

    Correct Answer
    D. 256
    Explanation
    An 8-bit stair step A/D converter has a resolution of 2^8 = 256 levels. Each level represents a unique counter state that the converter passes through during the conversion process. Therefore, the number of counter states the converter has to pass through before completing the conversion is 256.

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  • 42. 

    If an open capacitor is checked with an ohmmeter, the needle will:

    • A.

      Stay on zero

    • B.

      Stay on infinity

    • C.

      Move from zero to infinity

    • D.

      Move from infinity to zero

    Correct Answer
    B. Stay on infinity
    Explanation
    When an open capacitor is checked with an ohmmeter, it means that there is no electrical connection between the capacitor plates. In this case, the ohmmeter will measure infinite resistance, indicating that there is no continuity or path for the current to flow through. As a result, the needle on the ohmmeter will stay on infinity, indicating an open circuit.

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  • 43. 

    The minimum number of NAND gates required to implement the Boolean function 𝐴 + 𝐴𝐵̅ + 𝐴𝐵̅𝐶 is equal to

    • A.

      1

    • B.

      4

    • C.

      7

    • D.

      0

    Correct Answer
    D. 0
    Explanation
    The given Boolean function 𝐴 + 𝐴𝐵̅ + 𝐴𝐵̅𝐶 can be simplified using Boolean algebra. By applying De Morgan's theorem and the distributive property, we can rewrite the function as 𝐴(1 + 𝐵̅(1 + 𝐶)). Since the expression inside the parentheses is always equal to 1, the simplified function becomes 𝐴. This means that the function can be implemented using a single NAND gate, as the output of a NAND gate is the inverse of the logical AND operation. Therefore, the minimum number of NAND gates required is 0.

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  • 44. 

    If a charged capacitor is disconnected from a circuit, it will:

    • A.

      Immediately discharge

    • B.

      Recharge

    • C.

      Remain charged

    • D.

      Leak the charge

    Correct Answer
    D. Leak the charge
    Explanation
    When a charged capacitor is disconnected from a circuit, it will slowly lose its charge over time due to the leakage of electrons across the dielectric material. This leakage is caused by the imperfect insulation of the dielectric, which allows a small amount of current to flow through. As a result, the capacitor gradually loses its stored charge and eventually becomes discharged. Therefore, the correct answer is "leak the charge."

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  • 45. 

    For the given circuit output f is given by,

    • A.

      1

    • B.

      X

    • C.

      X'

    • D.

      0

    Correct Answer
    D. 0
    Explanation
    The given circuit is a 2-input AND gate. In an AND gate, the output is only 1 when both inputs are 1. In this case, one of the inputs is 0 (X') and the other input is X (which could be either 0 or 1). Since one of the inputs is always 0, the output of the AND gate will always be 0. Therefore, the answer is 0.

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  • 46. 

    The excess 3 code of decimal number 26 is

    • A.

      0100 1001

    • B.

      1011001

    • C.

      1000 1001

    • D.

      1001101

    Correct Answer
    B. 1011001
    Explanation
    The excess 3 code is a binary code that represents decimal numbers by adding 3 to the original number and converting it to binary. In this case, the decimal number 26 is converted to binary using the excess 3 code, resulting in the binary number 1011001.

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  • 47. 

    How many flip flops are required to construct a decade counter

    • A.

      10

    • B.

      4

    • C.

      2

    • D.

      8

    Correct Answer
    B. 4
    Explanation
    A decade counter is a type of counter that counts from 0 to 9 and then resets back to 0. Each digit in the counter can be represented by a flip flop, and since a decade counter has 10 states (0 to 9), it would require 4 flip flops to construct it. Each flip flop represents one digit of the counter, and together they count from 0000 to 1001, which is a total of 10 states.

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  • 48. 

    If a certain multiplexer can switch one of 32 data inputs to its output. How many different inputs does this MUX have?

    • A.

      30 data inputs & 5 select inputs

    • B.

      32 data inputs and 5 select inputs

    • C.

      32 data inputs and 4 select inputs

    • D.

      None

    Correct Answer
    B. 32 data inputs and 5 select inputs
    Explanation
    A multiplexer is a device that selects one of many inputs and forwards it to a single output. In this case, the given answer states that the multiplexer has 32 data inputs and 5 select inputs. This means that the multiplexer can choose from 32 different data inputs and the selection can be controlled using 5 select inputs. Therefore, the answer suggests that the multiplexer has a total of 32 different inputs.

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  • 49. 

    Minimum number of 2-input NAND gates required to realize AB'C

    • A.

      2

    • B.

      3

    • C.

      4

    • D.

      5

    Correct Answer
    D. 5
    Explanation
    To realize AB'C using NAND gates, we can break it down into smaller steps. First, we can take the complement of input B using a NAND gate. Then, we can take the product of A and B' using another NAND gate. Finally, we can take the product of the previous result and input C using a NAND gate. Therefore, a minimum of 5 2-input NAND gates are required to realize AB'C.

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  • 50. 

    A two bit Binary multiplier can be implemented using

    • A.

      2 input AND gates only

    • B.

      2 input XOR gates and 2 input AND gates only

    • C.

      Two 2 input NOR gates and one 2 input Xor gate

    • D.

      Xor gates and shift registers

    Correct Answer
    B. 2 input XOR gates and 2 input AND gates only
    Explanation
    A two-bit binary multiplier can be implemented using 2 input XOR gates and 2 input AND gates only. XOR gates are used to perform the addition operation, while AND gates are used for the multiplication operation. The XOR gates are responsible for adding the bits together, while the AND gates are used to determine when a carry is needed. By combining these two types of gates, the binary multiplier can perform the necessary calculations to multiply two binary numbers.

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