Digital electronics miscellaneous

Digital electronics miscellaneous

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Digital Electronics

Digital electronics miscellaneous
  1. Which of the following Boolean algebra rules is correct? (A) (B) (C) (D)








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    ● A·A = 1 → Incorrect
    ● A + AB = A + B → Incorrect
    ● A + AB = A + B →→ Correct
    ● A (A + B) = B → Incorrect

    Correct Option: C

    ● A·A = 1 → Incorrect
    ● A + AB = A + B → Incorrect
    ● A + AB = A + B →→ Correct
    ● A (A + B) = B → Incorrect

  1. The Boolean expression Y Z + X Z + X Z is logically equivalent to—








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    F = Y Z + X Z + X Z
    From K-map, we get F = X + Y Z
    Alternative method
    F = Y Z + X Z + XZ
    or F = Y Z + X ( Z + Z)
    or F = Y Z + X


    Correct Option: A

    F = Y Z + X Z + X Z
    From K-map, we get F = X + Y Z
    Alternative method
    F = Y Z + X Z + XZ
    or F = Y Z + X ( Z + Z)
    or F = Y Z + X


  1. In Boolean algebra if F = (A + B). ( A + C), then—








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    NA

    Correct Option: C

    NA

  1. Which one of the following statement is correct? For a 4- input NOR gate, when only two inputs are to be used, the best option for the used input is to—








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    The given four inputs NOR gate
    When only two inputs are used then rest inputs must be connected to ground, because if any input is connected to Vcc we will get output 0 irrespective of the applied input. However if inputs are open we get floating output. Hence option (A) is the best choice.


    Correct Option: A

    The given four inputs NOR gate
    When only two inputs are used then rest inputs must be connected to ground, because if any input is connected to Vcc we will get output 0 irrespective of the applied input. However if inputs are open we get floating output. Hence option (A) is the best choice.


  1. Minimum no. of NAND gates require to implement the logic Y = AB + A + B + C








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    Given Y = AB + A + B + C
    or Y = A + B + A + B C
    or Y = 1 + B + B C
    or Y = 1
    Therefore, no NAND gates required to implement the logic.

    Correct Option: A

    Given Y = AB + A + B + C
    or Y = A + B + A + B C
    or Y = 1 + B + B C
    or Y = 1
    Therefore, no NAND gates required to implement the logic.