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Logic gates X and Y have the truth table...

Logic gates `X` and `Y` have the truth tables shown below

`|(P,Q,R,P,R),(0,0,0,0,1),(1,0,0,1,0),(0,1,0,,),(1,1,1,,)|`
When the output of `X` of connected to the input of `Y`, the resulting combination is equivalent to a single.

A

`NOT`gate

B

`OR` gate

C

`NAND` gate

D

`AND` gate

Text Solution

Verified by Experts

The correct Answer is:
C
The truth table of the resulting logic circuit by connecting `X` of `Y` is as follows
`|(P,Q,R,R),(0,0,0,1),(1,0,0,1),(0,1,0,1),(1,1,1,0)|`
Hence, from the truth table, the combination is equivalent to a single `NAND` gate. (`OR X` is an `AND`gate and `Y` is a `NOT` gate, thus the combination is `NOT AND` gate, i.e., a `NAND` gate)
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Knowledge Check

  • Logic gates X and Y have the truth tables shown below {:(P,Q,R,P,R),(0,0,0,0,1),(1,0,0,1,0),(0,1,0,,),(1,1,1,,):} When the output of X is connected to the input of Y , the resulting combination is equivalent to a single.

    A
    `NOT` gate
    B
    `OR` gate
    C
    `NOR` gate
    D
    `NAND` gate

  • The truth table given below is for: |(A,B,X),(0,0,0),(0,1,0),(1,0,0),(1,1,1)|

    A
    `XOR`
    B
    `AND`
    C
    `XNOR`
    D
    `OR`

  • The truth table given below is for which gate? |(A,B,C),(0,0,1),(0,1,1),(1,0,1),(1,1,0)|

    A
    `XOR`
    B
    `OR`
    C
    `AND`
    D
    `NAND`

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