The combination of gates shown below yields

  • Option 1)

    NAND gate

  • Option 2)

    OR gate

  • Option 3)

    NOT gate

  • Option 4)

    XOR gate

 

Answers (1)

As we learnt in

NAND gate -

NOT + AND gate

- wherein

Y= \overline{A\cdot B}

A and B are input

Y is out put

 

 

X=\overline{\bar{a}\ .\ \bar{b}}=A+B (From D'morgan law)

Hence, This circuit represent OR gate.

Correct optio is 2.

 


Option 1)

NAND gate

This is an incorrect option.

Option 2)

OR gate

This is the correct option.

Option 3)

NOT gate

This is an incorrect option.

Option 4)

XOR gate

This is an incorrect option.

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