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To get output '1' at R, for the given logic gate circuit the input values must be:

  • Option 1)

    X=0, Y=1

     

  • Option 2)

    X=1, Y=1

  • Option 3)

    X=1, Y=0

  • Option 4)

    X=0, Y=0

Answers (1)

best_answer

 

D'morgan's Theorem -

1)     \overline{A+B }= \bar{A}\cdot \bar{B}

2)      \overline{A\cdot B }= \bar{A}+ \bar{B}

3)    \overline{\bar A+ \bar B }= {A}\cdot {B}

4)    \overline{\bar A\cdot \bar B }= {A}+{B}

- wherein

A and B are input.

we want

R=1

and P=\bar{X}+Y\\\\\\Q= \: \overline{\bar{Y}.X}\\\\R= output =\overline{P+Q}\\\\R=1\Rightarrow P+Q=0\\\\\Rightarrow \bar{X}+Y+\left ( \overline{\bar{Y}.X} \right )=0

So For

X=1,Y=0

 

R=1

 


Option 1)

X=0, Y=1

 

Option 2)

X=1, Y=1

Option 3)

X=1, Y=0

Option 4)

X=0, Y=0

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