The combination of NAND gates shown here under figure, are equivalent to

Question

The combination of NAND gates shown here under figure, are equivalent to (A) an OR gate and an AND gate respectively (B) an AND gate and a NOT gate respectively (C) an AND gate and an O R gate respectively (D) an OR gate and a NOT gate respectively

Tardigrade Admin · Accepted Answer

For first case, the Boolean expression is, Y= overline barA ⋅ barB=A+ B hence for OR gate and for second case, the Boolean expression is Y= overline barA ⋅ barB=A ⋅ B, and hence for AND gate.

Q. The combination of NAND gates shown here under figure, are equivalent to

an OR gate and an AND gate respectively

an AND gate and a NOT gate respectively

an AND gate and an O R gate respectively

an OR gate and a NOT gate respectively

For first case, the Boolean expression is, $Y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B}} = A + B$ hence for OR gate and for second case, the Boolean expression is $Y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B}} = A \cdot B$ , and hence for AND gate.

Q. The combination of NAND gates shown here under figure, are equivalent to

an OR gate and an AND gate respectively

an AND gate and a NOT gate respectively

an AND gate and an O R gate respectively

an OR gate and a NOT gate respectively

For first case, the Boolean expression is, Y=Aˉ⋅Bˉ=A+B hence for OR gate and for second case, the Boolean expression is Y=Aˉ⋅Bˉ=A⋅B, and hence for AND gate.

For first case, the Boolean expression is, $Y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B}} = A + B$ hence for OR gate and for second case, the Boolean expression is $Y = \overline{\overset{ˉ}{A} \cdot \overset{ˉ}{B}} = A \cdot B$ , and hence for AND gate.