question_answer

The combination of ?NAND? gates shown here under (figure) are equivalent to                  [Haryana CEET 1996]

A)            An OR gate and an AND gate respectively

B)            An AND gate and a NOT gate respectively

C)            An AND gate and an OR gate respectively

D)            An OR gate and a NOT gate respectively.

Correct Answer: A

Solution :

                   \[C=\overline{\bar{A}.\bar{B}}=\overline{\bar{A}+\bar{B}}=A+B\] (De morgan?s theorem) Hence output C is equivalent to OR gate. \[C=\overline{\overline{AB}.\overline{AB}}=\overline{\overline{AB}+\overline{AB}}=AB+AB=AB\] In this case output C is equivalent to AND gate.