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 OR gate respectively
• D. An OR gate and a NOT gate respectively.

Answer 1

Answer: (A)

An OR gate and an AND gate respectively

Answer 2

$C = \overline{\bar{A}.\bar{B}} = \overline{\bar{A} + \bar{B}} = A + B$ (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.