Question

The combination of 'NAND' gates shown here (in the figure below), 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)

$C=\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B$
(De Morgan's theorem)
Hence, output $C$ is equivalent to OR gate.

$C=\overline{\overline{A B} \cdot \overline{A B}}=\overline{\overline{A B}}+\overline{\overline{A B}}=A B+A B=A B$
In this case, output $C$ is equivalent to AND gate.