For the given digital circuit, write the truth table and identify the logic gate it represents :

• A

  OR-Gate
• B

  NOR-Gate
• C

  NAND - Gate
• Correct answer
  D

  AND-Gate

Solution

The given digital circuit is shown as follows:

Using Boolean law: \( A+A=A \), we get
\( \overline{A+A}=\bar{A} \) and \( \overline{B+B}=\bar{B} \)
Using de Morgan's theorem: \( \overline{A+B}=\bar{A} \cdot \bar{B} \)
\( \bar{A}+\bar{B}=\overline{\bar{A}} \cdot \overline{\bar{B}} \)
and using \( \bar{A}=A \), we get
\( \bar{A} \cdot \bar{B}=A \cdot B \)
Therefore, \( Y=A \cdot B \)
Truth table
$\begin{array}{|l|l|l|l|l|} \hline \mathbf{A} & \mathbf{B} & \overline{\boldsymbol{A}+\boldsymbol{A}} & \overline{\boldsymbol{B}+\boldsymbol{B}} & \mathbf{Y} \\ \hline 0 & 0 & 1 & 1 & 0 \\ \hline 0 & 1 & 1 & 0 & 0 \\ \hline 1 & 0 & 0 & 1 & 0 \\ \hline \end{array}$

Hence, the correct answer is option: 'D'.