KCET · Physics · Electrostatics
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
- D AND-Gate
Answer & Solution
Correct Answer
(D) AND-Gate
Step-by-step Solution
Detailed explanation
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
\(\begin{array}{|l|l|l|l|l|}
\hline \mathbf{A} & \mathbf{B} & \overline{\boldsymbol{A}+\boldsymbol{A}} & \overline{\boldsymbol{B}+\boldsymbol{B}} & \mathbf{Y} \\
\hline 0 & 0 & 1 & 1 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 0 & 0 & 1 & 0 \\
\hline
\end{array}\)

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
\(\begin{array}{|l|l|l|l|l|}
\hline \mathbf{A} & \mathbf{B} & \overline{\boldsymbol{A}+\boldsymbol{A}} & \overline{\boldsymbol{B}+\boldsymbol{B}} & \mathbf{Y} \\
\hline 0 & 0 & 1 & 1 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 0 & 0 & 1 & 0 \\
\hline
\end{array}\)
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