Question

The logic performed by the circuit shown in the figure is equivalent to: 

• A. \( AND \)
• B. \( NAND \)
• C. \( OR \)
• D. \( NOR \)

Hint:

De Morgan’s laws simplify logic circuits: \( \overline{A + B} = A \cdot B \) and \( \overline{A \cdot B} = A + B \). These transformations can turn complex circuits into simpler ones.

Answer 1

The Correct Option is A

The circuit performs the following logical operations:

1. The first two NOT gates take inputs \( a \) and \( b \), producing the complements \( \bar{a} \) and \( \bar{b} \) respectively.
\[ \text{Output of the first NOT gate: } \bar{a}, \quad \text{Output of the second NOT gate: } \bar{b}. \]
2. The OR gate then combines the complements:
\[ \text{Output of the OR gate: } \bar{a} + \bar{b}. \]
3. The result from the OR gate is then complemented by another NOT gate, yielding the final output:
\[ Y = \overline{\bar{a} + \bar{b}}. \]

Using De Morgan's law, we simplify:

\[ Y = a \cdot b. \]

Thus, the circuit implements the logic of an AND gate.

Truth Table:

\[ \begin{array}{|c|c|c|} \hline A & B & Y = A \cdot B \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{array} \]

The output matches the truth table of an AND gate.

Final Answer:

\[ \boxed{\text{AND Gate}} \]