Question:

The logic performed by the circuit shown in the figure is equivalent to: \begin{center} \includegraphics[width=0.4\textwidth]{37.png} % Replace with your image file \end{center}

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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.
Updated On: Jan 22, 2025
  • \( AND \)
  • \( NAND \)
  • \( OR \)
  • \( NOR \)
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The Correct Option is A

Solution and Explanation

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