Question:

What is the Boolean equation for a NAND gate with inputs \(A\) and \(B\)?}

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De Morgan's Theorem: \(\overline{AB} = \overline{A} + \overline{B}\) and \(\overline{A+B} = \overline{A}\cdot\overline{B}\). NAND is the complement of AND.
Updated On: Apr 8, 2026
  • \(Y = A + B\)
  • \(Y = \overline{A+B}\)
  • \(Y = \overline{A} + \overline{B}\)
  • \(Y = \overline{A \cdot B}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
A NAND gate produces the complement of an AND operation.
Step 2: Detailed Explanation:
By definition: NAND gate output is \(\overline{A \cdot B}\). By De Morgan's theorem, this is equivalent to \(\overline{A} + \overline{B}\), but the Boolean equation is \(\overline{A \cdot B}\).
Step 3: Final Answer:
Boolean equation: \(Y = \overline{A \cdot B}\).
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