Question

In the case of NAND gate, if A and B are inputs and Y is the output then

• A. \(Y = \overline{A+B}\)
• B. \(Y = A\cdot B\)
• C. \(Y = \overline{A\cdot B}\)
• D. \(Y = A+B\)

Hint:

NAND means NOT-AND, so put a bar over A times B.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A NAND gate is a logic gate whose name means NOT-AND. It is an AND gate followed by a NOT gate.

Step 2: Key Formula or Approach:
The AND gate has output \(A\cdot B\). Inverting it gives the NAND output: \[ Y = \overline{A\cdot B} \]

Step 3: Truth table:
For inputs (0,0), (0,1), (1,0), (1,1), the AND output is 0, 0, 0, 1. Inverting gives the NAND output 1, 1, 1, 0. The output is low only when both inputs are high.

Step 4: Check the options:
Option (A) \(\overline{A+B}\) is the NOR gate. Option (B) \(A\cdot B\) is the AND gate. Option (D) \(A+B\) is the OR gate. Only option (C) is \(\overline{A\cdot B}\), the NAND gate.

Final Answer:
The NAND gate output is \(Y = \overline{A\cdot B}\), option (C). \[ \boxed{Y=\overline{A\cdot B}} \]