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

In the case of NAND gate, if A and B are the inputs and $Y$ is the output then
  1. $\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}$
  2. $Y=\overrightarrow{A-B}$
  3. $Y=\overline{A+B}$
  4. $\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}$

Solution

$\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}$

Asked in: MHT CET 2023 (11 May Shift 1)

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