MHT CET · Physics · Semiconductors
In the case of NAND gate, if A and B are the inputs and \(Y\) is the output then
- A \(\mathrm{Y}=\mathrm{A} \cdot \mathrm{B}\)
- B \(Y=\overrightarrow{A-B}\)
- C \(Y=\overline{A+B}\)
- D \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Step-by-step Solution
Detailed explanation
\(Y=\overline{A \cdot B}\)
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