| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
A) \[Y=\overline{A}+\overline{B}\]
B) \[Y=\overline{A+B}\]
C) \[Y=\overline{A}+B\]
D) \[Y=A.B\]
Correct Answer: A
Solution :
The given truth table corresponds to NAND gate whose Boolean expression is, \[Y=\overline{A.B}\] or \[Y=\overline{A}+\overline{B}\] (De Morgan's theorem)
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