BITSAT 2009
- Exam
- BITSAT
- Year
- 2009
- Subject
- Physics
- Chapter
- Semiconductors
BITSAT 2009 · Physics · Semiconductors
The truth table given below correspond to the logic gate tabular...
The truth table given below correspond to the logic gate
$\begin{tabular}{cc|c}\mathrm{A} & \mathrm{B} & \mathrm{X} \\
\hline 0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{tabular}
Options
- A
- BNOR
- CAND
- DNAND
Correct answer
B. NOR
Step-by-step solution
The given gate is a NOR gate. The output is high, when all inputs are low.
Truth Table:
\begin{tabular}{|c|c|c|}
\hline \mathbf{A} & \mathbf{B} & \mathbf{Y} \\
\hline 0 & 0 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline \mu lticolumn{2}{|c|} { Logic Symbols } \\
\hline
\end{tabular}
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Boolean expression:\overline{\mathrm{A}+\mathrm{B}}=\mathrm{Y}