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If the output of a NAND gate is given as input to a NOT gate, the resultant gate is
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2206 Upvotes
Verified Answer
The correct answer is:
AND
Truth table of NAND gate. The output of NAND gate is given as input to a NOT gate. The resultant gate is AND gate.
\begin{array}{|c|c|c|}
\hline \multicolumn{2}{|c|}{ Inputs } & Output \\
\hline \mathbf{A} & \mathbf{B} & \mathbf{Y} \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\begin{array}{|c|c|}
\hline \begin{array}{r}
Input (= output of \\
NAND gate)
\end{array} & \begin{array}{c}
Output of NOT \\
gate = AND \\
gate
\end{array} \\
\hline 1 & 0 \\
\hline 1 & 0 \\
\hline 1 & 0 \\
\hline 0 & 1 \\
\hline
\end{array}
\begin{array}{|c|c|c|}
\hline \multicolumn{2}{|c|}{ Inputs } & Output \\
\hline \mathbf{A} & \mathbf{B} & \mathbf{Y} \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\begin{array}{|c|c|}
\hline \begin{array}{r}
Input (= output of \\
NAND gate)
\end{array} & \begin{array}{c}
Output of NOT \\
gate = AND \\
gate
\end{array} \\
\hline 1 & 0 \\
\hline 1 & 0 \\
\hline 1 & 0 \\
\hline 0 & 1 \\
\hline
\end{array}
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