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The output of following combination is same as that of
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Verified Answer
The correct answer is:
NOR gate
Considering the truth table for the given circuit,
\begin{array}{|c|c|c|c|c|}
\hline \mathbf{A} & \mathbf{B} & \frac{\mathbf{C}=}{\mathbf{A}+\mathbf{B}} & \frac{\mathbf{D}=}{\mathbf{A}+\mathbf{B}} & \begin{array}{c}
\mathbf{Y}= \\
\mathbf{C}+\mathbf{D}
\end{array} \\
\hline 0 & 0 & 1 & 1 & 1 \\
\hline 0 & 1 & 0 & 0 & 0 \\
\hline 1 & 0 & 0 & 0 & 0 \\
\hline 1 & 1 & 0 & 0 & 0 \\
\hline
\end{array}
$\therefore$ The output of the given combination is same as that of a NOR gate.
\begin{array}{|c|c|c|c|c|}
\hline \mathbf{A} & \mathbf{B} & \frac{\mathbf{C}=}{\mathbf{A}+\mathbf{B}} & \frac{\mathbf{D}=}{\mathbf{A}+\mathbf{B}} & \begin{array}{c}
\mathbf{Y}= \\
\mathbf{C}+\mathbf{D}
\end{array} \\
\hline 0 & 0 & 1 & 1 & 1 \\
\hline 0 & 1 & 0 & 0 & 0 \\
\hline 1 & 0 & 0 & 0 & 0 \\
\hline 1 & 1 & 0 & 0 & 0 \\
\hline
\end{array}
$\therefore$ The output of the given combination is same as that of a NOR gate.
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