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To get the truth table shown, from the following logic circuit, the Gate $\mathrm{G}$ should be
\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
Options:
\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
Solution:
1559 Upvotes
Verified Answer
The correct answer is:
AND
Truth table for $\mathrm{Y}$, with the possible values of $\mathrm{C}$ is,
\begin{array}{|c|c|c|}
\hline \mathbf{A} & \mathbf{C} & \mathbf{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 0 & 0 \\
\hline 1 & 0,1 & 1 \\
\hline 1 & 0,1 & 1 \\
\hline
\end{array}
For gate $\mathrm{G}$
\begin{array}{|c|c|c|c|}
\hline & A & B & C \\
\hline (I) & 0 & 0 & 0 \\
\hline (II) & 0 & 1 & 0 \\
\hline (III) & 1 & 0 & 0,1 \\
\hline (IV) & 1 & 1 & 0,1 \\
\hline
\end{array}
$\mathrm{G}$ is not a NOT gate as NOT gate takes only one input. (II) indicates $\mathrm{G}$ is not a OR gate as OR gate would give high output for the inputs in (II). Also, (II) indicates it is not a XOR gate as XOR would also give high output for inputs in (II). Hence, the given truth table is satisfied only by AND gate.
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