The truth table for the circuit given in the fig is

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Correct Answer

(D) \( {\begin{array}{*{20}{c}}
A&B&Y\\
0&0&1\\
0&1&1\\
1&0&0\\
1&1&0
\end{array}} \)

Step-by-step Solution

Detailed explanation

\(C=A+B\)

and \(y=\overline{A . C}\)

\(\begin{array}{l|l|l|l|l|}\hline A & {\mathrm{B}} & {\mathrm{C}=(\mathrm{A}+\mathrm{B})} & {\mathrm{A} \mathrm{C}} & {\mathrm{y}=\overline{\mathrm{A} . \mathrm{C}}} \\ \hline 0 & {0} & {0} & {0} & {1} \\ {0} & {1} & {1} & {0} & {1} \\ {1} & {0} & {1} & {1} & {0} \\ {1} & {1} & {1} & {1} & {0} \\ \hline\end{array}\)

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