Question 190 of 209

The truth table for the circuit given in the figure is:

A$\left|\begin{matrix}\begin{matrix}A & B & Y\end{matrix} \\ \begin{matrix}0 & 0 & 0\end{matrix} \\ \begin{matrix}0 & 1 & 0\end{matrix} \\ \begin{matrix}1 & 0 & 1\end{matrix} \\ \begin{matrix}1 & 1 & 1\end{matrix}\end{matrix}\right|$B$\left|\begin{matrix}\begin{matrix}A & B & Y\end{matrix} \\ \begin{matrix}0 & 0 & 1\end{matrix} \\ \begin{matrix}0 & 1 & 0\end{matrix} \\ \begin{matrix}1 & 0 & 0\end{matrix} \\ \begin{matrix}1 & 1 & 0\end{matrix}\end{matrix}\right|$C$\left|\begin{matrix}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{matrix}\right|$D$\left|\begin{matrix}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{matrix}\right|$

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Answer: (D) $\left|\begin{matrix}A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{matrix}\right|$

$C=A+B$
and $y=\bar{A.C}$
$\begin{matrix}A & B & C=(A+B) & AC & y=\bar{A.C} \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0\end{matrix}$

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