Q.
In the following circuit, the output $Y$ for all possible inputs $A$ and $B$ is expressed by the truth table
$A$ $B$ $Y$
A) $\left| \begin{array}{r}
0\ \ \ 0\ \ \ \ 0 \\
0\ \ \ 1\ \ \ 0 \\
1\ \ \ 0\ \ \ 0 \\
1\ \ \ \ 1\ \ \ 1
\end{array} \right|$
B) $\left| \begin{array}{r}
0\ \ \ 0\ \ \ \ 1 \\
0\ \ \ 1\ \ \ 1 \\
1\ \ \ 0\ \ \ 1 \\
1\ \ \ \ 1\ \ \ 0
\end{array} \right|$
C) $\left| \begin{array}{r}
0\ \ \ 0\ \ \ \ 1 \\
0\ \ \ 1\ \ \ 0 \\
1\ \ \ 0\ \ \ 0 \\
1\ \ \ \ 1\ \ \ 0
\end{array} \right|$
D) $\left| \begin{array}{r}
0\ \ \ 0\ \ \ \ 0 \\
0\ \ \ 1\ \ \ 1 \\
1\ \ \ 0\ \ \ 1 \\
1\ \ \ \ 1\ \ \ 1
\end{array} \right|$
Explanation
(d) Boolean expression of the given circuit is $Y = \overline{\overline{A + B} + \overline{A + B}} = A + B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.