Question

The truth table for the following logic circuit is

• A. $\left|\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$
• B. $\left|\begin{array}{ccc}A& B& Y\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}\right|$
• C. $\left|\begin{array}{ccc}A& B& Y\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$
• D. $\left|\begin{array}{ccc}A& B& Y\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 0\\ 1& 1& 1\end{array}\right|$

Answer 1

The correct option is A $\left|\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$

The truth table for the given logic circuit is

$A$	$B$	$\overline{A}$	$\overline{B}$	$\overline{A}\cdot B$	$A\overline{B}$	$Y=\overline{A}B+A\overline{B}$
$0$	$0$	$1$	$1$	$0$	$0$	$0$
$0$	$1$	$1$	$0$	$1$	$0$	$1$
$1$	$0$	$0$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$0$	$0$	$0$