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

Answer 1

Answer: (A)

Output, $Y=A\overline{B}+\overline{A}B$ $\begin{array}{cc}A=0& B=0\\ \overline{A}=1& \overline{B}=0\end{array}]Y=0$
$\begin{array}{cc}A=0& B=1\\ \overline{A}=1& \overline{B}=0\end{array}]Y=1$
$\begin{array}{cc}A=1& B=0\\ \overline{A}=0& \overline{B}=1\end{array}]Y=1$
$\begin{array}{cc}A=1& B=1\\ \overline{A}=0& \overline{B}=0\end{array}]Y=0$
Hence, option is correct.