Question

The truth table for the circuit given in the fig. is:

• A. $\begin{vmatrix}A&B&Y\\ 0&0&1\\ 0&1&1\\ 1&0&0\\ 1&1&0\end{vmatrix}$
• B. $\begin{vmatrix}A&B&Y\\ 0&0&0\\ 0&1&0\\ 1&0&1\\ 1&1&1\end{vmatrix}$
• C. $\begin{vmatrix}A&B&Y\\ 0&0&1\\ 0&1&0\\ 1&0&0\\ 1&1&0\end{vmatrix}$
• D. $\begin{vmatrix}A&B&Y\\ 0&0&1\\ 0&1&1\\ 1&0&1\\ 1&1&1\end{vmatrix}$

Answer 1

Answer: (A)

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