Question

Correct truth table for given circuit is.

• A. $\begin{array}{ccc}A& B& X\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 0\\ 1& 1& 0\end{array}$
• B. $\begin{array}{ccc}A& B& X\\ 0& 0& 1\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$
• C. $\begin{array}{ccc}A& B& X\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 1\\ 1& 1& 1\end{array}$
• D. $\begin{array}{ccc}A& B& X\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

Answer 1

The correct option is C $\begin{array}{ccc}A& B& X\\ 0& 0& 1\\ 0& 1& 0\\ 1& 0& 1\\ 1& 1& 1\end{array}$

Input at A goes to OR gate and input at B goes to NAND gate.

Similarly, whole truth table can be constructed.