Truth table for the given circuit will be

Truth table for the given circuit will be
  1. $$ \begin{array}{cc|c} \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} $$
  2. $$ \begin{array}{cc|c} \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array} $$
  3. $$ \begin{array}{cc|c} x & y & z \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array} $$
  4. $$ \begin{array}{cc|c} \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array} $$

Solution

Truth table of the circuit is as follows $ \begin{array}{|c|c|c|c|c|c|} \hline x & y & \bar{x} & a=x \cdot y & b=\bar{x} \cdot y & z=\overline{a \cdot b} \\ \hline 0 & 0 & 1 & 0 & 0 & 1 \\ \hline 0 & 1 & 1 & 0 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 0 & 1 \\ \hline 1 & 1 & 0 & 1 & 0 & 1 \\ \hline \end{array} $

Asked in: JEE Main 2018 (15 Apr Shift 2 Online)

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