Physics - Semiconductors Question with Solution | TestHub

Considering the truth table for the given circuit, \begin{array}{|c|c|c|c|c|} \hline \mathbf{A} & \mathbf{B} & \frac{\mathbf{C}=}{\mathbf{A}+\mathbf{B}} & \frac{\mathbf{D}=}{\mathbf{A}+\mathbf{B}} & \begin{array}{c} \mathbf{Y}= \\ \mathbf{C}+\mathbf{D} \end{array} \\ \hline 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 \\ \hline 1 & 1 & 0 & 0 & 0 \\ \hline \end{array} $\therefore$ The output of the given combination is same as that of a NOR gate.