The resultant gate and its Boolean expression in the given circuit is
The resultant gate and its Boolean expression in the given circuit is

- $\mathrm{NOR}, \overline{\mathrm{A}+\mathrm{B}}$
- $AND, A.B$
- $\text { OR, A } + B$
- $NAND, \overline{\mathrm{AB}}$
Solution
Truth table:
\begin{array}{|l|l|l|l|l|}
\hline A & B & C & D & Y \\
\hline 0 & 0 & 1 & 1 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 0 & 0 & 1 & 0 \\
\hline 1 & 1 & 0 & 0 & 1 \\
\hline
\end{array}
This is truth table of AND gate. $\therefore \mathrm{Y}=\mathrm{A} \cdot \mathrm{B}$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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