Semiconductor Electronics question for JEE Main (JEE Main 2026 (24 Jan, Shift 2)), with solution

Answer: (A) $\begin{array}{|c|c|c|}\hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array}$

- The top AND gate has both inputs connected to $A$: output $= A\cdot A = A$.
- The lower gate is a NAND of $A$ and $B$: output $\overline{AB}$. It passes through an AND gate with both inputs tied together, which leaves it unchanged: $\overline{AB}$.
- The final AND gate: $Y = A\cdot\overline{AB}$.

$A=0$: $Y = 0$ (for both $B$).

$A=1, B=0$: $\overline{AB} = 1$, $Y = 1$.

$A=1, B=1$: $\overline{AB} = 0$, $Y = 0$.

So $Y = A\overline B$: outputs $0, 0, 1, 0$. This is table (A).

Solution by Sreeraj P, M.Sc Physics