Q 12-14-134JEE MainJEE Main 2021 (24 Feb, Shift 2)Medium
The logic circuit shown above is equivalent to:
Answer: (C) see figure
$C = \overline{A + \overline B} = \overline A\cdot\overline{\overline B} = \overline A\cdot B$ (De Morgan's theorem).
This is an AND gate with $A$ passed through a NOT gate and $B$ fed directly, which is option (3).
Solution by Sreeraj P, M.Sc Physics