Q 12-14-057JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
For the circuit shown, the equivalent gate is
Answer: (A) OR gate
- Upper gate (NAND of $A$, $B$): $\overline{AB}$
- Lower gate (NOR of $A$, $B$): $\overline{A + B}$
- Output NAND:
$$Y = \overline{\overline{AB}\cdot\overline{A + B}} = AB + (A + B) = A + B$$
The circuit behaves as an OR gate.
Solution by Sreeraj P, M.Sc Physics