Q.
The circuit is equivalent to
A) NOR gate
B) OR gate
C) AND gate
D) NAND gate
Explanation
(a) The output of NOR gate gives $y^{'} = \overline{A + B}$
The output of NAND gate gives
$y^{''} = \overline{\overline{(A + B)}.\overline{(A + B)}}$
$= \overline{\overline{(A + B)}} + \overline{\overline{(A + B)}}$
$= (A + B) + (A + B) = A + B$
The output of NOT gate, $y = \overline{A + B}$
It is a boolen expression for NOR gate
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.