Q. A logic circuit provides the output $Y$ as per the following truth table: [table] A | B | Y ; 0 | 0 | 1 ; 0 | 1 | 0 ; 1 | 0 | 1 ; 1 | 1 | 0 [/table]
A) $\bar{B}$
B) $B$
C) $A.B+\bar{A}$
D) $A.\bar{B}+\bar{A}$
Explanation
[Solution by Claude - not from source] $Y=1$ when $B=0$ and $Y=0$ when $B=1$, independent of $A$. So $Y=\bar B$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.