Q.
Truth table for system of four NAND gates as shown in figure is
A) $\left| \begin{array}{r}
A\ \ \ B\ \ \ Y \\
0\ \ \ 0\ \ \ 0 \\
0\ \ \ \ 1\ \ \ 1 \\
1\ \ \ \ 0\ \ \ 1 \\
1\ \ \ \ 1\ \ \ 0
\end{array} \right|$
B) $\left| \begin{array}{r}
A\ \ \ B\ \ \ C \\
0\ \ \ 0\ \ \ 0 \\
0\ \ \ \ 1\ \ \ 0 \\
1\ \ \ \ 0\ \ \ 1 \\
1\ \ \ \ 1\ \ \ 1
\end{array} \right|$
C) $\left| \begin{array}{r}
A\ \ \ B\ \ \ Y \\
0\ \ \ 0\ \ \ 1 \\
0\ \ \ \ 1\ \ \ 1 \\
1\ \ \ \ 0\ \ \ 0 \\
1\ \ \ \ 1\ \ \ 0
\end{array} \right|$
D) $\left| \begin{array}{r}
A\ \ \ B\ \ \ Y \\
0\ \ \ 0\ \ \ \ 1 \\
0\ \ \ \ 1\ \ \ 0 \\
1\ \ \ \ 0\ \ \ 0 \\
1\ \ \ \ 1\ \ \ 1
\end{array} \right|$
Explanation
(a) [image:14_image209.png]
[table] A | B | C | $D = \overline{A.C}$ | $E = \overline{C.B}$ | Y ; 0 | 0 | 1 | 1 | 1 | 0 ; 0 | 1 | 1 | 1 | 0 | 1 ; 1 | 0 | 1 | 0 | 1 | 1 ; 1 | 1 | 0 | 1 | 1 | 0 [/table]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.