# The truth table for the circuit given in the fig. is : Option 1) $\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 1\\ 1& 0 & 0\\ 1& 1 & 0 \end{vmatrix}$ Option 2) $\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 1\\ 1& 0 & 1\\ 1& 1 & 1 \end{vmatrix}$ Option 3) $\begin{vmatrix} A & B &Y \\ 0 & 0 &0 \\ 0& 1 & 0\\ 1& 0 & 1\\ 1& 1 & 1 \end{vmatrix}$ Option 4) $\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 0\\ 1& 0 & 0\\ 1& 1 & 0 \end{vmatrix}$

NAND gate -

NOT + AND gate

- wherein

$Y= \overline{A\cdot B}$

A and B are input

Y is out put

By putting values of the $4$ cases, you can see that the option (1) is correct

Option 1)

$\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 1\\ 1& 0 & 0\\ 1& 1 & 0 \end{vmatrix}$

Option 2)

$\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 1\\ 1& 0 & 1\\ 1& 1 & 1 \end{vmatrix}$

Option 3)

$\begin{vmatrix} A & B &Y \\ 0 & 0 &0 \\ 0& 1 & 0\\ 1& 0 & 1\\ 1& 1 & 1 \end{vmatrix}$

Option 4)

$\begin{vmatrix} A & B &Y \\ 0 & 0 &1 \\ 0& 1 & 0\\ 1& 0 & 0\\ 1& 1 & 0 \end{vmatrix}$

Exams
Articles
Questions