In the logic circuit shown in the figure,if i | vedclass

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(A) Given inputs are $A = 0$ and $B = 1$.$1$. The input $A=0$ goes to the $NOR$ gate and also to a $NOT$ gate. The output of this $NOT$ gate is $1$.$2

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(A) Given inputs are $A = 0$ and $B = 1$.$1$. The input $A=0$ goes to the $NOR$ gate and also to a $NOT$ gate. The output of this $NOT$ gate is $1$.$2$. The input $B=1$ goes to a $NOT$ gate,so its output is $0$. This $0$ is fed to the $NAND$ gate and also to another $NOT$ gate,which outputs $1$.$3$. The $NOR$ gate receives $A=0$ and the output of the second $NOT$ gate,which is $1$. Thus,the $NOR$ gate output is $\overline{0+1} = \overline{1} = 0$.$4$. The $NAND$ gate receives the output of the first $NOT$ gate $(1)$ and the output of the $NOT$ gate connected to $B$ $(0)$. Thus,the $NAND$ gate output is $\overline{1 \cdot 0} = \overline{0} = 1$.$5$. Finally,the $AND$ gate receives the outputs of the $NOR$ gate $(0)$ and the $NAND$ gate $(1)$. The output $Y$ is $0 \cdot 1 = 0$.Therefore,$x = 0$.

In the logic circuit shown in the figure,if inputs $A$ and $B$ are $0$ and $1$ respectively,the output at $Y$ is $x$. The value of $x$ is...........

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