In the circuit shown in the following figure, | vedclass

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(A) Let the output of the circuit be $y$. The circuit consists of three $NAND$ gates $(G_1, G_2, G_3)$ and two $NOT$ gates.$1$. The input to the first

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$0$

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(A) Let the output of the circuit be $y$. The circuit consists of three $NAND$ gates $(G_1, G_2, G_3)$ and two $NOT$ gates.$1$. The input to the first $NOT$ gate is $1$,so its output is $0$. This $0$ is fed to the second $NOT$ gate,making its output $1$. Thus,the input $B$ to gate $G_2$ is $1$.$2$. The input to gate $G_3$ is $0$ (from the first $NOT$ gate) and $y$ (feedback from the output). The output of $G_3$ is $\overline{0 \cdot y} = \overline{0} = 1$.$3$. The gate $G_1$ has inputs $0$ and $y$. Its output is $\overline{0 \cdot y} = \overline{0} = 1$. This output $1$ is the input $A$ to gate $G_2$.$4$. Now,gate $G_2$ has inputs $A = 1$ and $B = 1$. Its output is $y = \overline{A \cdot B} = \overline{1 \cdot 1} = \overline{1} = 0$.$5$. Since the output $y$ is $0$,the circuit is consistent. Therefore,the value of $y$ is $0$.

In the circuit shown in the following figure,the value of $y$ is:

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