In the circuit shown in the following figure, | vedclass

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(A) Let the output be $Y$. The circuit consists of three $NAND$ gates $(G_1, G_2, G_3)$ and two $NOT$ gates.$1$. The input $1$ passes through a $NOT$

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$0$

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(A) Let the output be $Y$. The circuit consists of three $NAND$ gates $(G_1, G_2, G_3)$ and two $NOT$ gates.$1$. The input $1$ passes through a $NOT$ gate,becoming $0$. This $0$ is fed to the input of $G_3$ and also passes through another $NOT$ gate to become $1$,which is the input $B$ for gate $G_2$.$2$. Gate $G_3$ has inputs $0$ and $Y$ (feedback from output). Since one input is $0$,the output of $G_3$ is $NAND(0, Y) = 1$. This $1$ is fed back to $G_1$.$3$. Gate $G_1$ has inputs $0$ and $Y$. Its output is $NAND(0, Y) = 1$. This $1$ is the input $A$ for gate $G_2$.$4$. Gate $G_2$ has inputs $A = 1$ and $B = 1$. Its output is $Y = NAND(1, 1) = 0$.Thus,the final output $Y$ is $0$.

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

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