In the circuit shown in the following figure, | vedclass

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(A) Let the output be $Y$. The circuit consists of $NAND$ gates. $1$. The input $1$ passes through a $NOT$ gate,becoming $0$. This $0$ is fed to the b

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let the output be $Y$. The circuit consists of $NAND$ gates. $1$. The input $1$ passes through a $NOT$ gate,becoming $0$. This $0$ is fed to the bottom $NAND$ gate and also through another $NOT$ gate to input $B$ of gate $G_2$,so $B = 1$. $2$. The output $Y$ is fed back to the input of gate $G_1$. Gate $G_1$ has inputs $0$ and $Y$. Its output is $A = \overline{0 \cdot Y} = \overline{0} = 1$. $3$. Gate $G_2$ has inputs $A = 1$ and $B = 1$. Its output is $Y = \overline{A \cdot B} = \overline{1 \cdot 1} = 0$. $4$. The bottom $NAND$ gate has inputs $0$ and $0$ (from the first $NOT$ gate),so its output is $\overline{0 \cdot 0} = 1$. However,the feedback loop and the structure define $Y$ as the output of $G_2$. Thus,the value of $Y$ is $0$.

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

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