In the digital circuit shown in the figure,fo | vedclass

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(B) $1$. The inputs to the first $NAND$ gate are $1$ and $1$. The output of the $NAND$ gate is $\overline{1 \cdot 1} = \overline{1} = 0$.$2$. This out

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$P=0, Q=0$

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(B) $1$. The inputs to the first $NAND$ gate are $1$ and $1$. The output of the $NAND$ gate is $\overline{1 \cdot 1} = \overline{1} = 0$.$2$. This output $0$ is fed to the $AND$ gate and the $NOT$ gate.$3$. The inputs to the $AND$ gate are $0$ and $0$ (since the output of the $NAND$ gate is $0$ and it is connected to both inputs of the $AND$ gate). Thus,$P = 0 \cdot 0 = 0$.$4$. The inputs to the $OR$ gate are the outputs of the two $NOT$ gates connected to the initial inputs $1$ and $1$. The inputs to the $OR$ gate are $\overline{1} = 0$ and $\overline{1} = 0$. The output of the $OR$ gate is $0 + 0 = 0$.$5$. The $NOR$ gate receives two inputs: one from the $NOT$ gate (which inverts the $NAND$ output $0$ to $1$) and one from the $OR$ gate output $0$. The output of the $NOR$ gate is $Q = \overline{1 + 0} = \overline{1} = 0$.$6$. Therefore,$P = 0$ and $Q = 0$.

In the digital circuit shown in the figure,for the given inputs,the $P$ and $Q$ values are:

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