In the logic circuit given below,if X=1 and Y | vedclass

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(A) Given inputs are $X=1$ and $Y=1$.$1$. The input $X$ passes through a $NOT$ gate,so the input to the $OR$ gate is $\bar{X} = 0$.$2$. The input $Y$

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$P=1, Q=1, R=0$

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(A) Given inputs are $X=1$ and $Y=1$.$1$. The input $X$ passes through a $NOT$ gate,so the input to the $OR$ gate is $\bar{X} = 0$.$2$. The input $Y$ is directly connected to the $OR$ gate,so the input is $1$.$3$. The output $P$ of the $OR$ gate is $P = \bar{X} + Y = 0 + 1 = 1$.$4$. The input $X$ is directly connected to the $NAND$ gate,so the input is $1$.$5$. The input $Y$ passes through a $NOT$ gate,so the input to the $NAND$ gate is $\bar{Y} = 0$.$6$. The output $Q$ of the $NAND$ gate is $Q = \overline{X \cdot \bar{Y}} = \overline{1 \cdot 0} = \overline{0} = 1$.$7$. Finally,$P$ and $Q$ are inputs to a $NOR$ gate to give output $R$.$8$. $R = \overline{P + Q} = \overline{1 + 1} = \overline{1} = 0$.Thus,$P=1, Q=1, R=0$.

In the logic circuit given below,if $X=1$ and $Y=1$,then the values of $P, Q$ and $R$ are:

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