The circuit diagram shows a logic combination | vedclass

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(C) The circuit consists of an $AND$ gate,a $NOR$ gate,a $NOT$ gate,and a final $NOR$ gate.$1$. The $AND$ gate takes inputs $P$ and $Q$. Its output is

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$0, 1, 0$

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(C) The circuit consists of an $AND$ gate,a $NOR$ gate,a $NOT$ gate,and a final $NOR$ gate.$1$. The $AND$ gate takes inputs $P$ and $Q$. Its output is $X = P \cdot Q$.$2$. The first $NOR$ gate takes inputs $R$ and $S$. Its output is $\overline{R + S}$.$3$. This output passes through a $NOT$ gate to give $Y = \overline{\overline{R + S}} = R + S$.$4$. The final $NOR$ gate takes inputs $X$ and $Y$ to give $Z = \overline{X + Y}$.Given new inputs: $P = 0, Q = 1, R = 0, S = 1$.- Calculate $X$: $X = P \cdot Q = 0 \cdot 1 = 0$.- Calculate $Y$: $Y = R + S = 0 + 1 = 1$.- Calculate $Z$: $Z = \overline{X + Y} = \overline{0 + 1} = \overline{1} = 0$.Thus,the states of outputs $X, Y$ and $Z$ are $0, 1, 0$ respectively.

The circuit diagram shows a logic combination with the states of outputs $X, Y$ and $Z$ given for inputs $P, Q, R$ and $S$ all at state $1$. When inputs $P$ and $R$ change to state $0$ with inputs $Q$ and $S$ still at $1$,the states of outputs $X, Y$ and $Z$ change to:

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