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 $NOR$ gate.$1$. The output $X$ is the output of an $AND$ gate with inputs $P

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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 $NOR$ gate.$1$. The output $X$ is the output of an $AND$ gate with inputs $P$ and $Q$. Thus,$X = P \cdot Q$.$2$. The lower part consists of a $NOR$ gate with inputs $R$ and $S$,followed by a $NOT$ gate. The output of the $NOR$ gate is $\overline{R+S}$. Passing this through a $NOT$ gate gives $Y = \overline{\overline{R+S}} = R+S$.$3$. The final output $Z$ is the output of a $NOR$ gate with inputs $X$ and $Y$. Thus,$Z = \overline{X+Y}$.$4$. Given inputs: $P=0, Q=1, R=0, S=1$.$5$. Calculating $X$: $X = P \cdot Q = 0 \cdot 1 = 0$.$6$. Calculating $Y$: $Y = R + S = 0 + 1 = 1$.$7$. Calculating $Z$: $Z = \overline{X+Y} = \overline{0+1} = \overline{1} = 0$.Therefore,the new 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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