In the given logic circuit diagram,A,B,and C | vedclass

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(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate.The inputs to the $NOR$ gate are $B$ and $C$,so its output is $\overline{B+C}

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when $A=1, B=0, C=0$

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(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate.The inputs to the $NOR$ gate are $B$ and $C$,so its output is $\overline{B+C}$.This output is then fed into an $AND$ gate along with input $A$.Therefore,the final output $Y$ is given by the Boolean expression: $Y = A \cdot \overline{(B+C)}$.For the output $Y$ to be ' $HIGH$ ' $(Y=1)$,both $A$ must be $1$ and $\overline{(B+C)}$ must be $1$.$\overline{(B+C)} = 1$ implies that $(B+C) = 0$,which means both $B=0$ and $C=0$.Thus,$Y=1$ when $A=1, B=0$,and $C=0$.

In the given logic circuit diagram,$A$,$B$,and $C$ are the inputs,and $Y$ is the output. The output $Y$ is ' $HIGH$ ' (i.e.,$Y=1$):

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The circuit diagram shows a logic combination. Initially, for inputs $P, Q, R$ and $S$ all at state $1$, the outputs are $X=1, Y=1$ and $Z=0$. When inputs $P$ and $R$ change to state $0$ while inputs $Q$ and $S$ remain at state $1$, what are the new states of outputs $X, Y$ and $Z$?

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