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(B) The given circuit consists of an $OR$ gate,an $AND$ gate,and a final $OR$ gate.Let the inputs be $A$ and $B$. The third input is fixed at $1$.The

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$A=0, B=0$

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(B) The given circuit consists of an $OR$ gate,an $AND$ gate,and a final $OR$ gate.Let the inputs be $A$ and $B$. The third input is fixed at $1$.The output of the first $OR$ gate is $(A + B)$.The output of the $AND$ gate is $(B \cdot 1) = B$.The final output $Y$ is the $OR$ operation of these two outputs:$Y = (A + B) + B$Using Boolean algebra laws,specifically the idempotent law $(B + B = B)$,we simplify the expression:$Y = A + (B + B) = A + B$We want the output $Y$ to be $0$. For an $OR$ gate,the output is $0$ only if all inputs are $0$.Therefore,$A + B = 0$ implies $A = 0$ and $B = 0$.Thus,the output is $0$ when $A = 0$ and $B = 0$.

The output $(Y)$ of the logic circuit given below is $0$ only when:

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