Refer to the logic circuit given below. For t | vedclass

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(C) The circuit consists of a $NOT$ gate, an $OR$ gate, an $AND$ gate, and a $NOR$ gate.Let the output of the $NOT$ gate be $A'$. Thus, $A' = \overlin

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$0, 0$ respectively

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(C) The circuit consists of a $NOT$ gate, an $OR$ gate, an $AND$ gate, and a $NOR$ gate.Let the output of the $NOT$ gate be $A'$. Thus, $A' = \overline{A}$.The inputs to the first $OR$ gate are $A'$ and $B$. So, its output is $X = A' + B = \overline{A} + B$.The inputs to the $AND$ gate are $A'$ and $B$. So, its output is $Z = A' \cdot B = \overline{A} \cdot B$.The final gate is a $NOR$ gate with inputs $X$ and $Z$. Thus, the final output is $Y = \overline{X + Z} = \overline{(\overline{A} + B) + (\overline{A} \cdot B)}$.Using Boolean algebra, $Y = \overline{\overline{A} + B + \overline{A} \cdot B} = \overline{\overline{A} + B} = A \cdot \overline{B}$.For $(A=1, B=1)$: $Y = 1 \cdot \overline{1} = 1 \cdot 0 = 0$.For $(A=0, B=1)$: $Y = 0 \cdot \overline{1} = 0 \cdot 0 = 0$.Wait, re-evaluating the circuit: The final gate is a $NOR$ gate. Let's re-calculate.$X = \overline{A} + B$, $Z = \overline{A} \cdot B$.$Y = \overline{X + Z} = \overline{(\overline{A} + B) + (\overline{A} \cdot B)} = \overline{\overline{A} + B} = A \cdot \overline{B}$.For $(A=1, B=1)$, $Y = 1 \cdot 0 = 0$.For $(A=0, B=1)$, $Y = 0 \cdot 0 = 0$.Looking at the options, there might be a misinterpretation of the final gate. If the final gate is an $OR$ gate, $Y = X + Z = \overline{A} + B + \overline{A} \cdot B = \overline{A} + B$. For $(1, 1)$, $Y = 0 + 1 = 1$. For $(0, 1)$, $Y = 1 + 1 = 1$. This doesn't match. If the final gate is a $NOR$ gate, the result is $(0, 0)$. Thus, option $C$ is correct.

Refer to the logic circuit given below. For two inputs $(A=1, B=1)$ and $(A=0, B=1)$, the output $(Y)$ will be . . . . . . .

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