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' = \overline{A}

Vedclass · Accepted Answer

$0, 0$ 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.

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

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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