If three logic gates are connected as shown i | vedclass

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(B) The circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The upper $AND$ gate receives inputs $A$ and $B

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

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(B) The circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The upper $AND$ gate receives inputs $A$ and $B$,so its output is $Y_1 = A \cdot B$.The lower $OR$ gate receives inputs $B$ and $A$,so its output is $Y_2 = B + A$.The final $NAND$ gate receives $Y_1$ and $Y_2$ as inputs,so the final output is $Y = \overline{Y_1 \cdot Y_2} = \overline{(A \cdot B) \cdot (A + B)}$.Using Boolean algebra: $Y = \overline{(A \cdot B) \cdot A + (A \cdot B) \cdot B} = \overline{(A \cdot B) + (A \cdot B)} = \overline{A \cdot B}$.This is the truth table for a $NAND$ gate:If $A=0, B=0$,then $Y = \overline{0 \cdot 0} = 1$.If $A=0, B=1$,then $Y = \overline{0 \cdot 1} = 1$.If $A=1, B=0$,then $Y = \overline{1 \cdot 0} = 1$.If $A=1, B=1$,then $Y = \overline{1 \cdot 1} = 0$.Comparing this with the given options,option $B$ is correct.

If three logic gates are connected as shown in the figure,then the correct truth table of the circuit is

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

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

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

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