Consider the circuit given below. Choose the | vedclass

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(D) The given circuit consists of a $NAND$ gate followed by a $NOT$ gate (since the second gate is an $OR$ gate with both inputs tied together, it act

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(D) The given circuit consists of a $NAND$ gate followed by a $NOT$ gate (since the second gate is an $OR$ gate with both inputs tied together, it acts as a $NOT$ gate). Thus, the circuit is equivalent to a $NAND$ gate followed by a $NOT$ gate, which is an $AND$ gate. The output $Y$ is given by $Y = A \cdot B$. Truth table for the $AND$ gate:| $A$ | $B$ | $Y = A \cdot B$ ||---|---|---|| $0$ | $0$ | $0$ || $0$ | $1$ | $0$ || $1$ | $0$ | $0$ || $1$ | $1$ | $1$ |Based on the logic $Y = A \cdot B$, the output $Y$ is high $(1)$ only when both inputs $A$ and $B$ are high $(1)$. Otherwise, the output is low $(0)$.

Consider the circuit given below. Choose the sketch depicting the output $Y$ of this circuit having inputs $A$ and $B$ as given below.

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Choose the correct logic circuit for the given truth table having inputs $A$ and $B$.
Inputs Inputs Outputs
$A$ $B$ $Y$
$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $1$

The logic circuit below has the truth table, same as that of

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Inputs	Inputs	Outputs
$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$