For the given circuit,what will be the truth | vedclass

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(C) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate (represented by the second gate with shorted inputs). $1$. The output of the f

Vedclass · Accepted Answer

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

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(C) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate (represented by the second gate with shorted inputs). $1$. The output of the first $NOR$ gate is $C = \overline{A+B}$. $2$. The second gate is a $NOT$ gate (since both inputs are connected together),so its output is $Y = \overline{C}$. $3$. Substituting $C$,we get $Y = \overline{(\overline{A+B})} = A+B$. $4$. This is the Boolean expression for an $OR$ gate. $5$. The truth table for an $OR$ gate is: - If $A=0, B=0$,then $Y=0$. - If $A=0, B=1$,then $Y=1$. - If $A=1, B=0$,then $Y=1$. - If $A=1, B=1$,then $Y=1$. This matches option $C$.

For the given circuit,what will be the truth table for the output $Y$ based on all possible inputs $A$ and $B$?

Similar Questions

The truth table given below is for which gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$C: 1, 1, 1, 0$

The combination of $NAND$ gates is shown in figures $(I)$ and $(II)$. For the given inputs,the outputs in both the combinations are respectively.

For the given combination of logic gates,the inputs $A, B$ and $C$ are as follows. If $A=B=C=0$ and $A=B=1, C=0$,then the logic states of output $D$ are respectively:

What is the truth table for the given logic gate combination?

The truth table of the given circuit diagram is:

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

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

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

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