The output Y of the given network is :- | vedclass

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(C) The given circuit consists of a $NOR$ gate,a $NOT$ gate (implemented by a $NAND$ gate with shorted inputs),and an $AND$ gate.$1$. The output of th

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$B$

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(C) The given circuit consists of a $NOR$ gate,a $NOT$ gate (implemented by a $NAND$ gate with shorted inputs),and an $AND$ gate.$1$. The output of the $NOR$ gate with inputs $A$ and $B$ is $\overline{A+B}$.$2$. This output is fed into a $NAND$ gate whose inputs are shorted,acting as a $NOT$ gate. Thus,the output becomes $\overline{\overline{A+B}} = A+B$.$3$. This signal $(A+B)$ and the input $B$ are fed into an $AND$ gate.$4$. Therefore,the final output $Y = (A+B) \cdot B$.$5$. Using Boolean algebra: $Y = A \cdot B + B \cdot B = A \cdot B + B = B(A+1) = B \cdot 1 = B$.

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$

The output $Y$ of the given network is :-

Similar Questions

Which of the following circuits correctly represents the following truth table?
$A$ $B$ $C$
$0$ $0$ $0$
$0$ $1$ $0$
$1$ $0$ $1$
$1$ $1$ $0$

The output $Y$ for the inputs $A$ and $B$ of the circuit is given by the truth table of the shown circuit:

The combination of $NAND$ gates shown in the figure is equivalent to:

To get an output $Y = 1$ in the given circuit,which of the following inputs will be correct?

The output $D$ for the given circuit is:

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