The following logic gate circuit is equivalen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the inputs be $A$ and $B$. The first two $NAND$ gates act as $NOT$ gates because their inputs are shorted. Thus,the outputs of these gates are

Vedclass · Accepted Answer

$NOR$ Gate

Vedclass · Answer

(A) Let the inputs be $A$ and $B$. The first two $NAND$ gates act as $NOT$ gates because their inputs are shorted. Thus,the outputs of these gates are $\bar{A}$ and $\bar{B}$ respectively.These are fed into a third $NAND$ gate,which produces the output $Y' = \overline{\bar{A} \cdot \bar{B}} = A + B$ (by De Morgan's Law).This output $Y'$ is then fed into a final $NAND$ gate acting as a $NOT$ gate,resulting in the final output $Y = \overline{Y'} = \overline{A + B}$.The Boolean expression $Y = \overline{A + B}$ corresponds to a $NOR$ gate.Truth table:$A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$1$$0$

The following logic gate circuit is equivalent to which of the following gates?

Similar Questions

To get an output $1$ from the circuit shown in the figure,the input must be

Which of the following Boolean expressions is not correct?

What will be the input of $A$ and $B$ for the Boolean expression $\overline{(A+B) \cdot(A \cdot B)}=1$?

The output of the given circuit diagram is

If $X$ and $Y$ are the inputs, the given circuit works as . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module