An AND gate is followed by a NOT gate in seri | vedclass

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

Question

(B) $1$. An $AND$ gate takes two inputs $A$ and $B$ and produces an output $X = A \cdot B$.$2$. This output $X$ is then passed through a $NOT$ gate.$3

Vedclass · Accepted Answer

$\overline{A \cdot B}$

Vedclass · Answer

(B) $1$. An $AND$ gate takes two inputs $A$ and $B$ and produces an output $X = A \cdot B$.$2$. This output $X$ is then passed through a $NOT$ gate.$3$. $A$ $NOT$ gate inverts the input,so the final output $Y = \overline{X}$.$4$. Substituting the value of $X$,we get $Y = \overline{A \cdot B}$.$5$. This combination of an $AND$ gate followed by a $NOT$ gate is known as a $NAND$ gate.

An $AND$ gate is followed by a $NOT$ gate in series. With two inputs '$A$' and '$B$',the Boolean expression for the output '$Y$' will be

Similar Questions

In the Boolean algebra,$(\overline {\bar A \cdot \bar B} ) \cdot A$ is equal to:

The truth table for the given circuit is:

The following truth table corresponds to which logic gate?
$A: 0, 0, 1, 1$
$B: 0, 1, 0, 1$
$X: 0, 1, 1, 1$

If $A = 100101$ and $B = 110110$ are applied to $X-NOR$ gate,then the output will be

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have

Vedclass Test Series

Exam Paper Generator

Online Exam Module