The output Y of the logic circuit given below | vedclass

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

Question

(B) The input $A$ passes through a $NOT$ gate, resulting in $\bar{A}$.This $\bar{A}$ is fed into an $AND$ gate along with input $B$, producing an outp

Vedclass · Accepted Answer

$\bar{A}$

Vedclass · Answer

(B) The input $A$ passes through a $NOT$ gate, resulting in $\bar{A}$.This $\bar{A}$ is fed into an $AND$ gate along with input $B$, producing an output of $\bar{A} \cdot B$.This result $(\bar{A} \cdot B)$ and the original $\bar{A}$ are then fed into an $OR$ gate.Therefore, the final output $Y$ is given by:$Y = \bar{A} + (\bar{A} \cdot B)$Using the absorption law of Boolean algebra, which states that $X + (X \cdot Y) = X$, we can simplify the expression:$Y = \bar{A} \cdot (1 + B)$Since $(1 + B) = 1$, we get:$Y = \bar{A} \cdot 1 = \bar{A}$

The output $Y$ of the logic circuit given below is:

Similar Questions

In the given logic circuit diagram,$A$,$B$,and $C$ are the inputs,and $Y$ is the output. The output $Y$ is ' $HIGH$ ' (i.e.,$Y=1$):

The combination of the gates shown will produce:

Give the truth table for the circuit shown in the figure.

The given truth table is correct for:
$P: 1, 1, 0, 0$
$Q: 1, 0, 1, 0$
$R: 1, 0, 0, 0$

The logic performed by the circuit shown in the figure is equivalent to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module