The truth table for the circuit shown in the | vedclass

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

Question

(D) The circuit consists of a $NAND$ gate and an $OR$ gate feeding into a final $OR$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gat

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The circuit consists of a $NAND$ gate and an $OR$ gate feeding into a final $OR$ gate. Let the inputs be $A$ and $B$. The output of the $NAND$ gate is $\overline{AB}$. The output of the $OR$ gate is $A+B$. The final output $Y$ is the $OR$ combination of these two: $Y = (A+B) + \overline{AB}$. Using Boolean algebra: $Y = A + B + \bar{A} + \bar{B}$. Since $A + \bar{A} = 1$ and $B + \bar{B} = 1$,the expression simplifies to $Y = 1 + 1 = 1$. Thus,for all input combinations $(A, B)$,the output $Y$ is always $1$. Therefore,the correct truth table is option $D$.

The truth table for the circuit shown in the figure is

Similar Questions

Logic gates are constructed by

In the given circuit, the binary inputs at $A$ and $B$ are both $1$ in one case and both $0$ in the next case. The respective outputs at $Y$ in these two cases will be

Converting the binary number $1011.01$ into decimal gives ....

$A$ typical integrated circuit $(IC)$ consisting of logic gates $\leq 1000$ is termed as

What is a logic gate?

Vedclass Test Series

Exam Paper Generator

Online Exam Module