For the logic circuit shown in the figure,the | vedclass

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

Question

(C) $1$. The first gate is a $NOR$ gate with both inputs connected to $A$. The output of this $NOR$ gate is $\overline{A + A} = \bar{A}$.$2$. The seco

Vedclass · Accepted Answer

$Y = \bar{A}$

Vedclass · Answer

(C) $1$. The first gate is a $NOR$ gate with both inputs connected to $A$. The output of this $NOR$ gate is $\overline{A + A} = \bar{A}$.$2$. The second gate is a $NOT$ gate. The input to this gate is $\bar{A}$,so its output is $\overline{\bar{A}} = A$.$3$. The third gate is a $NAND$ gate with both inputs connected to the output of the $NOT$ gate (which is $A$). The output of this $NAND$ gate is $\overline{A \cdot A} = \bar{A}$.$4$. Therefore,the final Boolean expression for the circuit is $Y = \bar{A}$.

For the logic circuit shown in the figure,the Boolean expression is:

Similar Questions

The Boolean equation of a logic gate is $Y = \bar{A} \cdot B + A \cdot \bar{B}$. Identify the logic gate and describe its construction using $OR$,$AND$,and $NOT$ gates.

The Boolean equation $A + \bar{A} = .......$

The given truth table is of:
$A$ $X$
$0$ $1$
$1$ $0$

The expression for $Y$ in the following circuit is

The truth table for the given circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module