For the combination of logic gates shown in t | vedclass

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

Question

(D) The circuit consists of two $OR$ gates whose outputs are fed into a $NAND$ gate. Let the inputs be $A$ and $B$. The output of both $OR$ gates is $

Vedclass · Accepted Answer

$NOR$

Vedclass · Answer

(D) The circuit consists of two $OR$ gates whose outputs are fed into a $NAND$ gate. Let the inputs be $A$ and $B$. The output of both $OR$ gates is $X = A + B$. These are the inputs to the $NAND$ gate. The final output $Y$ is given by $Y = \overline{X \cdot X} = \overline{X} = \overline{A + B}$. This is the Boolean expression for a $NOR$ gate. The truth table is as follows:$A$$B$$X = A + B$$Y = \overline{X \cdot X}$$0$$0$$0$$1$$0$$1$$1$$0$$1$$0$$1$$0$$1$$1$$1$$0$

For the combination of logic gates shown in the figure,the equivalent logic gate is

Similar Questions

Three logic gates are connected as shown in the figure. If the inputs are $A=1, B=0$ and $C=1$, then the values of $y_1, y_2$ and $y_3$ respectively are

The truth table for the given logic circuit is

If two inputs of a $NAND$ gate are shorted,the resulting gate is

Identify the logic gate represented by the given circuit and match the values of $A, B, Y$ in the bracket to verify.

$A$ logic circuit provides the output $Y$ as per the following truth table:

$A$ $B$ $Y$

$0$ $0$ $1$

$0$ $1$ $0$

$1$ $0$ $1$

$1$ $1$ $0$

The expression for the output $Y$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module