The truth table for the given logic circuit i | vedclass

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

Question

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the $NOR$ gate is $\overline{A+B}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.Let the inputs be $A$ and $B$.The output of the $NOR$ gate is $\overline{A+B}$.The output of the $AND$ gate is $A \cdot B$.These two outputs are fed into a $NAND$ gate.Therefore,the final output $Y$ is given by:$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$Using De Morgan's Law,$\overline{X \cdot Z} = \overline{X} + \overline{Z}$:$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$$Y = (A+B) + (\overline{A} + \overline{B})$$Y = (A + \overline{A}) + (B + \overline{B})$Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:$Y = 1 + 1 = 1$Thus,for any combination of inputs $(A, B)$,the output $Y$ is always $1$.

The truth table for the given logic circuit is:

Similar Questions

The following configuration of gates is equivalent to

In the given logic circuit,$A=1$ and $B=0$. The values of $Y_1$ and $Y_2$ are respectively:

$A$ logic gate circuit has two inputs $A$ and $B$ and output $Y$. The voltage waveforms of $A, B$ and $Y$ are shown below. Identify the logic gate circuit.

The binary number $1000101.101$ is converted to decimal as:

What are the logic gates shown below equivalent to?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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

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

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

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