The truth table corresponding to the circuit | vedclass

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

Question

(B) From the given circuit diagram,the output of the $OR$ gate is $(A + B)$.This output is then fed as an input to the $AND$ gate along with input $A$

Vedclass · Accepted Answer

$A$$B$$C$$0$$0$$0$$0$$1$$0$$1$$0$$1$$1$$1$$1$

Vedclass · Answer

(B) From the given circuit diagram,the output of the $OR$ gate is $(A + B)$.This output is then fed as an input to the $AND$ gate along with input $A$.Therefore,the final output $C$ is given by the Boolean expression: $C = A \cdot (A + B)$.Using the distributive law of Boolean algebra,$C = (A \cdot A) + (A \cdot B)$.Since $A \cdot A = A$,we have $C = A + (A \cdot B)$.Using the absorption law,$C = A(1 + B) = A \cdot 1 = A$.Thus,the output $C$ is equal to input $A$.Let's verify this with the truth table:$A$$B$$A+B$$C = A \cdot (A+B)$$0$$0$$0$$0$$0$$1$$1$$0$$1$$0$$1$$1$$1$$1$$1$$1$Comparing this with the given options,option $B$ matches the truth table.

The truth table corresponding to the circuit given below is

Similar Questions

Which logic gate is represented by the following combination of logic gates?

The logic gates in which all the inputs must be low to get a high output are

The output $W$ for the given logic circuit is:

Which gate provides the same operation as the circuit shown below?

For the logic circuit shown,the output waveform at $Y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$	$B$	$C$
$0$	$0$	$0$
$1$	$0$	$0$
$0$	$1$	$0$
$1$	$1$	$1$

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$

$A$	$B$	$C$
$0$	$0$	$1$
$1$	$0$	$0$
$0$	$1$	$0$
$1$	$1$	$0$

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$1$