The truth table corresponding to the circuit | vedclass

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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

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

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

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

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