Digital-Grandtest1-Q1

+2 votes

Given A\bar B + \bar A B = C, A\bar C + \bar A C = ?

(A). B

(B).\bar B

(C).AC

(D).\bar A\bar C

asked Jun 19 in Digital by gbmentor (14,290 points)
reshown Jun 20 by gbmentor

3 Answers

0 votes
 
Best answer
Answer : option A) -> B

Solution :

Put given C= AB' +A'B

Hence, AC'+A'C=  A(AB'+A'B)' + A'(AB'+A'B)

->Now by rule : (X+Y)'= X'Y'

Hence = A( (AB')'.(A'B)' )  + A'(AB'+A'B)

-> Again similar rule :  (XY)' = X'+Y'     (DeMorgan's)      

Hence =  A((A'+B).(A+B')) + A'(AB'+A'B)

-> Rule: X (Y+Z)= XY + XZ

Hence = A(A'B'+AB)+A'(AB'+A'B')

           = AA'B+AAB+A'AB'+A'A'B

 -> Since AA' =0

Hence = AB+A'B    ,Taking B out= B(A+A')   = B.1  =B
answered Jun 21 by gbmentor (14,290 points)
selected Jun 21 by gbmentor
+1 vote
A B \bar A \bar B A \bar B \bar A B C \bar C A\bar C \bar A C A\bar C+\bar AC
0 0 1 1 0 0 0 1 0 0 0
0 1 1 0 0 1 1 0 0 1 1
1 0 0 1 1 0 1 0 0 0 0
1 1 0 0 0 0 0 1 1 0 1

 So the option is A

answered Jun 20 by 23rishiyadavpro20 (6,560 points)
+1 vote
Here c is nothing but XOR gate

And complement of XOR is XNOR

-->  A.XNOR+A'.XOR

--> A(A'B'+AB)+A'(AB'+A'B')

--> AA'B+AAB+A'AB'+A'A'B

--> 0+AB+0+A'B=AB+A'B

--> B(A+A')=B.1=B

 

SO OPTION A) IS CORRECT.
answered Jun 20 by tsnikhilsharmagate2018 (17,270 points)
edited Jun 20 by tsnikhilsharmagate2018
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