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+18 votes

What is the minimum number of gates required to implement the Boolean function (AB+C) if we have to use only 2-input NOR gates?

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asked in Digital Logic by Veteran (59.4k points)
edited by | 4k views

5 Answers

+44 votes
Best answer

given boolean function is 

 f = AB + C

   = (A+C) . (B +C)  

   =((A+C)' +(B+C)')' 

therefore 3 NOR gate required .


answered by Loyal (5.9k points)
reshown by
+4 votes
Answer should be 3 nor gates...A.B + C can be written as (A+C).(B+C) which is a OR-AND realization and can be made with 3 NOR gates as OR- AND is similar to a NOR-NOR realization...
answered by (61 points)
edited by
+2 votes

Option : B 


answered by (57 points)
0 votes
answer - D

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

two NOR gates for complementing A and B

one for computing (A' + B')'

one to compute ((A' + B')' + C)'

one to negate the last result
answered by Loyal (9.2k points)
0 votes

Almost same solution,

answered ago by Active (2k points)

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