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+22 votes

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

  1. $2$
  2. $3$
  3. $4$
  4. $5$
asked in Digital Logic by Veteran (52k points)
edited by | 5.2k views

5 Answers

+53 votes
Best answer

Given boolean function is 

 f$ = AB + C$

   $= (A+C) . (B +C)$ 

   $=((A+C)' +(B+C)')'$

Therefore, $3$ NOR gate required .

Correct Answer: $B$

answered by Active (5k points)
edited by
Silly question but
Where is this rule written AB+C = (A+c)(B+C)

I just can figure out by looking at the question also never found this rule before anywhere.

It is a rule you can check it by using truth table or look into morris mano postulate 4, distributive (b)

beautiful explanation.
+6 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 (81 points)
edited by
+3 votes

Option : B 


answered by (67 points)
+3 votes

Almost same solution,

answered by Boss (10.1k 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 (8.7k points)

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