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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. $5$
edited | 4.3k views

Given boolean function is

f$= AB + C$

$= (A+C) . (B +C)$

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

Therefore, $3$ NOR gate required .

edited
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...
edited

Option : B

Solution:

+1 vote

Almost same solution,

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