44 votes 44 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? $2$ $3$ $4$ $5$ Digital Logic gatecse-2009 digital-logic min-no-gates normal + – Kathleen asked Sep 22, 2014 • edited Jun 19, 2018 by Pooja Khatri Kathleen 20.7k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments abhiarns commented Jun 14, 2020 reply Follow Share @mohan123 Thanks 1 votes 1 votes dutta18 commented Jan 13, 2021 reply Follow Share Can you just tell me how did you convert AB+C to (A+B)(B+C) ? Is it SOP to POS ? 0 votes 0 votes mohan123 commented Feb 7, 2021 reply Follow Share firstly u write the wrong AB+C to (A+B)(B+C) correct AB+C to (A+C)(B+C) Distributive Law states that the multiplication of two variables and adding the result with a variable will result in the same value as the multiplication of addition of the variable with individual variables. For example: A + BC = (A + B) (A + C) 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes Option : B Solution: Krishankant Ray 6 answered May 29, 2017 Krishankant Ray 6 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 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 ankitrokdeonsns answered Oct 19, 2014 ankitrokdeonsns comment Share Follow See all 0 reply Please log in or register to add a comment.
