4 votes 4 votes ____ number of gates are required to implement the Boolean function $(AB+C)$ with using only $2$-input NOR gates. $2$ $3$ $4$ $5$ Digital Logic nielit-2018 digital-logic boolean-algebra + – Arjun asked Dec 7, 2018 • recategorized Oct 24, 2020 by Krithiga2101 Arjun 1.4k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply OneZero commented Dec 7, 2018 reply Follow Share is it 4? 0 votes 0 votes Shubhgupta commented Dec 7, 2018 reply Follow Share should be 3. $AB + C = (A+C)(B+C) = \overline{\overline{(A+C)}+\overline{(B+C)}}$ 0 votes 0 votes OneZero commented Dec 8, 2018 reply Follow Share @Shubhgupta The answer to (A+B).(B+C) = AB + AC + BC + C. 0 votes 0 votes Shubhgupta commented Dec 8, 2018 reply Follow Share It's (A+C)(B+C) = AB+AC+CB+C = AB +(A+B+1)C and we know 1+ anything =1, so = AB+C moreover its a distributive law. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes 3 NOR GATES can implement the given function. but the official key says 2. https://gateoverflow.in/1298/gate2009-6 Asim Siddiqui 4 answered Feb 17, 2019 Asim Siddiqui 4 comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes NOR is Complement of OR AB+C = (A+C)(B+C) ← Distribution of + over = ((A+C)’+(B+C)’)’ 1st NOR- (A+C)’. Let X = (A+C)’ 2nd NOR- (B+C)’. Let Y = (B+C)’ 3rd NOR- (X+Y)’ topper98 answered Mar 18, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.