1 1 vote The following expression was to be realized using 2-input AND and OR gates, but by mistake all 2-input AND gates were taken as 2-input NAND gates, $$(a.b).c + (\bar a.c).d + (\bar b.c).d + a.d$$ What is the function finally realized ? $1$ $\bar a + \bar b + \bar c + \bar d$ $\bar a + b + \bar c + \bar d$ $\bar a + \bar b + c + \bar d$ Digital Logic digital-logic go-digital-logic-1 circuit-output two-marks + – Bikram 1.4k views answer comment Share Follow Print See 1 comment 1 1 comment reply Arpit Dhuriya commented Sep 26, 2016 reply Follow flag I think 1 should be the answer ((not b nand c) nand d) + ((a nand b) nand c) + ((not a nand c) nand d)+ (a nand d) reduces to not a + b + not c + not b +not d = 1 ? 1 1 replyShare Please log in or register to add a comment.
Best answer 5 5 votes I have not indicated the k map properly but answer is 1 Kashyap Avinash answered Oct 31, 2016 • selected Oct 31, 2016 by Digvijay Pandey Kashyap Avinash comment Share Follow See all 2 Comments 2 2 Comments reply Hira Thakur commented Oct 27, 2017 reply Follow flag plz see this also:https://gateoverflow.in/3471/gate2007-it-38 0 0 replyShare pallaviamu commented Aug 17, 2018 reply Follow flag How can you assume NOT gate is available and directly take complement for eg in (a'c) term you are writing it as (a'b)' ?? 0 0 replyShare Please log in or register to add a comment.
1 1 vote right answer is A Wanted answered Jan 5, 2017 • edited Jan 5, 2017 by Wanted Wanted comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Actually Different place have different Questions if Question is above then ans is -A ✅ But in Some places this is the 👇 Question then Here is the Explanation 👇 Prashant-G answered Jun 13 Prashant-G comment Share Follow 0 reply Please log in or register to add a comment.
