27 votes 27 votes The following circuit implements a two-input AND gate using two $2-1$ multiplexers. What are the values of $X_1, X_2, X_3$? $X_1 = b, X_2 = 0, X_3 = a$ $X_1 = b, X_2 = 1, X_3 = b$ $X_1 = a, X_2 = b, X_3 = 1$ $X_1 = a, X_2 = 0, X_3 = b$ Digital Logic gateit-2007 digital-logic normal multiplexer + – Ishrat Jahan asked Oct 29, 2014 • edited Jul 12, 2019 by ajaysoni1924 Ishrat Jahan 7.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Overflow04 commented Dec 22, 2022 reply Follow Share @GateCse why a,b is considered as a.b not (a+b) 0 votes 0 votes Hira Thakur commented Oct 27, 2023 reply Follow Share if we put the value on expression $F(a,b)=\bar{X_3.}X_2+X_3(\bar X_1.b+X_1.a)$ option 1) : $ab$ option 2): $a+\bar b$ option 3): $a+b$ option 4): $b$ 0 votes 0 votes Please log in or register to add a comment.
Best answer 34 votes 34 votes Answer: A $F = (bX_1' + aX_1)X_3 + X_2X_3'$ Put $X_1 = b, X_2 = 0, X_3 = a$ to get $F = ab.$ Rajarshi Sarkar answered Apr 14, 2015 • selected Apr 14, 2015 by Arjun Rajarshi Sarkar comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Prasanna commented Dec 22, 2015 reply Follow Share +1 Thank You! 0 votes 0 votes Purple commented Nov 21, 2016 reply Follow Share If in option B if X2 was 0 then B also could have been right? 0 votes 0 votes Tendua commented Nov 21, 2016 reply Follow Share Right pepper 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes F = (bX1' + aX1)X3 + X2X3' If we put X1 = b X2 = 0 X3 = a Then we get, F = ab varunrajarathnam answered May 2, 2021 varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.