1 1 vote The given figure shows a ripple counter using positive edge triggered flip-flops. If the present state of the counter is $Q_{2}Q_{1}Q_{0}=011$, then its next state $\left(Q_{2}Q_{1}Q_{0}\right)$ will be$010$$100$$111$$101$ Sequential Circuits gate2005-ec digital-circuits sequential-circuits flip-flops numerical-answers counters + – admin 2.4k views answer comment Share Follow Add Sync Questions Print See all 2 Comments 2 2 Comments reply yadavmayank742 commented Oct 20, 2023 reply Follow flag The question was asked for “next state (Q₂Q₁Q₀)” not ”next state (Q₂Q₁Qₙ)”, plz update the description. 0 0 replyShare GO Classes Support commented Sep 22 reply Follow flag Watch the Detailed Video Solution by clicking the button below..!Watch Detailed Video Solution 0 0 replyShare Please log in or register to add a comment.
1 1 vote PFA The answer in the attached image : Thus, the sequence is 000 001 010 011 100 101 110 111 and repeat – Asynchronous 3-bit Up Counter. We are given to start at 011 hence the next state is 100 viz option B. yadavmayank742 answered Oct 20, 2023 yadavmayank742 comment Share Follow 0 reply Please log in or register to add a comment.
