1 votes 1 votes If the Initial state $(Q_{2}\,Q_{1}\,Q_{0})$ of the counter is $101$,then the state $(Q_{2}\,Q_{1}\,Q_{0})$ after $4$ clock pulses is Digital Logic made-easy-test-series digital-logic digital-counter + – naveen81 asked Jan 30, 2017 edited Mar 4, 2019 by adeebafatima1 naveen81 486 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes this is the solution Smriti012 answered Jan 31, 2017 selected Jul 25, 2017 by sourav. Smriti012 comment Share Follow See all 2 Comments See all 2 2 Comments reply Ajay Prakash Verma commented Oct 13, 2017 reply Follow Share i think in 3rd clock the o/p will be 101 and in 4th it will be 011 so 011 will be answer. 0 votes 0 votes akshat sharma commented Oct 13, 2017 i edited by akshat sharma Oct 13, 2017 reply Follow Share 010 will be the answer at 3 clock cycle 100 and at 4 clock cycle : -> 010 NOTE: In question there is one typo (Q0)' will be i/p to the J . take Refrence from Smriti012 Answer. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes initally 101 011 after 1st clk.. 000 after 2nd clk.. 100 after 3rd clk.. 010 after 4th clk.. <-Answer Correct me if I am wrong! Smriti012 answered Jan 31, 2017 Smriti012 comment Share Follow See 1 comment See all 1 1 comment reply lokesh biradar commented Jan 31, 2017 reply Follow Share How you got 011 in first clk cycle. if possible please explain me. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 001 after 4 clock Neeraj Chandrakar answered Jul 25, 2017 Neeraj Chandrakar comment Share Follow See all 0 reply Please log in or register to add a comment.