The following arrangement of master-slave flip flops has the initial state of P, Q as 0, 1 (respectively). After three clock cycles the output state P, Q is (respectively),
(A) 1, 0
(B) 1, 1
(C) 0, 0
Explanation: Given P = 0 , J = 1 and k = 1
- JK flipflop toggles the input in ‘11’ state. Therefore, output of first flipflop at P is ‘1’.
- Initial value of P is input for D flipflop. So, D = 0 .Therefore, output of second flipflop at Q is ‘0’.
Thus, option (A) is the answer.
Please comment below if you find anything wrong in the above post.