## 5 Answers

Here, clocks are applied to both flip flops simultaneously. Outputs for $3$ cycles will proceed as follows:

- When $11$ is applied to $JK$ flip flop it toggles the value of $P.$ So, output at $P$ will be $1.$
- Input to D flip flop will be $0$ (initial value of P). So, output at $Q$ will be $0.$

$JK$ flip flop it toggles the value of $P.$ So, output at $P$ will be $0.$

- Input to D flip flop will be $1$ (current value of P) . So, output at $Q$ will be $1.$
- $JK$ flip flop it toggles the value of $P.$ So, output at $P$ will be $1.$
- Input to D flip flop will be $0$ (initial value of P) so output at $Q$ will be $0.$

So, answer is A.

now when clock is aplied J= 1 and K=1 then (it do complement of P) i.,e output at p = 1 and at same time FF D has input 0 so it change Output Q as ) only **[ on D-FF input = output]**

**SO output will be p= 1 and Q= 0 **

**[ keep in mind both FF work at same time dont wait for 1st complete then second will start].**

### 2 Comments

-> As there is Master Slave configuration, 1st flipflop will respond to the positive edges and the 2nd flipflop to the negative edges.

-> As the given Flipflops are synchronous,(clock is simultaneously given to both FFs) $Q$ responds according to immediate values of $P$.

-> Thus, after 3 cycles, as shown in the diagram, $P = 1$ and $Q=0$