0 votes 0 votes In the circuit, the present value of $Z$ is $1$. Neglecting the delay in the combinatorial circuit, the values of $S$ and $Z$, respectively, after the application of the clock will be $S=0, Z=0$ $S=0, Z=1$ $S=1, Z=0$ $S=1, Z=1$ new gate2024-ee + – Arjun asked Feb 16 • edited Feb 25 by makhdoom ghaya Arjun 15.9k points answer comment Share See all 0 reply Please log in or register to add a comment.
