00:01
So having this question, the path effort f is given by, sorry, the path effort, path effort, f is given by g into b, where b is the branching effort.
00:19
In this case, the load capacitance at final output node in an inverter, which is 25 .6 ,4 times the size of first inverter.
00:27
So b is equals to 25 .64.
00:32
Then we calculate the branching effort.
00:36
Here, the branching effort, the branching effort, effort, effort, b is equals to c load by c in, and that is 25 .6 .4...