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Evaluating Power Supplies with Different Compensation Settings
No trap here, answer is easy to find!
We have : \text{Aring}=\pi\times(R^2-r^2)
Since the length of the half string is 1 meter, we have:
1+r^2=R^2
Therefore :
A\text{ring=\textbackslash pi}
Right MikeLh , congrats!
Ring area is PI ~ 3.14...
r^2 + 1 = R^2 -> R^2 - r^2 = 1Aring = PI*R^2 - PI*r^2 = PI * (R^2 - r^2) = PI
1 . From Pitagora R^2 - r^2 = 1. The area is pi*(R^2 - r^2)=pi
Yes, you get it AlexTru , the area is simply Pi.... Amazing isn't it?
Bingo Celin , you found it, congratulations!
Pi
Bingo bradstev22 , you get it!