The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114 degrees.  From this, and using Coulson's Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized.  Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the "s" character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2=3.74}$, or the C-C bond is $\ce{sp^{3.74}}$ hybridized.  We see that the C-C bond in cyclopropane is very high in $\ce{P}$ character.  It is this high $\ce{P}$ content that allows cycloprane to behave in a similar fashion to an olefin in terms of stabilizing adjacent charge, absorbing bromine, etc.  By the way, an x-ray study of a cyclopropane derivative (Acta Cryst., 20, 80 (1966)) shows significant electron density only exterior to the cyclopropane ring, there is NO significant electron density interior to the ring, consistent with this analysis.

Why does cyclopropane give bromine water test?
I have a graphics card for a laptop whose driver kills itself if it runs for longer than two minutes with no external monitor attached. To solve that, I decided to make a sort of dummy plug that would be detected as a monitor by PC. 

I have managed to find out what impedances video card  expects for most of the pins, but for H-Sync and V-Sync, Google brought up numerous conflicting answers, mostly from people who are guessing. 
