Quote:
Originally posted by Tom M
Is this something that can't be answered? |
No, it's just that it is something that only a few people are going to be interested in reading about... finding the answer isn't too difficult, though I suspect its kinship with those hated, "two trains leave separate stations...," type of word problem is glazing over everyone's eyes that could answer it, just like in middle school.
90R is the characteristic impedance of your cable, you say? That sounds an awful like the coax used for RCA cables; which is to say, it wasn't exactly manufacturered nor characterized for constant impedance. At any rate, that's for you to worry about, not me
Rearrange the bits of the equation algebraically. Don't let the square root throw you off. Keep reminding yourself how your 6/7/8th grade teacher was right- you really did end up needing to know that silly crap one day!
One one side you have Z, the characteristic impedance of the cable; on the other side you have the mess (L/C)^0.5. First you square both sides, giving you: Z^2 = L/C. Now you can multiply both sides by the inverse of L, giving you Z^2/L=1/C. Invert to make it less cumbersome et voilla, you get:
L/(Z^2) = C
Or, in this case,
(0.22*10^-6)/8100 = C
Or 27pF.
This is about right for a piece of RCA cable between 10 and 14" long, I'd guess.
edited: grammar in my attempt at humor in the first paragraph
