# Coaxial cable with Zo = 50 ohms. Please check my answer.

Discussion in 'Homework Help' started by electrogirl, Mar 1, 2011.

1. ### electrogirl Thread Starter Member

May 2, 2010
47
0
see attachment for the given problem

solution for a,
Zo = 138/sqrt(Er) (log10 D/d)
50 = 138/(sqrt(8.85x10^-12) ( log10 N)

10^1.078X10^-6 = N
N = 1.000002482
D = nd
D = 0.1260003128 ''
outside diameter of outside conductor = D + 0.05''

= 0.176 ''

how do you solve for b?

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
659
Er is not 8.85x10^-12. That is E0, the permittivity of a vacuum (basically the same as for air).
When you get the answer for D = 0.1260003128 '', and d is given as 0.126", you need to apply a sanity test: Is it reasonable that the dielectric thickness (D-d)/2 is 3.12e-7"?

Er is relative permittivity (dielectric constant). Your problem does not state the value for Er.
The relative permittivity of air is 1. Coax dielectrics typically range from Er=1 to around 3 or 4, if memory serves me.

Where did you get 0.025" for the outside conductor thickness?

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