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?
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    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?
     
    electrogirl likes this.
  3. bertus

    Administrator

    Apr 5, 2008
    15,646
    2,345
    electrogirl likes this.
Loading...