Radial resistance of coaxial cable

[azaharak]

One of the textbooks I've stumbled across states that the radial resistance of a coaxial cable ( current running from inner cylinder to outer cylinder via silicon in between) is given byR = [resistivity * natural log of (b/a)] / [2*pi()*(length of cable)]

where b is the outer radii and a is the inner radiiThe derivation was given as dR= resistivity*dr / A and then integrated from a to b.My question/issue is the dependence of the area on the radial length, I'm not so sure about the correctness of the differential form above.

If you follow through the chain rule with A a function of r

dR/dr= resistivity/Area + (d/dA)*(resistivity*r/Area)*(dA/dr)

which becomes

dR/dr= resistivity/area -[resistivity*r/(Area^2)] * (2*pi()*L)

which becomes

dR/dr=resistivity/area - resistivity/area =0

meaning that the resistance is a constant value which should be

R = resistivity / (2*Pi*L ) where L is the length of the cable (constant)
Help! Thank you very much.Thank you!

 

[gsal]

I did not quite follow what you were trying to do; but I never saw you write the area as a function of r, in the first place.

What you need to do is to write what is the cross-sectional area the a radial current 'sees' at a given radiur r...don't call it A, write the formula...it should simply depend on r, pi, and L; then,
integrate the dR from a to b

 

[francisyes]

http://buphy.bu.edu/~duffy/semester2/c09_resistance_calc.html

Hope this helps