Calculate the total resistance by integration using the conductivity equation

[Dr-LucienSanchez]

Homework Statement

The conductivity of a non-homogeneous rod (length l, cross sectional area A) varies linearly from the value $g_a$ at the end to the value $g_b$ at the end b; that is, at any point on the cross section of the rod lying at the distance x from a we have g (x) (below).

Divide the rod into discs of thickness dx, (i) compute the resistance of a typical disc, (ii) and show by integration that the total resistance of the rod is R (below)

Relevant Equations

Conductance at point x: (*) g(x) = g_a + (( g_b-g_a ) x ) / l
Total Resistance: (**) R = ( l/A ) * ( 1 / (g_b-g_a) ) * ln( g_b/g_a )

Between ends a and b of a conducting rod of length l and area A:
Average conductivity for non-homogeneous cylindrical conductors: (***) g_ave = (l / A)*G(ab)
Resistivity is the reciprocal of the conductivity i.e: (****) R(ab) = (l/gA), where g is the electric conductivity

(i) Dividing the rod into thicknesses of dx we get discs of area A with lengths=dx so using (****) we have the resistance of a typical disc (between point x' and x'+dx) as:

(1) $R(x'dx)=\frac{dx}{g(x)A}$

(ii) Using (1) and (*) and the integrating from a to b of the entire rod we get:

$R(ab)=\int_{g_a}^{g_b}\frac{dx}{g_a+\frac{(g_b-g_a)x}{l}A}=\frac{l}{A}\int_{g_a}^{g_b}\frac{dx}{g_a l + (g_b-g_a)x}=\frac{l}{(g_b-g_a)A}ln(g_al+(g_b-g_a))$

Now my solution has a pesky $g_al$ which if it was not there then the total resistance would be the same as (**). However it is there which is why i need help.

I suspect that I have set up the problem incorrectly which leads me to integrate the wrong equation or I am missing something else.

Can anyone assist me to figure out where I have tripped up?
FYI this problem is from "Electromagnetic Fields and Waves" by Vladimir Rojansky page 11.

 

[Baluncore]

Welcome to PF.
What do you not understand about the question?

 

[Dr-LucienSanchez]

Welcome to PF.
What do you not understand about the question?

I guess I am unsure why I have the $g_al$ in the log of my final answer when it is not in (**). Because it is there leads me to think that I have not set up the integral correctly by the way I compute the resistance of a typical disc in (i).
So I guess I don't understand how to set up the resistance of a typical disc so that I can form and solve the integral for the total resistance.

 

[Baluncore]

What happens if you instead interpolate by; Gx = ( ( Ga * ( L - x ) + x * Gb ) / L )
then r = integral from x=0 to L; 1 / ( A * Gx ) .dx

 

[Baluncore]

It appears you are integrating from Ga to Gb ? rather than x=0 to Length.

Conductance varies linearly so we define a straight line y = m*x+c
At x=0 we have Gx=Ga, at x=L we have Gx=Gb. Then m = (Gb-Ga)/L and c = Ga.
Table of integrals gives; Integral dx/(m*x+c) = (1/m)* Log(m*x+c)
We integrate from x=0 to x=L.
When x=L we have; (1/m)*Log(m*L+c)
When x=0 we have; (1/m)*Log(c)
Difference is; (1/m)* (Log(m*L+c) - Log(c))
But 1/m = L/(Gb-Ga); also (m*L+c) = ((Gb-Ga)/L)*L+Ga = Gb; and c = Ga.
So; L/(Gb-Ga) * (Log(Gb) - Log(Ga)) = (L/(Gb-Ga))*Log(Gb/Ga)
Don't forget to include the constant rod area A;
R = L / ( A * ( Gb - Ga ) ) * Log( Gb / Ga )
which agrees with numerical integration.

 

[Dr-LucienSanchez]

It appears you are integrating from Ga to Gb ? rather than x=0 to Length.

looks like I derped out on my limits, thanks for spotting that.

Conductance varies linearly so we define a straight line y = m*x+c
At x=0 we have Gx=Ga, at x=L we have Gx=Gb. Then m = (Gb-Ga)/L and c = Ga.
Table of integrals gives; Integral dx/(m*x+c) = (1/m)* Log(m*x+c)
We integrate from x=0 to x=L.
When x=L we have; (1/m)*Log(m*L+c)
When x=0 we have; (1/m)*Log(c)
Difference is; (1/m)* (Log(m*L+c) - Log(c))
But 1/m = L/(Gb-Ga); also (m*L+c) = ((Gb-Ga)/L)*L+Ga = Gb; and c = Ga.
So; L/(Gb-Ga) * (Log(Gb) - Log(Ga)) = (L/(Gb-Ga))*Log(Gb/Ga)
Don't forget to include the constant rod area A;
R = L / ( A * ( Gb - Ga ) ) * Log( Gb / Ga )
which agrees with numerical integration.

Thanks for explaining this it helped me a lot