Potential in a non uniform resistor.

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I need to calculate potential in a nonuniform conductor. The complete
problem is
2 dimensional, but even 1D is causing me trouble.


In the 1D problem, you have a 1D conductor of length 3l. The first l
has a
conductivity of k and the remaining 2l of the conductor has a
conductivity of
2k. The two sections have the same resistance. If a voltage is applied
across
this conductor, the both of the sections will drop half the voltage,
and the
voltage change in each section will be linear.


Here are the the equations:

Let J be current density, E be the electric field and V be the
potential.

Since there is no free charge:

div J = 0.

The current is related to the field via the isotropic but not constant
conductivity k:

J = k E.

And the field is the gradient of the potential V:

E = grad V.

Putting this together gives:

div (k grad V) = 0.

In 1D (dimension x) this is:

dk / dx * dv/dx + k * d^2 v / dx^2 = 0

d^2 v / dx^2 = - dk / dx * dv/dx

Taking a finite difference approximation gives:

k(i) * ( v(i-1) -2 v(i) + v(i+1) ) / D^2 = -(k(i+1) - k(i-1))/2D * (v(i
+1) - v(i-1))/2D

where D is the grid spacing, and i is the index which is the descrete
position.

This comes out as:

v(i) = = (v(i+1) * ( k(i) + (k(i+1) - k(i-1))/4) + v(i-1)*(k(i) - (k(i
+1)-k(i-1))/4))/ (2 * k(i))

which can be iterated. This is the relaxation solution for the
differential equation above.



If I solve the above equation, for the problem above, I seem to get
the right
answer (the middle voltage gets very close to half the total voltage,
butit
takes a long time to converge). However, if I make the conductivitites
significantly different, then the equations become unstable and I
can't get an
answer.

Can anyone help?


-Ed

I have a simple solver in matlab. Here is the code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%
xs = [1 : 90 ]; %Positions
vs = 0 * xs; %Voltages

rs = 0 * xs; %Conductivities

rs(1:30) = 5;
rs(31:90) = 10;



while 1

vs(1) = 1; %Boundary conditions: 1 volt at the left edge
vs(90) = 0; %Boundary conditions: 0 volts at the right edge edge

v2 = vs;

for i=2:89
v2(i) = (vs(i+1) * ( rs(i) + (rs(i+1) - rs(i-1))/4) + vs(i-1)*(rs(i)
- (rs(i+1)-rs(i-1))/4))/ (2 * rs(i));
end

hold off
plot(xs, vs, 'g');
hold on
plot(xs, v2, 'r');



vs = v2;

end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%

 

[[email protected]]

My apologies. Google Groups seems to have wrapped my posting badly.
Here is the same post with better formatting:


I need to calculate potential in a nonuniform conductor. The complete
problem is 2 dimensional, but even 1D is causing me trouble.


In the 1D problem, you have a 1D conductor of length 3l. The first l
has a conductivity of k and the remaining 2l of the conductor has a
conductivity of 2k. The two sections have the same resistance. If a
voltage is applied across this conductor, the both of the sections
will drop half the voltage, and the voltage change in each section
will be linear.


Here are the the equations:

Let J be current density, E be the electric field and V be the
potential.

Since there is no free charge:

div J = 0.

The current is related to the field via the isotropic but not constant
conductivity k:

J = k E.

And the field is the gradient of the potential V:

E = grad V.

Putting this together gives:

div (k grad V) = 0.

In 1D (dimension x) this is:

dk / dx * dv/dx + k * d^2 v / dx^2 = 0

d^2 v / dx^2 = - dk / dx * dv/dx

Taking a finite difference approximation gives:

k(i) * ( v(i-1) -2 v(i) + v(i+1) ) / D^2 =
-(k(i+1) - k(i-1))/2D * (v(i+1) - v(i-1))/2D

where D is the grid spacing, and i is the index which is the descrete
position.

This comes out as:

v(i) = = (v(i+1) * ( k(i) + (k(i+1) - k(i-1))/4) +
v(i-1)*(k(i) - (k(i+1)-k(i-1))/4))/ (2 * k(i))

which can be iterated. This is the relaxation solution for the
differential equation above.



If I solve the above equation, for the problem above, I seem to get
the right answer (the middle voltage gets very close to half the total
voltage, butit takes a long time to converge). However, if I make the
conductivitites significantly different, then the equations become
unstable and I can't get an answer.

Can anyone help?


-Ed

I have a simple solver in matlab. Here is the code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
xs = [1 : 90 ]; %Positions
vs = 0 * xs; %Voltages

rs = 0 * xs; %Conductivities

rs(1:30) = 5;
rs(31:90) = 10;



while 1

vs(1) = 1; %Boundary conditions: 1 volt at the left edge
vs(90) = 0; %Boundary conditions: 0 volts at the right edge edge

v2 = vs;

for i=2:89
v2(i) = (vs(i+1) * ( rs(i) + (rs(i+1) - rs(i-1))/4) + ...
vs(i-1)*(rs(i) - (rs(i+1)-rs(i-1))/4))/ (2 * rs(i));
end

hold off
plot(xs, vs, 'g');
hold on
plot(xs, v2, 'r');



vs = v2;

end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%