6 replies to this topic

[mmiguel]

A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm

A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms

What is the total resistance of the whole object for current flowing along the x-axis?

Edited by mmiguel, 03 September 2012 - 10:20 AM.

[mmiguel]

A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm

A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms

What is the total resistance of the whole object for current flowing along the x-axis?

Let me know if you want any hints

[bonanova]

Spoiler for approach

Resistance R of a regular homogeneous solid is given by its resistivity rho multiplied by its length L (in the direction of current flow) divided by its cross-sectional area A:

R = rho x L / A. The unit of resistivity is ohm-cm.
R of a unit cube is thus numerically equal to rho.

If the material were homogeneous, we would slice the object into circular "steaks" of thickness (length) dL and area A(x) = pi r(x)², compute, and add the resistances. The material is not homogeneous. Its resistivity depends on distance r from the x-axis. So there is an added step. Instead of "steaks" we have a collection of "onion rings" where each ring has a resistance commensurate with the same length dL, but area A = 2 pi r dr and resistivity rho[r]. The resistances of the rings are added in parallel. That is their conductances are added and the reciprocal of the sum is taken to get the resistance of the onion slice. The resistances of the onion slices are then summed to get the resistance of the object.

You could do this numerical procedure with a program to arbitrary accuracy using aribitrarily small values of dL and dr. Or, you could perform a double integral to get the answer exactly.

That's the procedure. The answer is left to another solver.

[Rob_Gandy]

A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm

A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms

What is the total resistance of the whole object for current flowing along the x-axis?

Spoiler for approach

Resistance R of a regular homogeneous solid is given by its resistivity rho multiplied by its length L (in the direction of current flow) divided by its cross-sectional area A:

R = rho x L / A. The unit of resistivity is ohm-cm.
R of a unit cube is thus numerically equal to rho.

If the material were homogeneous, we would slice the object into circular "steaks" of thickness (length) dL and area A(x) = pi r(x)², compute, and add the resistances. The material is not homogeneous. Its resistivity depends on distance r from the x-axis. So there is an added step. Instead of "steaks" we have a collection of "onion rings" where each ring has a resistance commensurate with the same length dL, but area A = 2 pi r dr and resistivity rho[r]. The resistances of the rings are added in parallel. That is their conductances are added and the reciprocal of the sum is taken to get the resistance of the onion slice. The resistances of the onion slices are then summed to get the resistance of the object.

You could do this numerical procedure with a program to arbitrary accuracy using aribitrarily small values of dL and dr. Or, you could perform a double integral to get the answer exactly.

That's the procedure. The answer is left to another solver.

Spoiler for So...

Does R(p) give me rho?

[mmiguel]

Spoiler for So...

Does R(p) give me rho?

p is radius from x axis

R(p) is resistance in ohms as a function of radius from x-axis

[bonanova]

Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units.

It thus makes sense to speak of a material having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an object, it's meaningful to speak of resistance.

Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.

Spoiler for So ...

To analyze objects that are not electrically homogeneous [resistivity varies with position in the material], it us usual to break the object up into small, regular, near-homogeneous regions [cubes or near-cubes] and combine the resistances of these regions appropriately to get the resistance of the object.

[mmiguel]

Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units.

It thus makes sense to speak of a material having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an object, it's meaningful to speak of resistance.

Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.

Spoiler for So ...

To analyze objects that are not electrically homogeneous [resistivity varies with position in the material], it us usual to break the object up into small, regular, near-homogeneous regions [cubes or near-cubes] and combine the resistances of these regions appropriately to get the resistance of the object.

I was attempting to recreate a problem from an EE final I had 1.5 years ago based on memory, but I realized after I submitted that I had some minor problems in the statement. One which causes the actual calculation to be slightly more difficult than I anticipated (harder integral), and now.... units.

I would argue that the differential volume elements have shape and are objects, but this would not be helpful, because integration involves more than just summing stuff together - it also involves a change in units due to multiplication of differential elements of whatever units belong to the variable of integration. Thus if you intend to use something as part of an integrand, such that after integration it becomes what you originally had, what's in the integrand is actually a density of the original quantity vs. being portions of the original quantity.

Spoiler for

The way this is intended to be used is that it should be integrated in sheets perpendicularly to the x-axis. This double integral involves multiplying it by differential area components.
However, resistances do not sum when added in parallel. It is actually conductances that must be added together here.
The total conductance of a cross-sectional element (infinitesimal in the x-axis) is the integral of the reciprocal of R(p). The units of this are area/units of R.
The reciprocal of the entire double integral we just calculated is then taken to get another form of resistive "density". This may then be integrated along the x-axis. This is equivalent to multiplying each individual resistive slice value by length and summing together. The units of resistive slices are (units of R)/area.
Thus (units of R)/area*length = Ohms = units of R/length;
Thus units of R = Ohms * length = Ohm*cm.
These are units of resistivity.

Really, the reason I posted this problem, was because I found it very interesting that one needs to do two reciprocals in the middle of 3 integrals in order to get things to combine right. I was trying to come up with a similar math problem outside of the context of EE, where this is required, but couldn't come up with anything quick enough, so I just went with this one.

Anyway, R(p) is more of a resistive density (per area and times length). This is a well known, widely used concept called resistivity as bonanova pointed out.

In essence, bonanova is correct.

Edited by mmiguel, 06 September 2012 - 07:48 AM.