Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section

Edith Mayer 2022-05-17 Answered
Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section?
Before I explain my query, I would like to clarify that I am a ninth-grader who got this question while studying the formula R 1 A where A is the area of cross-section.
I have often asked this question to my teachers and they always give me the classic "corridor and field example". They told me that if 5 people walk in a corridor, they will find it harder to get across than if they were to be walking through a field- the same goes for electrons passing through a conductor. My counter-argument would be that if the width of the conductor increases, so will the number of positive ions (my textbook says that positive ions in conductors hinder the flow of current) and hence, more the resistance.
I would really appreciate it if the answer could be explained to me in simple terms as I'm not well versed with the more complex formulae involved in this concept. If not, do let me know of the concepts I should read about (preferably the specific books) to understand the solution better.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Eliezer Olson
Answered 2022-05-18 Author has 16 answers
Sorry for my poor english. My native language is french.
We can also ask the question about the conductance: why is it proportional to the surface.
The reason is that we are working with a model in which the current density is distributed uniformly over the cross section of the conductor. It is as if we had a lot of identical conductors in parallel with each other. So, for the same voltage, if we multiply the area by N, we also multiply the current by N. This would not be the case if the current distribution was not uniform. For example, if we take into account the skin effect.
Another interesting example is that of the Hagen-Poiseuille's law in hydraulics: the flow rate is proportional to the pressure difference but the hydraulic conductance is proportional to the square of the area: it is not proportional to the area ! This is because the velocity profile is parabolic (and not uniform). Doubling the pipe area is not the same as taking two pipes.
Not exactly what you’re looking for?
Ask My Question
Brooklynn Hubbard
Answered 2022-05-19 Author has 3 answers
The following model is not entirely true, but it's a much truer model than people walking down a corridor. I hope it helps.
Start by considering a resistor that just one atom wide, but many atoms long, in a long string.
Each atom has an electron cloud. In a semiconductor, the outermost electrons in the electron cloud are weakly bound to their atoms. It doesn't take a very strong push to separate the electron from its atom, but it's not free, either. And pushing an electron free from its atom doesn't make it free to keep moving that electron.
As soon as the electron is pushed out of the first atom it will see another atom, sitting there with all of its electron orbitals filled. It needs to push one of that atom's electrons out of its orbital and take its place. Then as soon as that electron is pushed over, it encounters an atom, and so on.
The end result is that if we want to push an electron out at one end, we need to push an electron in at the other end, with an amount of force equal to the total amount of force required to push all the electrons one atom over. We pop one electron in at one end of the resistor, it pops an adjacent electron one atom over, that one pops the next electron one atom over, and so on, until we get to the end of the resistor and an electron pops out.
In a circuit, the circuit voltage sets the amount of force that the circuit can provide, per electron.
So, let's say that the circuit voltage can push ten times harder than is needed to push one electron through one line of atoms in one second.
Then the circuit can shove ten electrons a second into one end of the resistor, and ten electrons a second will fall out of the other end of the resistor. Current is flowing... but not very much.
What if we add a second line of atoms in parallel with the first? The circuit pushes per electron, so it can push another ten electrons per second through the second line of atoms. Two lines of atoms means two times the current.
So on for each additional line of atoms: each line of atoms gets us another ten electrons per second getting shoved in on one end and another ten electrons per second falling out on the other end of the resistor.
The cross-sectional area of the resistor is proportionate to the number of lines of atoms we can shove electrons in one end of so that electrons fall out the other end.
So we can say: A I, the cross-sectional area is proportionate to the current.
And since R = V / I
R 1 / A
We can go farther with this model and reach the standard formula for resistance.
What happens if we add more atoms to each line, in series, so that the line of atoms is longer? Each atom opposes our push by the same small amount, and is the same size. So, the resistance must be proportionate to length.
R L and R 1 / A
so
R L / A
Now we just need a proportionality constant related to how hard each electron is to push one atom over. For Resistance, the proportionality constant is represented by ρ (rho), the material's Resistivity.
R = ρ L / A
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-05-07
The figure shows four charges at the corners of a square of side L. What magnitude and sign of charge Q will make the force on charge q zero?
Express your answer with the appropriate units.
photo
asked 2022-05-17
What happens to pipe length when the pipe diameter changes?
Intuitively, when the diameter of pipe is decreased, there will be more friction loss, more water pressure, and a higher flow rate. Is there a direct relationship/equation derived to see the affect of the pipe lengths?
Deriving from Hagen-Poiseuille's equation, we get:
  Q = Δ P π r 4 8 μ L
where :
Q = flow rate
Δ   P = change in fluid pressure
r = radius of pipe,
μ = dynamic viscosity of fluid,
L = length of pipe
To keep similar flow rates, are we able to use Poiseuille's derived formula to find the new lengths of pipe with a change in r (pipe radius)?
asked 2022-05-09
Matching Bernoulli equation and Hagen Poiseuille law for viscous fluid motion
I thought that Bernoulli equation could be used only in the case of non viscous fluid. But doing exercises on 2500 Solved Problems In Fluid Mechanics and Hydraulics (Schaum's Solved Problems Series) I found that this procedure is followed.
In the case of viscous laminar flow, Bernoulli equation is written as
(1) z 1 + v 1 2 2 g + p 1 ρ g = z 2 + v 2 2 2 g + p 2 ρ g + h L
Where h L is the head loss (due to viscosity) calculated using Hagen Poiseuille law.
(2) h L = ρ g 8 η L v ¯ R 2
Is this a correct way to solve exercises involving viscosity?
Furthermore are there limitation to this use of Bernoulli equation (in case of viscosity)?
In particular
If the flow is not laminar, I cannot use ( 2 ), but can I still write ( 1 ) in that way?
Is ( 1 ) valid only along the singular streamline or between different ones (assuming the fluid irrotational)?
asked 2022-05-08
What determines the dominant pressure-flow relationship for a gas across a flow restriction?
If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to
Δ P = K 2 Q 2 + K 1 Q
where Δ P is the pressure drop and Q is the volumetric flow
and what I've observed is that if the restriction is orifice-like, K 2 >> K 1 and if the restriction is somewhat more of a complex, tortuous path, K 1 >> K 2 and K 2 tends towards zero.
I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the K 1 component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?
asked 2022-05-20
How does Newtonian viscosity not depend on depth?
I am currently trying to self-study basic Classical Mechanics and right now I am trying to understand fluid dynamics. I have read about Newtonian viscosity: as my book says, the force exerted on a viscous fluid depends on the coefficient of viscosity, the cross-section and the velocity gradient, but, to my amazement, not on the fluid's height, or in any other way on its total amount.
Usually I'd be more than willing to trust such a result even if not proven, and the way it is used in deriving something like Poiseuille's law, for example, almost makes sense to me. But I still, to my best efforts, fail to understand or even get an intuition on what should be the most basic case of application, i.e. a fluid that can be approximated as many rectangular plates flowing on top of each other. If I try to apply the same method I would use for a cylindrical tube, I get that considering a section of any height h, if the velocity gradient in the rectangle is constant at every depth, the force I need to apply to that section is the same no matter the value of h. In particular, dividing the rectangle in an arbitrarily large number of sections of arbitrarily small value of h, keeping the gradient the same I get that the total force I need to apply to maintain it is arbitrarily large.
Also, it sounds unintuitive that keeping the same gradient and halving the amount of fluid I would need the same force (and double the pressure). In addition to this, even with the tube, if instead of applying the viscosity law to a full cylinder I try applying it to a cylindrical ring, I get the same problems.
Can someone explain? Am I doing this wrong (probably)? Does the viscosity law somehow only work for the "top" section of the liquid? If so, what does top mean? How can the fluid "know" what is the top and what isn't? Are different pressures really needed depending on the amount of fluid to maintain the same gradient? What's an intuitive explanantion of this? Is there something else I'm missing that would make the explanation work?
asked 2022-05-15
Is there a contradiction between the continuity equation and Poiseuilles Law?
The continuity equation states that flow rate should be conserved in different areas of a pipe:
Q = v 1 A 1 = v 2 A 2 = v π r 2
We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.
Another way I was taught to describe flow rate is through Poiseuilles Law:
Q = π r 4 Δ P 8 η L
So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:
v A = v π r 2 = π r 4 Δ P 8 η L
Therefore:
v = r 2 Δ P 8 η L
Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.
What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.
asked 2022-05-18
Proving that the water leaving a vertical pipe is exponential (decay)
How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :
A e k x
I know that Torricelli's law is:
2 g h
But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.

New questions