-2
$\begingroup$

What causes a potential drop in a resistor or load? Why a does wire with neglible resistance have the same potential across it?

$\endgroup$
  • $\begingroup$ Kvl in a loop comes, that integral of E. dl is zero in close loop, $\endgroup$ – yuvraj singh Oct 22 at 11:22
  • $\begingroup$ Do you mean "same" potential or "some" potential. $\endgroup$ – Bob D Oct 22 at 11:53
  • $\begingroup$ You yourself answer a wire having negligible resistance has same potential across it. $\endgroup$ – VK_fan Oct 22 at 14:35
  • $\begingroup$ It's same not some. $\endgroup$ – aditya siroutiya Oct 23 at 2:22
-4
$\begingroup$

There are many ways to solve it.

Let's start with Ohm's law, $V=iR$. This law may also be formulated in a more insightful, microscopic form, especially in the simplest microscopic version of Ohm's law (due to Gustav Kirchhoff):

$$ \vec J = \sigma \vec E $$

Here, $\vec J$ is the current density, $\sigma$ is the conductivity of the material, and $\vec E$ is the electric field, i.e. the potential (voltage) drop per unit distance (including the direction of the decrease).

These arguments can be visualized so easily as follows. Imagine the circuit. The switch is open. This means, the charges could not flow through the circuit and complete the circuit. This is due to the gap in between the switch. So the applied energy is not converted to current even though the resistance is still there. Hence voltage will be there, where you applied it.

Now, in the second case, there is no resistance. This means the supplied energy (the voltage) is completely utilized by the charges to flow through the circuit. There is nothing left so as to be detected by any external circuit (a voltmeter probably). Hence one cannot see any voltage difference between the conductors. You can see energy somewhere if it is there. But if it's being utilized, then you will not see it there, but it's there in some other form - here it is the kinetic energy of the charges.

The energy concept here belongs to phonons. Actually, the electrons scatter off phonons in the resistor and the electron kinetic energy gets converted to energy of the phonons. The energy that goes into the phonons gets distributed around the resistor and ends up as heat.

$\endgroup$
  • $\begingroup$ How energy is converted into current?Please don't edit the question until you answer my query $\endgroup$ – VK_fan Oct 22 at 14:33
  • $\begingroup$ Have you heard about phonons. $\endgroup$ – yuvraj singh Oct 22 at 16:16
  • $\begingroup$ But how phonons convert energy into current.In fact electric current is a tensor while energy is scalar.Are you saying that tomatoes and oranges are interchangeable? $\endgroup$ – VK_fan Oct 22 at 17:39
  • $\begingroup$ @unique What actually happens is the electrons scatter off phonons in the resistor and the electron kinetic energy gets converted to energy of the phonons. The energy that goes into the phonons gets distributed around the resistor and ends up as heat. $\endgroup$ – yuvraj singh Oct 23 at 2:40
3
$\begingroup$

What causes a potential drop in a resistor or load?

Perhaps the following mechanical analogy will help.

Consider a block sliding down an incline plane with friction. Let the component of the force of gravity acting down the block exactly equal the kinetic friction force acting up the plane so that the block is sliding down at constant velocity. The block loses gravitational potential energy sliding down the block. Since the velocity is constant there is no increase in kinetic energy and all of the gravitational potential energy is converted to friction heating of the block/plane contact surface.

The rough electrical analogy is:

  1. The incline plane is analogous to our resistor.

  2. Its coefficient of kinetic friction is analogous to the resistivity of our resistor.

  3. The block is analogous to our electric charge.

  4. The block moving at constant velocity is analogous to our current.

  5. The loss of gravitational potential energy is analogous to the loss in electrical potential energy.

  6. The friction heat generated at the surface of the inclined plane is analogous to the energy dissipated in the resistor due to resistance heating.

Why a wire having negligible resistance have same potential across it?

I'm assuming you mean "some" potential across it. A wire (unless super cooled) will always have some resistance, i.e., it will never be zero. So if there is current in the wire there will be some potential drop (loss in electrical potential energy) but it will be lower than the potential drop across a higher resistance in series with the wire.

Carrying the above mechanical analog further

  1. Let the block (charge) encounter another inclined plane (resistor) after coming to the bottom of the first. The two incline planes (resistors) are in "series".

  2. This second incline plane is very shallow so that its vertical height is much less than the first incline. That means the loss of gravitational potential energy (electrical potential energy) when the block reaches the bottom of the second incline will be much less than the first.

  3. To keep the block going at the same constant velocity (same current) as it did on the first incline (the two planes (resistors) being in "series"), the coefficient of kinetic friction (electrical resistivity) of the second plane is much less than the first. This second incline is our low resistance wire.

These analogies are not exact and are only intended to give you a better feel as to what is happening.

Hope this helps.

$\endgroup$
  • $\begingroup$ I believe the OP did actually mean same potential across it. A wire of negligible resistance would be the ideal circuit element where you can assume potential drop across the wire is 0. (see also them repeating this question with a textbook picture physics.stackexchange.com/questions/509662/…) $\endgroup$ – JMac Oct 23 at 3:16
0
$\begingroup$

In a DC circuit there is an excess of electrons on the negative terminal and a deficiency on the positive terminal of the power supply. Throughout the circuit the electrons arrange themselves between these extremes in a manner that maintains a constant flow of current. In regions of high resistance (where the mean free path between collisions with atoms is low) the charge gradient (and electric field strength) must be high to maintain the required average drift velocity (achieved by acceleration between collisions). The potential energy of the electrons drops more rapidly in regions where the field is strong. They are losing this energy in the collisions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.