8
$\begingroup$

Problem

Current is the amount of charge that is flowing through a component per unit of time.

For a given voltage, Ohm's law tells us that if we increase the resistance, then the current must decrease.

But what's actually happening to decrease the current?

My reasoning so far - is it correct?

More resistance (if we're talking about something of the same size for simplicity) is more 'stuff' in the way (higher resistivity), so more collisions. More collisions means it takes longer for the charge to 'get through' the component. The charge is moving slower and so the current is lower.

Or, is it that the speed is always the same and that somehow if we have more resistance then it means there is just less charge able to flow, hence a lower current. If so, why?

$\endgroup$
4
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html $\endgroup$
    – evil999man
    Apr 12, 2014 at 11:06
  • 2
    $\begingroup$ The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed. $\endgroup$
    – Alexander
    Apr 12, 2014 at 12:13
  • $\begingroup$ @Alexander can you expand on your answer? (I'd like to assume that we'd only be changing the material type and not the size, and that all materials are Ohmic) $\endgroup$
    – User 17670
    Apr 12, 2014 at 12:17
  • $\begingroup$ You probably mean resistivity instead of resistance. Changing the resistance can already be done,trivially, by taking a different length / cross section ratio. $\endgroup$
    – my2cts
    Jul 2, 2021 at 13:18

6 Answers 6

9
$\begingroup$

The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed.

In a basic Drude model for electrical transport both, $n$, the charge carrier density and $\tau$, the time between collisions determine the resistance:

$$\mathbf{J} = \left( \frac{n q^2 \tau}{m} \right) \mathbf{E}.$$ Here, $\mathbf{J}$ is the current density and $\mathbf{E}$ the applied electrical field. The term in parentheses characterizes the material properties, i.e. its resistance.

By changing to a different material you can influence both, $n$ and $\tau$, so both cases are possible.

$\endgroup$
0
$\begingroup$

Your understanding is correct. From $V=IR$ if voltage stays the same while resistance is increased, the current should be decreased.

But if you have heard of another equation $I = \frac{Q}{t}$

If current(I) is increased and the charge $Q$ is fixed (Charge is fixed if the power supply is from power cells like battery). Time will be decreased.

Which means that

The charges are moving faster! Because numbers of charge are fixed by the size of battery.

Imagine a four lanes road. Then in 300 meters ahead, there is a construction for road maintenance for three lanes. Making only one lane available to pass. Cars from 4 lanes must combine into one lane, making traffic jams and cause everyone to get home slower.

4 lanes road = Normal wire

Construction = Resistors

1 lane road and traffic jams = Problems cause by resistance.

Cars = Electrons

Home= + terminal of battery

Like some phone battery have 10000mAh. Which actually is the value of $Q$ but in milliamperes * hours not ampere * seconds. So 1 mAh is equal to 5/18 Coulomb.

I hope you understand my explanation.

$\endgroup$
10
  • 1
    $\begingroup$ What do you mean by 'capacity'? As far as I'm aware, batteries don't store charge, they only move it. Are you saying it has the capacity to move 10000 mAh? $\endgroup$
    – User 17670
    Apr 12, 2014 at 11:40
  • $\begingroup$ Thanks for the response though! But, my question is about fixed voltage - I want to know 'on a microscopic level, what is resistance doing to reduce current' $\endgroup$
    – User 17670
    Apr 12, 2014 at 11:42
  • $\begingroup$ @User17670 But In either way you are asking about Currents , right? $\endgroup$ Apr 12, 2014 at 11:44
  • $\begingroup$ @User17670 Batteries do have charges in itself.A fixed amount of +charges and -charges.These to charges combine into Q(Culombs). But the charges are neither deplete or destroyed. -charges(electrons) runs to +charges. When all electrons are all run to positive poles of the batteries, It's called "battery empty". If sorry to confuse you with the word "capacity". I will edit that. $\endgroup$ Apr 12, 2014 at 11:48
  • 2
    $\begingroup$ "if voltage is increased while resistance stay the same , the current should be decreased." I think you messed that up. You probably meant "if resistance is increased while voltage stays the same". $\endgroup$ Apr 12, 2014 at 12:09
0
$\begingroup$

Current is produced in a metal when the free electrons in the metals acquire a drift velocity due to an electric field. But when these free electrons travel through the metal, their path is hindered by other atoms and particles and their electomagnetic pull. More the resistance, higher is this hindrance and lesser is the drift velocity. Hence, the current moves through the metal more slowly (which directly corresponds to less number of electrons passing through a given point in the wire per unit time). So yes, your reasoning is correct.

$\endgroup$
0
$\begingroup$

For an Ohmic conductor (not a semiconductor) the number of conduction electrons is constant. It is the drift velocity that increases as a larger potential difference means a larger acceleration between scattering events. You can change the resistivity by changing the temperature. This will change the scattering probability, hence the time interval of acceleration hence the drift velocity.

Of course you can also change to a different material, which will have a different resistivity for example because it has a different conduction electron density and mobility.

$\endgroup$
0
$\begingroup$

For a given voltage, Ohm's law tells us that if we increase the resistance, then the current must decrease.

But what's actually happening to decrease the current?

If we increase the potential, the current increases, rest all the parameters are constant - which are resistance, material, temperature etc. Drift velocity of the charges, here, have increased due to increased potential.

Drift velocity is $v_d = \frac{e.V.d.\tau}{m}$, where

$e$ - charge

$V$ - Potential

$d$ - distance moved by the charge in time $\tau$

$m$ - mass of the charge

So here we can say that Charges move with greater speed now

Now to change the resistance, let us just change resistivity i.e changing the material and rest all the parameters are same except Free Charge Density of the material, it depends on the material.

Current, $I = n.e.A.v_d$, here $(n.e)$ is the charge density, the effective charge that is flowing per unit volume.

So here the amount freely flowing charge changes and also the drift velocity changes because $\tau$, the effective time between consecutive collisions, may also change.

$\endgroup$
8
  • $\begingroup$ The second point is not correct. In both cases it is the drift velocity that increases. The number of charges remains constant for an Ohmic conductor. $\endgroup$
    – my2cts
    Jul 2, 2021 at 12:53
  • $\begingroup$ Do the charge density remain same for all different ohmic materials ? I am changing the resistance by changing the "material". $\endgroup$ Jul 2, 2021 at 12:56
  • $\begingroup$ Okay, I get it now. But the charge density changes too. $\endgroup$ Jul 2, 2021 at 13:08
  • $\begingroup$ If you change the 'material' (why the quotes?) then of course there is no reason why the resistivity should be the same. $\endgroup$
    – my2cts
    Jul 2, 2021 at 13:15
  • $\begingroup$ quotes to emphasis that I have changed material to change resistance, hence the charge density changes. this is what I meant. $\endgroup$ Jul 2, 2021 at 13:36
0
$\begingroup$

My answer is based on the similarities between electricity and fluid behaviour:

Imagine that current is the fluid flow. You can change current either by

  1. changing resistance or
  2. changing the potentional.

If you increase the pressure in a hose then the water will flow through it more quickly. But if you change its resistance, e.g. if you increase the cross section of the hose and you keep the same pressure as before, then there will be more water but with the same rate of flow (the same velocity).

So I suppose that if you change the resistance BY INCREASING THE CROSS SECTION, you change the amount of electrons, but if you change potential, you change the velocity of electrons. Correct me if I am false.

However if you increase the resistance by increasing the length of the resistor, or by heating the element, then I don't know.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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