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?

  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html $\endgroup$ – evil999man Apr 12 '14 at 11:06
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    $\begingroup$ 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 '14 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 '14 at 12:17

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.

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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.

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  • $\begingroup$ 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 '14 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 '14 at 11:42
  • $\begingroup$ @User17670 But In either way you are asking about Currents , right? $\endgroup$ – Poomrokc The 3years Apr 12 '14 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. $\endgroup$ – Poomrokc The 3years Apr 12 '14 at 11:48
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    $\begingroup$ "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". $\endgroup$ – CodesInChaos Apr 12 '14 at 12:09

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.

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