# Effect of Resistance on Current

#### dschmidt12

I'm in an elementary E&M physics class and thought I understood voltage and current pretty well until we got to resistance. I have a few questions concerning the effect of resistance on current that I hope somebody will be able to answer...I know I am missing something fundamental in my thought process.

1. How is a certain potential difference applied to/maintained in a resistor? Why doesn’t the potential difference change depending upon the current that is passing through it rather than the other way around?

2. I know that electrical energy is converted into thermal energy when passing through a resistor, thus leading into the reducing of the current through the resistor, but how does this make sense when the current leaving a resistor must be the same as the current entering a resistor?

3. How does one mathematically calculate the current in an ideal battery? Since you can’t use V=IR, is there any other way to do it besides using P=IV or any of the other more complex ways (such as in using drift velocity, number of charge carriers per volume, etc.)?

Thanks to anyone who might be able to help clear up my questions.

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#### Gerenuk

1. How is a certain potential difference applied to/maintained in a resistor? Why doesn’t the potential difference change depending upon the current that is passing through it rather than the other way around?
A potential difference means there is an electric field inside (field=dragforce/charge=potential/distance). Therefore somewhere outside there is an inbalance of electrons and vacant electron places, so that an electric field between them is created. This electric field drags electrons through the medium. These electron collide and lose the drag energy they have just gains, but start feeling the electric field again and go on for the next collision further down the way :)
In theory, if you somehow manage to change the electric current, then the potential difference will also change.... since you can only change the current by the potential difference :) Unless you manage to change the resistance of course. Maybe its philosophical to ask which one is the cause. A "current source" would adjust its voltage until the desired current is reached.
The potential difference also tells you how much energy an electron gains when traversing it (potential=energy/charge). But the electron doesn't keep that energy - it loses it as heat to the resistor.

2. I know that electrical energy is converted into thermal energy when passing through a resistor, thus leading into the reducing of the current through the resistor, but how does this make sense when the current leaving a resistor must be the same as the current entering a resistor?
The resistor reduces the velocity of the electron momentarily, but at the same time the field gives extra velocity to the electron. Effectively the mean velocity will settle at an equilibrium between these two processes, so the potential and field in fact determine the velocity of the electrons.
The current however depends on speed of the electrons *and also the density* of the electrons. The density adjusts to make the current constant (otherwise charge would pile up and urge for a change).
Imagine a car driving through city traffic followed by a highway (without an emerging traffic jam). The speed is different (different electric field and potential difference), but the throughput (current) is the same as cars on the highway have a larger distance between each other.

3. How does one mathematically calculate the current in an ideal battery? Since you can’t use V=IR, is there any other way to do it besides using P=IV or any of the other more complex ways (such as in using drift velocity, number of charge carriers per volume, etc.)?
The current depends on *all* parts of the circuit. In particular the attached resistors and so on. From the voltage and the circuit layout you know what current is leaving the battery. Or do you mean something else?

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#### dschmidt12

Thank you so much for all your help! I think I'm starting to understand all this, although I still feel that I have some remaining questions...

So, since the resistor has a potential difference, it also has an electric field associated with it. But this electric field is different than the electric field that is associated with the potential difference between both terminals of the battery (or other source of potential), right? Does this mean that a charged particle traveling through the resistor feels the effects of both the electric field produced by the battery and the electric field produced by the resistor? Also, does this mean that one end of the resistor was previously charged in order to create this difference in charge and thus a potential difference?

Also, from your answer I get the impression that the energy dissipated by a resistor was produced by the collision of electrons in the resistor. Firstly, does this occur because the current density is greater and the electrons have a higher chance of colliding? (By this I mean, if you had something without resistance, would the electrons not collide at all?) Also, I thought I remembered that the energy dissipated by the resistor was equal to the power produced by the battery. Is this true, and, if it is, how does this make sense if the electrons could conceivably keep colliding with each other, gaining energy back from the electric field, and collide with each other again?

I'm sorry that I have so many questions...this particular subject just seems to confuse me a lot more than a lot of other ones we've been covering. :)

#### sophiecentaur

Gold Member
The energy that the battery gives each charge is transferred as the charge flows around the circuit. It can be transferred in a resistor or in a motor or some electronic device. By the time the charge has gone right round the circuit, the energy has 'all gone'. In a simple, ideal, circuit, the electric field across the resistor will be equal to the battery volts divided by the length of the resistance path (field is measured in volts per metre). Very little energy is transferred in moving the charges through the wires so there is very little voltage drop but there is a finite electric voltage gradient along the whole path.

It helps NOT to think of the voltage as a force. That's a beginner's approach and can lead to many further misconceptions. In simple circuits the resistance of a component is best thought of as just the ratio of V and I, after a suitable settling time after switch on.

#### Gerenuk

I think that's very clever questions. That's the way to understand physics.
Electricity seems to be a simple phenomenon, and yet I was unsuccessful in trying to find out in different physics forums what exactly happens when DC electricity flows. I was interested in where what electric/magnetic field exists, what the charge density is and by what mechanism electrons propagate.
The only insight I had was that skin effect is important. Maybe now I can think about it again :) It can't be that hard to at least find the static state scenario...
...if there is a billiard effect and why electrons don't flow between opposite poles of different voltage sources.

Does this mean that a charged particle traveling through the resistor feels the effects of both the electric field produced by the battery and the electric field produced by the resistor? Also, does this mean that one end of the resistor was previously charged in order to create this difference in charge and thus a potential difference?
I think there is only this *one* field that can be calculated from the potential difference and the distance. The resistor doesn't need to be charged as the electric field from outside has no problems going into the resistor.
The resistor doesn't produce an own contribution.

Also, from your answer I get the impression that the energy dissipated by a resistor was produced by the collision of electrons in the resistor. Firstly, does this occur because the current density is greater and the electrons have a higher chance of colliding? (By this I mean, if you had something without resistance, would the electrons not collide at all?)
What do you mean by "current density is *greater*"? Before application of a voltage electrons also collided with impurities, but they were in equilibrium with the material and were happy scattering in all directions. After an electric field is applied the electrons are pushed to a *unidirectional motion*, however again collision will scatter the electrons and destroy the directionality (i.e. current flow). The scattering hasn't changed much, but the current flow is hampered.

Also, I thought I remembered that the energy dissipated by the resistor was equal to the power produced by the battery. Is this true, and, if it is, how does this make sense if the electrons could conceivably keep colliding with each other, gaining energy back from the electric field, and collide with each other again?
Yes, the battery provides exactly the power to make up for the resitance losses.
Not absolutely sure what you mean.
The electrons collide with impurities in the resistor material and there create vibrations in the resistor which is effectively heat. On their way electrons keep gaining energy from the electric field and losing it again to the resistor material.
If electrons collide amongst each other they of course don't lose energy from the "electron system".

#### sophiecentaur

Gold Member
When discussing electron flow due to an electric current it is important to realise that the average speed of flow of elctrons is only in the order of mm per second compared with a very much higher mean velocity due to the thermal motion. Also, it isn't 'collisions' with atoms as much as 'interactions with the structure' which accounts for the energy transfer into thermal energy.

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