# Electric Circuit Questions

1. Sep 3, 2009

### Red_CCF

Hi, I just have a couple of circuit questions here

1. When we increase current in a circuit, does the charges flow faster or does more charges flow?

2. Does a resistor require electrons to pass through them or do they just require the potential energy of the electrons? Let's say if I add energy (perhaps delivered by photons) into a motor or lightbulb, would they run?

3. When a switch to a circuit is open, is the resistance across the switch infinity?

4. When an electrons moves in a circuit, does some of its potential energy convert into kinetic energy before its energy is used up on the resistors? If so how do we determine at a certain point on the circuit how much kinetic and potential energy it has?

2. Sep 3, 2009

### davidrit

It can be either depending no the circuit element. Generally, in a wire the number of electrons is constant and increased current results from increasing the net velocity of the electrons. In other devices like transistors and tubes the number of electrons can be the primary way of varying current.
Where will you put in the photons and what exactly do you expect them to run?
It all depends on the type of switch of course but generally when a switch is open the resistance may be sufficiently high that it has a negligible effect (many gigaohms). Usually you could consider it to be infinite but in some cases it may have a significant effect. Also, the impedance at high frequencies could be much lower due to stray capacitance.
Yes, the electronics increase in kinetic energy and lose some of that energy when they collide with the ions in the lattice. In most circuits it would be difficult to determine the kinetic and potential energy of a single electron (and in fact impossible below a certain precision) but you could determine the mean kinetic and potential energy of many electrons.

If you read up on the Drude model of conduction and Fermi Dirac statistics it may give you a better understanding of these things.

3. Sep 3, 2009

### user111_23

1. When electrons move faster, more electrons pass a point in the conductor in one second, so technically, both increase. But the of number electrons is always the same, however.

2. Like the above person said, where would you put them? Photons don't run motors, the flow of electrons is what determines the speed of the shaft. But since it's a resistor, it also uses some energy from the charges.
As for the light bulb, electrons that are moving collide with the stationary atoms in the resistor, which produces heat, and depending on the resisting material, produces light. Sorry if I confused you.

3. Well it's not infinite, but the resistance of the air across the switch is very, very high.

4. Like the above person said, you can't determine the energy of an electron, only many. But yes, PE of the charges does convert into KE throughout the circuit.

4. Sep 3, 2009

### Red_CCF

Thanks for the replies.

What I'm wondering with my 2nd question is that whether some resistors require just energy to run and that charged particles going through them are not actually necessary given that the energy is provided. So with a lightbulb, if I take out the filament and add heat to it, would it glow? But with your description of motors, I take it that we need electrons at a certain current to drive the motor so in this case we do need particles going through the motors. If so can I understand it like this: some resistors require electrons as well as energy to move them while others just require the energy?

With regards to my 4th question, if we were to consider a bunch of electrons, how do we calculate its kinetic and potential energy at a certain point on the circuit? Does an electron have nearly all kinetic energy before it goes through a resistor?

Thanks for the help

5. Sep 3, 2009

### Integral

Staff Emeritus
Resistors are passive elements they do not "Run". When a potential is applied current flows and heat is produced.

The color of a glowing body is a measure of its temperature. So if you heat a body it will glow. Current passing through a filament heat it up to a given temperature therefore the filament glows with a characteristic color. It does not matter how you heat it, it will glow at the color which corresponds to the temperature. It does not matter how you heat the filament the result is the same.

6. Sep 3, 2009

### Red_CCF

So does a motor work in the same way? When electrons pass through them energy is given off to run the motors or does the motor actually need particles passing through it to go?

7. Sep 3, 2009

### Staff: Mentor

The kinetic energy of the electrons exists, but it is negligible for any practical circuit.

8. Sep 3, 2009

### Red_CCF

Okay so the energy that an electron loses going past a resistor is mostly in the form of potential energy? This is a part that I don't get because I thought potential energy is stored, so how is it lost/transferred?

9. Sep 4, 2009

### Staff: Mentor

Last edited by a moderator: May 4, 2017
10. Sep 4, 2009

### mikelepore

You seem to be visualizing potential energy being converted into kinetic energy and then being supplied to the resistor. That's not what happens. There is a potential difference _across_ the resistor. It's like in a gravity problem, you have the top of a hill and the bottom of a hill, therefore a ball rolls downhill. The two terminals of the resistor have different potentials, therefore the charge flows through it. Each unit of charge loses its potential energy in the act of flowing through the resistor.

11. Sep 4, 2009

### Integral

Staff Emeritus
In a purely resistive load, like an incandescent lamp, ALL energy is lost to heat. In general energy lost to heat is not recoverable, therefore it is not considered potential energy.

It is different in a motor since energy is stored in the field due to the motors inductance. For the most part energy stored in a magnetic field is recoverable. Evidence of this is the rotational kinetic energy of the armature.

Electrons in the current flow of the windings have two energy sinks, there is heat loss due to dc resistance of the motor windings and energy "lost" to creation and maintenance of the magnetic field. Work done building the magnetic field impedes the motion of the electrons. When the current drops, electrons get a push as the fields collapse. So inductors want to maintain existing current and resist any change in current.

It appears that it is the energy stored in the magnetic field which has you puzzled. Unfortunately the best way to understand this interaction is mathematically. If you are serious about understanding electric fields you must have the mathematical sophistication to understand Maxwell's equations.

12. Sep 6, 2009

### uzair_ha91

1. Because I=Q/t, therefore when more current flows, that means more charges flow in the same interval of time or the same number of charges flow in a smaller interval of time.

2. Any resistance is a conductor, thus electrons can flow across it. When we apply a potential difference across it i.e. connect it with a battery and close the circuit, the electrons in the whole circuit start flowing at the same time (just like when we open the tap, the water which is already present in the whole pipe starts flowing at the same time throughout the pipe). The electrons in the resistance, however, are simply facing a "resistance". So as they try to flow and passby this "resistance", they use up their energy (that is the emf of the battery being supplied to the electrons) which results in a disspation of heat. ((Energy provided by battery=>Potential difference=>Energy used up in resistance))

(When we apply voltage across a conductor, all the electrons acquire net velocity in the direction opposite to the electric field called the drift velocity, so we can say that the energy prvided by the battery is indeed K.E. and that being dissipated is K.E. into heat energy.
The potential differnce implies that the electrons before they pass through the resistance have the "potential" to do some work which they do. The analogy of droppig a ball from a height and then taking it back up to that height is appropriate here)

3. When a switch is open, I=0 and V=0, therefore: R=V/I=0.