What factors determine wattage of electrons flowing through a wire?

[LouisL]

Homework Statement

Find equations that help you conceptualize what factors influence the wattage of electrons in a wire.

Relevant Equations

Not sure, which is partially why I am asking this question. I know Ohms law is V=IR, where V is Voltage, I is current and R is resistance. I am also thinking that the kinetic enegy equation, E=1/2 mv^2 plays a role too, since Energy is proportional to velocity squared and in this case the average velocity of the electrons in the wire. But I think I need some more in depth insight.

Preface: I am new to physics and am trying to learn on my own, though I hope to take a class soon. I looked for a forum for beginners but didn't see one, but this seemed to be the closest forum. So I hope I can get some help with my question here. If it is the wrong forum, feel free to transfer my question to the correct one.

My intuition is telling me that the factors that affect wattage (or energy/time given in Joules/sec) are the speed of the electrons passing through a given point of a wire per second and the # of electrons passing through a given point per second. I am probably missing some factors here. I assume Temperature surrounding the wire and cross sectional area of the wire probably play a role too.

Here is a simple visual I have in my mind and maybe it is too simplistic: I am imagining 2 metal wires (both in the same room, so that the Temperatue is the same for each wire and also let's assume they have the same cross sectional area). The 1st wire has many slow moving electrons passing a given point every second; and for comparison, a second wire of a different metal has half as many electrons passing a given point per second but moving twice as fast. My intuition is telling me that the 2nd wire has 2x the energy or wattage of the first wire.

I am already feeling confused and I think this can be cleared up by getting some equations to give me a feel for what factors of the electrons influence their energy going through a wire.

Please be easy with me. I am a novice and am trying to learn.

Thanks.

 

[Delta2]

The power is $$P=I^2R=(nSve)^2\rho\frac{l}{S}$$ where $n$ density of free electrons in the wire, $S$ cross section ,$l$ length, $v$ the velocity of free electrons and $\rho$ the resistivity of the wire. (and $e$ the charge of free electron). So I guess you can figure out by yourself which factors affect the power.

 

[jbriggs444]

I am also thinking that the kinetic energy equation, E=1/2 mv^2 plays a role too, since Energy is proportional to velocity squared and in this case the average velocity of the electrons in the wire.

The kinetic energy of the electrons turns out to be utterly irrelevant. Although this is not an unreasonable guess about the nature of electrical power flow, it turns out to be completely wrong.

As a general rule that holds at least until one starts talking about transistors and diodes, it is best to ignore the existence of electrons entirely. They have nothing to contribute to a model of electricity or current. But since all students insist on thinking about electrons in spite of this fact...

The average velocity of an electron in a wire that is carrying a current is typically on the order of magnitude of one millimeter per second. Wikipedia gives an example of 1 amp in a 2 millimeter copper wire and computes a drift velocity of 0.023 millimeters per second. [The thermal velocity of the electrons is much much higher. We are talking here about the average drift velocity as the electrons slowly move against the conventional current flow]

Electrons have an extremely low mass. Combined with an extremely low drift velocity and the fact that it is the square cube of the drift velocity which figures into power, this means that the kinetic energy of electrons does not contribute significantly to the transmitted power.

The energy is actually transmitted in the electrical field.

The intuition that I use to wrap my head around this is the fluid analogy. Electricity acts a lot like a fluid [which can tolerate insanely high pressure and and has insanely low viscosity]. Electrical potential (or Voltage) acts a lot like fluid pressure. If you push an electrical current (a fluid flow) across a pressure drop (a potential difference) then you get power dissipated equal to potential difference times current (pressure drop times flow rate).

In other words, it works just like a hydrostatic drive.

 

[PeroK]

Homework Statement:: Find equations that help you conceptualize what factors influence the wattage of electrons in a wire.
Relevant Equations:: Not sure, which is partially why I am asking this question. I know Ohms law is V=IR, where V is Voltage, I is current and R is resistance. I am also thinking that the kinetic energy equation, E=1/2 mv^2 plays a role too, since Energy is proportional to velocity squared and in this case the average velocity of the electrons in the wire. But I think I need some more in depth insight.

Preface: I am new to physics and am trying to learn on my own, though I hope to take a class soon. I looked for a forum for beginners but didn't see one, but this seemed to be the closest forum. So I hope I can get some help with my question here. If it is the wrong forum, feel free to transfer my question to the correct one.

My intuition is telling me that the factors that affect wattage (or energy/time given in Joules/sec) are the speed of the electrons passing through a given point of a wire per second and the # of electrons passing through a given point per second. I am probably missing some factors here. I assume Temperature surrounding the wire and cross sectional area of the wire probably play a role too.

Here is a simple visual I have in my mind and maybe it is too simplistic: I am imagining 2 metal wires (both in the same room, so that the Temperatue is the same for each wire and also let's assume they have the same cross sectional area). The 1st wire has many slow moving electrons passing a given point every second; and for comparison, a second wire of a different metal has half as many electrons passing a given point per second but moving twice as fast. My intuition is telling me that the 2nd wire has 2x the energy or wattage of the first wire.

I am already feeling confused and I think this can be cleared up by getting some equations to give me a feel for what factors of the electrons influence their energy going through a wire.

Please be easy with me. I am a novice and am trying to learn.

Thanks.

Perhaps you need to learn a bit more about electromagnetism before attempting this question.

If we consider the simple model of a charged particle moving at a constant velocity, then (by Newton's first law) there is no change in energy or power input or output.

The power of a current, therefore, does not relate to the simple linear motion of charged particles.

PS from what sources are you trying to learn about EM (electromagnetism)?

 

[jbriggs444]

If we consider the simple model of a charged particle moving at a constant velocity, then (by Newton's first law) there is change in energy or power input or output.

I think you meant to say "no change" here.

But the model that I read from OP does not involve constant velocity. It involves a high velocity side and a low velocity side. You send the high velocity pellets through some sort of magic whiz-bang and get slow velocity pellets out, harvesting energy as a result.

This sort of design actually works for fluids. Although many hydraulic turbines in low pressure operation run at nearly equal inlet and outlet flow velocities, some hydraulic turbines in high pressure environments use high inlet velocity and harvest the kinetic energy.

The hydraulic turbines can also be categorized into two types: impulse and reaction turbines based on the change of the water pressure while passing through the rotor. For impulse turbines, the pressure of the water remains the same when passing through the rotor. However, when the water passes through the nozzle, the energy in the water is transformed into kinetic energy. The high-speed jet then strikes the bucket shaped vanes mounted on a rotating shaft and consequently transforming the kinetic energy of the water to a rotation movement through the runner. Pelton turbine is the most common among the impulse turbines category. This type of turbine is suitable for high head 200–4000 m and at low water flow rate, ranging from 0.5 to 20m3/s with net power reaching 200 MW, and its efficiency can reach 90% [22]. For the reaction turbines, water pressure varies depending on the variation of the flow path shape, which comprises fixed guide vanes, adjustable guide vanes, and rotating blades.

Newton's laws do have something to say about such a design. If you take in a high velocity unidirectional flow and spit out a low velocity unidirectional flow, a net force must result. You'd better bolt your turbine to the floor. [Or operate it in reverse and bolt it to the jet-ski].

We do not observe any net thrust from a resistor in an electrical circuit. This is powerful evidence against a simple mechanical model.

[If the kinetic energy is embodied in some kind of omni-directional thermal thing then the second law of thermodynamics has something to say]

 

[LouisL]

The power is $$P=I^2R=(nSve)^2\rho\frac{l}{S}$$ where $n$ density of free electrons in the wire, $S$ cross section ,$l$ length, $v$ the velocity of free electrons and $\rho$ the resistivity of the wire. (and $e$ the charge of free electron). So I guess you can figure out by yourself which factors affect the power.

The power is $$P=I^2R=(nSve)^2\rho\frac{l}{S}$$ where $n$ density of free electrons in the wire, $S$ cross section ,$l$ length, $v$ the velocity of free electrons and $\rho$ the resistivity of the wire. (and $e$ the charge of free electron). So I guess you can figure out by yourself which factors affect the power.

Thank you! I will be thinking about this.

 

[LouisL]

The kinetic energy of the electrons turns out to be utterly irrelevant. Although this is not an unreasonable guess about the nature of electrical power flow, it turns out to be completely wrong.

As a general rule that holds at least until one starts talking about transistors and diodes, it is best to ignore the existence of electrons entirely. They have nothing to contribute to a model of electricity or current. But since all students insist on thinking about electrons in spite of this fact...

The average velocity of an electron in a wire that is carrying a current is typically on the order of magnitude of one millimeter per second. Wikipedia gives an example of 1 amp in a 2 millimeter copper wire and computes a drift velocity of 0.023 millimeters per second. [The thermal velocity of the electrons is much much higher. We are talking here about the average drift velocity as the electrons slowly move against the conventional current flow]

Electrons have an extremely low mass. Combined with an extremely low drift velocity and the fact that it is the square cube of the drift velocity which figures into power, this means that the kinetic energy of electrons does not contribute significantly to the transmitted power.

The energy is actually transmitted in the electrical field.

The intuition that I use to wrap my head around this is the fluid analogy. Electricity acts a lot like a fluid [which can tolerate insanely high pressure and and has insanely low viscosity]. Electrical potential (or Voltage) acts a lot like fluid pressure. If you push an electrical current (a fluid flow) across a pressure drop (a potential difference) then you get power dissipated equal to potential difference times current (pressure drop times flow rate).

In other words, it works just like a hydrostatic drive.

Thank you! I will be thinking about this.
Regarding your statement: it is best to ignore the existence of electrons entirely. They have nothing to contribute to a model of electricity or current. But since all students insist on thinking about electrons in spite of this fact...
This makes no sense to me. How can electrons have nothing to do with the model of electricity or current, since e*some* properties of electrons cause current! My goal is to find out "which properties those are." I believe you that it is not kinetic energy, as you seem really knowledgeable about this subject. I like that you brought up the concept of "thermal velocity", something I am not familiar with. I will read about it now.
Maybe my original question would have been better stated: What properties do electrons in a wire have that determine their wattage? I will plug away and study some more!

 

[LouisL]

Perhaps you need to learn a bit more about electromagnetism before attempting this question.

If we consider the simple model of a charged particle moving at a constant velocity, then (by Newton's first law) there is no change in energy or power input or output.

The power of a current, therefore, does not relate to the simple linear motion of charged particles.

PS from what sources are you trying to learn about EM (electromagnetism)?

Thank you! I will heed your advice! I actually do not know a good source to learn about electomagnetism. If you know a good source or website that is good at explaining electromagnetism to beginners, I would appreciate you sharing that with me. Thanks again.

 

[jbriggs444]

How can electrons have nothing to do with the model of electricity or current, since e*some* properties of electrons cause current!

For determining the number of watts dissipated by a one ohm resistor subject to a one volt potential difference, electrons have nothing to offer. Electrons have no properties relevant to a basic understanding of electrical current. You can imagine an electrical current as a flow of some abstract, continuous, massless, positively charged fluid and lose nothing.

Electrons do have a finite charge. This hypothetical fluid is not quite continuous. The Millikan oil drop experiment allows us to determine the charge on a single electron. But the charge is so low that the continuous approximation is used for almost all practical purposes.

Electrons do have a non-zero mass. One could measure this by looking at the deflection in the electron beam in a cathode ray tube. But the mass is so low that we ignore it for almost all practical purposes.

Electrons have a negative charge. This fact is determinable by the above two experiments. This also manifests if one is using vacuum tubes (you have to heat the cathode to get a current to flow. Heating the anode does nothing). Or doing DC arc welding (more energy gets deposited at the anode). But for normal circuits involving resistors, capacitors, inductors and wires, the direction of electron flow is utterly irrelevant. It does not enter into any of the equations.