How electric circuits really work

[Mr Davis 97]

I've been studying circuits, but I can't find any resources that really answer my specific questions. So far I understand that a battery creates a potential difference between the two terminals, one of which is positive while the other is negative. There is a buildup of positive charge on the positive side and buildup on the negative side, and this creates an electric field which can accelerate electrons from a location of high potential to one of low potential. When you connect a conducting wire between positive and negative terminals, this creates a path of least resistance which electrons can flow through, where the electrons are accelerated by the electric field that the battery creates. Hopefully all of this correct so far. My main question arises when it comes to voltage drops. What really is a voltage drop? If we connect one resistor to the circuit, a circuit with 12 V, then there must be a voltage drop across the resistor of 12 V. However, why is this the case? And also, how are the electrons still able to move if on the other side of the resistor there is 0 V, since there is a voltage drop?

I know that most of this is probably wrong, so I need someone to really help me understand how circuits and specifically voltage drops work.

 

[David Lewis]

There is a voltage drop because some of the energy stored in the electric field is converted to heat by the resistor. After the current passes through the resistor, the voltage will be slightly more than zero. Typically we ignore voltage drop in the connecting wires to simplify circuit analysis.

 

[houlahound]

Or by analogy consider the electrons leaving the resistor at zero volts after falling down a 12V hill once they reach "ground" level there is nothing to cause them "frictional " losses so they coast back to the negative "ground" side of the battery and get pumped back up the 12V hill again taking energy from the battery.

I suck at analogy.

 

[cnh1995]

My main question arises when it comes to voltage drops. What really is a voltage drop?

Voltage drop is the drop in potential when current passes through a resistor. Consider two resistances of 5 ohm connected in series with a 10V battery. Voltage drop across each bulb is 5V. So, the supply voltage gets divided between the series resistors to have the same current flowing through each. If you assign potentials to various points like +ve terminal as 10V and -ve terminal as 0V, you'll see as you move from one terminal of resistor to the other in the direction of current, a drop of 5V in potential takes place. We assume zero drop in the wires but practically there is some voltage present(negligibly small).

 

[Mr Davis 97]

I'm still not seeing it. Is there analogy that can be used with gravitational potential and gravitational potential energy? Those are pretty understandable, so an analogy with it would help. For example, with GPE, if I start at 10 feet and fall, my GPE is continuously being converted into kinetic energy, and thus motion. For EPE this does not seem to happen. Even when a current moves across a resistor for a voltage drop (such that there is no more EPE), the charges just keep moving at the same rate (same current) as before the voltage drop. This I don't understand.

 

[houlahound]

For completion should not the voltage loss inside the battery be mentioned so all the numbers add up ie internal resistance.

 

[houlahound]

Are you thinking that current is like projectiles??

A single electron does not travel around the circuit, they hardly move anywhere.

 

[Drakkith]

What really is a voltage drop? If we connect one resistor to the circuit, a circuit with 12 V, then there must be a voltage drop across the resistor of 12 V. However, why is this the case?

To understand voltage drop, you must first understand what voltage is. Voltage is a difference in electric potential energy between two points per unit of charge. So if you have an electron out in space in an electric field, where the voltage is 10 volts per meter, then the electron will experience an acceleration and gain 1.6×10⁻¹⁸ J of energy kinetic energy over this meter, losing 1.6×10⁻¹⁸ J of potential energy in the process. If the electron is accelerated in the same field over 2 meters, then the difference in voltage is 20 volts (10 v/m x 2 m) and the electron gains double the kinetic energy as before.

If you have a particle with a charge of -100 (the same charge as 100 electrons) then the particle will gain 1.6×10⁻¹⁶ J of kinetic energy over one meter in that same field, with an equivalent loss of potential energy.

Obviously conditions are a little different inside a circuit. The electric field created by the battery accelerates electrons along the entire length of the circuit. Since the circuit isn't empty space, electrons are continuously interacting with the circuit components as they move, and as they do so they lose some of the energy given to them by the battery (if they didn't, they'd just accelerate to some velocity and. The net effect of lots of electrons undergoing constant acceleration and interacting with the circuit is that the energy is lost entirely as heat, with most of the energy lost as electrons pass through the resistor. This loss of energy is known as voltage drop. It represents the amount of energy lost per unit of charge that moves through a component.

And also, how are the electrons still able to move if on the other side of the resistor there is 0 V, since there is a voltage drop?

The voltage drop across the resistor is not quite 12 volts. There is some voltage drop in the wires and in the battery itself.

 

[Mr Davis 97]

To understand voltage drop, you must first understand what voltage is. Voltage is a difference in electric potential energy between two points per unit of charge. So if you have an electron out in space in an electric field, where the voltage is 10 volts per meter, then the electron will experience an acceleration and gain 1.6×10⁻¹⁸ J of energy kinetic energy over this meter, losing 1.6×10⁻¹⁸ J of potential energy in the process. If the electron is accelerated in the same field over 2 meters, then the difference in voltage is 20 volts (10 v/m x 2 m) and the electron gains double the kinetic energy as before.

If you have a particle with a charge of -100 (the same charge as 100 electrons) then the particle will gain 1.6×10⁻¹⁶ J of kinetic energy, with an equivalent loss of potential energy.

Obviously conditions are a little different inside a circuit. The electric field created by the battery accelerates electrons along the entire length of the circuit. Since the circuit isn't empty space, electrons are continuously interacting with the circuit components as they move, and as they do so they lose some of the energy given to them by the battery (if they didn't, they'd just accelerate to some velocity and. The net effect of lots of electrons undergoing constant acceleration and interacting with the circuit is that the energy is lost entirely as heat, with most of the energy lost as electrons pass through the resistor.
The voltage across the resistor is not quite 12 volts. There is some voltage drop in the wires and in the battery itself.

If electric potential energy is expended after a voltage drop, then why is there still current after the voltage drop? The electrons don't have anymore electrical potential energy to convert to kinetic energy, right? (This might be a stupid question but bear with me)

 

[Drakkith]

If electric potential energy is expended after a voltage drop, then why is there still current after the voltage drop? The electrons don't have anymore electrical potential energy to convert to kinetic energy, right? (This might be a stupid question but bear with me)

If the battery provides 12 volts, then you might have 11.999 volts of voltage drop at the point that the wire connects to the battery terminal. The essential idea here is that the amount of voltage drop in a circuit is always equal to the applied voltage (otherwise you'd violate conservation of energy).

 

[houlahound]

Cos an electron at the start of the circuit moves an electron at the end of the circuit, the average velocity is the same everywhere.

 

[cnh1995]

I'm still not seeing it. Is there analogy that can be used with gravitational potential and gravitational potential energy? Those are pretty understandable, so an analogy with it would help. For example, with GPE, if I start at 10 feet and fall, my GPE is continuously being converted into kinetic energy, and thus motion. For EPE this does not seem to happen. Even when a current moves across a resistor for a voltage drop (such that there is no more EPE), the charges just keep moving at the same rate (same current) as before the voltage drop. This I don't understand.

Ok. Current is a result of electric fields established in various parts of the circuit. Consider a simple series circuit with V=10V, R1=6 ohm and R2=4 ohm. From voltage division principle, voltage across R1=6V and voltage across R2=4V. This gives a current of 1A through both of them. What does this tell you?
The current is same everywhere. In the wires, there is a little resistance, so to maintain the current, almost no voltage is required(hence, we assume it to be 0). Inside the 6 ohm resistor, to maintain the same current, larger electric field is required, hence, more voltage is present across it.

 

[Mr Davis 97]

If I had a circuit with just a battery and no other elements, how are electrons attracted to the positive terminal if there is no voltage drop along the way? If the charges never lose their potential energy, then how do they get to the negative terminal?

 

[cnh1995]

If I had a circuit with just a battery and no other elements, how are electrons attracted to the positive terminal if there is no voltage drop along the way? If the charges never lose their potential energy, then how do they get to the negative terminal?

That is called a short circuit and ideally, the current is infinite. You can see that the electrons will go on accelerating, increasing the current at every moment sinnce there is no resistance. Practically, there is a small finite resistance in the wires and the source has its own resistance which will limit the current to a finite but very high value.

 

[houlahound]

Until the wire disintegrates at high temperature.

 

[cnh1995]

If you want to understand the mechanism of conduction at the electronic level, 'surface charge feedback' theory will be helpful. Google "matter and interactions" by Chabay-Sherwood. Also there are videos available on youtube.

 

[davenn]

A single electron does not travel around the circuit, they hardly move anywhere.

in an AC circuit, yes, they just oscillate about a "location

In a DC circuit, they do slowly make their way around a circuit ... google electron driftDave

 

[houlahound]

Curious, I don't have the smarts to do field theory but could a simple circuit be analysed as a quantised field, would there be any point to?

The idea of indistinguishable electrons with an individual drift velocity bumping into things and repelling each other via a 1/r^2 Coulomb type law just seemed a reach for classic physics simplicity.

Not sure what I am actually asking. I guess I find it more intuitive to think ofva whole circuit in an energised state with a total energy E partioned amongst the energy levels of the matter in the components which have discrete energy levels. Sounds better than all these little projectiles buzzing around exchanging energy and momentum.

Just ranting.

 

[Drakkith]

Curious, I don't have the smarts to do field theory but could a simple circuit be analysed as a quantised field, would there be any point to?

Not for this thread. If you want to know more about that, I suggest making a new thread so we don't confuse the OP.

 

[Khashishi]

A voltage difference is just the integral of the electric field along a path connecting two points. Charges, such as electrons, feel a force due to the electric field. But in a wire, the electrons are rather crowded in the wire and repel each other, so they push each other along like a train going down a hill. And because of friction, the electrons reach a steady drift velocity. A great analogy is water flowing though a pipe downhill.

 

[David Lewis]

Mr Davis 97 wrote: "If I had a circuit with just a battery and no other elements,
how are electrons attracted to the positive terminal if there is no voltage drop along the way?
If the charges never lose their potential energy, then how do they get to the negative terminal?"

David Lewis wrote: Keep in mind conventional current flows from positive to negative in the circuit,
but from negative to positive in the battery.

Now a battery with no other elements implies there is no circuit.
If that's what you mean then the only voltage drop arises inside the cell itself
as a consequence of a small self-discharge current.
(The separator between electrodes is an imperfect insulator.)

Drakkith wrote: "The electric field created by the battery accelerates electrons along the entire length of the circuit. The net effect of lots of electrons undergoing constant acceleration and interacting with the circuit is that the energy is lost entirely as heat."

David Lewis wrote: The electrons accelerate when you close the switch. Initially most of the voltage drop is caused by inductive impedance (energy is stored in a magnetic field) and radiation resistance (energy converted to radio waves). As the electrons reach a constant speed, the voltage drop transfers across the resistor and (to a smaller degree) internal impedance of the battery.

The voltage drop across the connecting wires is more than zero, but considered too small to worry about in most cases.

 

[houlahound]

Not for this thread. If you want to know more about that, I suggest making a new thread so we don't confuse the OP.

If I could compose a coherent question I would. Any leads?

 

[houlahound]

Ignoring the voltage drop across the wires is misleading, a standard multimeter has no problem measuring a non zero resistance across its own connecting leads when they are shorted.

Has anyone seen a transient analysis of a circuit building up its current when a switch is closed. Could 5Spice do it??

 

[David Lewis]

For simplicity we're analyzing cases where the voltage drop of connecting wires has no meaningful effect on circuit operation. Note that a standard multimeter can have a hard time measuring the voltage drop of a current-carrying piece of wire because the meter itself becomes part of the circuit. You sometimes have to go to a voltmeter that connects the test leads to the insulated gate of a field effect transistor.

 

[houlahound]

Any links to such a DMM and their specs, I want to do some measurements for fun testing the formula relating resistance to surface area, length and resistivity.

 

[David Lewis]

Measuring resistance is easier. The ohm-meter connects a battery to the test resistance and measures the current.* (The test resistance is connected in series with the meter.) The resistance of the test leads can be ignored because you zero out the display before you make your measurement.

Measuring the voltage drop of a good conductor is tricky because the test resistance is now in parallel with the meter. So you want to look at the internal impedance of the voltmeter (expressed in ohms/volt). The higher the better.

*A precision ohm-meter connects a current source to the test resistance and measures the voltage.

 

[sophiecentaur]

People often seem to assume that there is something more basic about Fields than about Potential and there isn't. Analysing a circuit in terms of the potentials is just as valid and it tells you just as much as if you want to talk about Fields. The problem with trying to analyse a circuit in terms of fields is that the fields are very hard to plot and the actually are pretty irrelevant for most purposes.
How can I say that? Well - imagine a 12V battery, connected to a 12V, 1W bulb. That information is all you need to know to predict the operation of the circuit.
Now look at the fields in a few places. The battery is 5cm across the terminals so the field across the terminals is 2.4V/cm. IS that worth knowing? Route the wires so they are separated by 1mm at some part of the circuit. The field will be 50V/cm at that point. How does that help in the scheme of things and does it affect the circuit operation? The filaments of two bulbs are, say, 5mm long and 10mm long (same resistance (/wattage). What is essentially different about the two bulbs?
You will find papers that attach great significance to the field layouts for various circuits and their message is perfectly valid but is it useful in any case except where the actual field is relevant (a Neon Bulb or a sharp spike in a HV circuit)?
V = -∫Edl over any path. (see the wiki link and loads of others)
Because the field is Conservative, the line integral from a point at one potential and another, is independent of the path taken so you really don't need all the hassle of choosing a path and having to know the fields all along it. The Potential Difference is the Potential Difference - job done.

 

[Dale]

What really is a voltage drop?

I would echo the comments by sophiecentaur and Khashishi, a voltage drop is just another way to write an E field. If the voltage drops by 1 V over a distance of 1 m then there is an E field which exerts a force of 1 N on a charge of 1 C. A lot of times you don't care about the geometry, so it is more convenient to work with voltage in analyzing the fields than to use the fields themselves.

 

[tthomson]

A voltage drop across a circuit provides the potential to do work on electrons in the circuit. In doing the work, the electrons increase their kinetic energy, which is passed to the surrounding atoms in the circuit components, causing them to heat up (Kinetic theory of heat). By inference this action reduces the kinetic energy of the electrons. The voltage difference is relative. e.g +12 volts and ground (0 volts), but could just as easily be -1000 Volts and -1012 Volts.

The quantity of energy converted to heat is determined by the resistance of the circuit components (R). Where R = V/I and E=VIt i.e. the energy converted to heat is the Voltage drop times the current flowing multiplied by the time in seconds that the circuit is operating for and the rate of energy conversion (Power) is VI.

A 12 Volt 60Watt bulb would draw 5 Amps when a voltage drop of 12 Volts is applied across it. The energy it uses is 60 Joules every second. (A Watt is a Joule per second).

I hope that makes sense to you.
TT

 

[marlowgs]

There is a chemical reaction in the battery that wants to take place but it needs the electrons from the other electrode to complete it. If you provide a path that the electrons can flow through (a conductor) the chemical reaction will complete. The voltage drop is the amount of energy (joules) per unit of electrons (coulomb) that will be given off by the chemical reaction if you provide a path for the electrons to complete the reaction.

 

[bobkolker]

I've been studying circuits, but I can't find any resources that really answer my specific questions. So far I understand that a battery creates a potential difference between the two terminals, one of which is positive while the other is negative. There is a buildup of positive charge on the positive side and buildup on the negative side, and this creates an electric field which can accelerate electrons from a location of high potential to one of low potential. When you connect a conducting wire between positive and negative terminals, this creates a path of least resistance which electrons can flow through, where the electrons are accelerated by the electric field that the battery creates. Hopefully all of this correct so far. My main question arises when it comes to voltage drops. What really is a voltage drop? If we connect one resistor to the circuit, a circuit with 12 V, then there must be a voltage drop across the resistor of 12 V. However, why is this the case? And also, how are the electrons still able to move if on the other side of the resistor there is 0 V, since there is a voltage drop?

I know that most of this is probably wrong, so I need someone to really help me understand how circuits and specifically voltage drops work.

Here is a "must read" Please read http://science.uniserve.edu.au/school/curric/stage6/phys/stw2002/sefton.pdf

Sefton demolishes the "garden hose" analogy and shows power is transmitted as a field effect. The electric and magnetic fields produce the Poynting Vector field which is really what carries the power from the energy source to the resistance.

See also http://cq-cq.eu/Galili_Goihbarg.pdf and http://amasci.com/elect/poynt/poynt.html

http://amasci.com/elect/poynt/poynt.html

http://www.furryelephant.com/conten...tric-current/surface-charges-poynting-vector/

 

[sophiecentaur]

In doing the work, the electrons increase their kinetic energy,

That is a very weak model / analogy. The total mass of free electrons will be in the order of 1/(tens of thousands) of the mass of the conductor and their mean drift speed is around 1mms^-1. How much KE is that?
Your argument is a bit like explaining how a bicycle chain works by saying its KE is the main part of the energy it transfers to the wheels.

 

[Stephen Tashi]

There is a buildup of positive charge on the positive side and buildup on the negative side, and this creates an electric field which can accelerate electrons from a location of high potential to one of low potential.

There are "levels" of understanding physical phenomena. If you want to understand an electric circuit at the level of atomic physics then you have quite a job ahead of you. Even if you take the model of a electron as a small charged classical mass, the detailed physics of an electrical circuit is not simple. For example, if the electric field accelerating electrons was the only force on an electron then it should accelerate while in the wire and have greater velocity at the "downstream" end of the wire than at the upstream end of the wire.

You should seek a level of understanding that matches what you want to do. If you want to build everyday electric circuits then you don't need to understand what's going on at the atomic level - and maybe nobody really understands the atomic level. If you want to design vacuum tubes, or design radio antennas, you need a different level of understanding.

 

[sophiecentaur]

The way that 'modern' Science education goes doesn't help at all with this. For some reason, the curriculum seems to want students to 'explain' in their heads, every electrical phenomenon in terms a very badly defined 'electrons'.
Particles Rule and Confusion Reigns. Bring back Charge and Current and leave it at that until students have an actual clue what sub atomic particles are like. (But I have been harping on about this ever since forums were available on the Internet - and even before.)

 

[bobkolker]

The way that 'modern' Science education goes doesn't help at all with this. For some reason, the curriculum seems to want students to 'explain' in their heads, every electrical phenomenon in terms a very badly defined 'electrons'.
Particles Rule and Confusion Reigns. Bring back Charge and Current and leave it at that until students have an actual clue what sub atomic particles are like. (But I have been harping on about this ever since forums were available on the Internet - and even before.)

lots of charged particles moving in a stream IS a current.