# B How electric circuits really work

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1. Mar 31, 2016

### 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.

2. Mar 31, 2016

### 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.

3. Mar 31, 2016

### 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.

4. Mar 31, 2016

### cnh1995

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).

5. Mar 31, 2016

### 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.

6. Mar 31, 2016

### houlahound

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

7. Mar 31, 2016

### houlahound

Are you thinking that current is like projectiles??

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

8. Mar 31, 2016

### Drakkith

Staff Emeritus
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.

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

9. Mar 31, 2016

### Mr Davis 97

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)

10. Mar 31, 2016

### Drakkith

Staff Emeritus
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).

11. Mar 31, 2016

### 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.

12. Mar 31, 2016

### cnh1995

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.

13. Apr 1, 2016

### 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?

14. Apr 1, 2016

### cnh1995

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.

15. Apr 1, 2016

### houlahound

Until the wire disintegrates at high temperature.

16. Apr 1, 2016

### 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.

17. Apr 1, 2016

### davenn

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 drift

Dave

18. Apr 1, 2016

### 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.

Last edited: Apr 1, 2016
19. Apr 1, 2016

### Drakkith

Staff Emeritus
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.

20. Apr 1, 2016

### 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.