# Voltage Problem

1. Sep 4, 2008

### DocZaius

I have been learning the concepts of voltage, current, and resistance. I am aware of Ohm's law and have read a lot of materials on it, but none of what I have read addresses a few questions that have been popping up in my head.

Many educational sources like to compare the situation to a water pressure system. The problem with that analogy is that with water, knowing how much volume per second is displaced is enough to know how much work that water can do. With electricity however, you must not merely know the flow rate (amps) but must also know at what voltage those amps are! I don't understand why merely knowing how many electrons are traveling down the circuit's line isn't enough. Isn't that all the information you should need? I mean, each electron has a definite charge, and if X electrons are moving down the circuit and they all have this definite charge...why must more be known? The way I've been interpreting it is that voltage causes electron flow which in turn causes power. Yet the way it's presented seems to keep the concept of voltage entangled with the subsequent current, and thus necessary to make calculations regarding power.

Which leads to another question I have. If voltage represents the difference in charge from one point to another and thus the electric potential energy, why doesn't voltage decrease over the lifetime of a battery sitting on a circuit? I would imagine all those electrons moving from the negatively charged area to the positively charged area would slowly but surely bring the system to an electric equilibrium, thereby dropping voltage to 0. Yet nowhere do I see this phenomenon mentioned. Does this happen and if not, why not?

Thanks for any help.

Last edited: Sep 4, 2008
2. Sep 4, 2008

### stewartcs

You will need to know how much electrical potential energy is being used to create the current flow. Look at the dimensions of voltage, current and watts and see what you can infer.

What do you mean by sitting on a circuit? If a battery is connected in a complete closed circuit, its electrical potential energy will eventually decrease to zero volts (once the chemical energy is transformed into heat or light or whatever).

If it is in an open circuit, it will also eventually decrease to zero volts due to the internal resistance of the battery.

CS

3. Sep 4, 2008

### DocZaius

But once I know the current flow created by that electrical potential energy, why do I still need to take into account the voltage that created it? What I don't understand is why can't we tell everything we need to know about power from simply how many electrons are flowing in the circuit.

If, say, 10 trillion electrons pass through a cross section of the circuit every second, I know the current and thus the amps. This account does not factor voltage in, but why should it is my question. Can we not know how much work that current can provide from the information given?

Imagine a battery connected to a closed circuit. The + terminal is connected to the - terminal. Since as a result of the closed circuit, the negatively charged area is connected to the positively charged area and electrons are moving from the negatively charged area to the positively charged area, wouldn't voltage drop to 0 even without a sizeable resistance? We are essentially moving electrons from one side of the battery to the other, are we not? As the positive area receives electrons, it would brings its charge down and as the negative area delivers electrons, it would bring its charge up, culminating in a state of equilibrium, no?

4. Sep 4, 2008

### cmos

Are you sure you're completely correct with the characterization of the water system? From my limited work with fluids (and by intuition) I would think that you would also need to have some knowledge of the pipe geometry. This is usually characterized by a pipe conductance or pipe throughput. With this knowledge you can determine the pressure of the system.

Assuming I am right, this has a direct comparison to electric circuits. Only knowing the current (C/s), you have no knowledge of the geometries of the circuit elements involved. For a resistor, this is characterized by the resistance. But say you didn't know how resistive the element is. All you would have to do is apply a known voltage and measure the current. The power is then P=VI as you have eluded it.

It goes without saying that the currents you get will be governed by Ohm's law.

Why do you think you have to periodically replace batteries? :tongue2:

Just as a side note, the voltage doesn't actually drop to 0, but rather to a value that can no longer drive most commercially designed circuits.

5. Sep 4, 2008

### Staff: Mentor

I didn't read the whole thread. The reason that you don't work with circuits by counting electrons is because there are no practical instruments that you work with day to day that are calibrated to read electrons/s. You will work with voltmeters and ammeters and oscilloscopes.

Last edited: Sep 4, 2008
6. Sep 4, 2008

### Staff: Mentor

Okay, I went back and read more (sorry if one of the other responders has made this point already). The issue with just knowing how many e- pass a point in a circuit is that you don't know what the "resistance" to that current flow is. It takes a lot more energy to squeeze n e- per second past a high resistance than it does to squeeze them past a low resistance. Hence the equation:

P = I^2 R

Does that make sense? And the water analogy is best left behind pretty early in your EE education, IMO.

7. Sep 5, 2008

### DocZaius

If I were to connect a copper wire from the the + terminal to the - terminal of a battery, would that battery be drained as quickly as if I did the same thing, but with a significant resistance connected to that wire?

Also, does the voltage of a battery steadily decrease over its lifetime? It was implied in a response above, that it does. But I would assume not, since many components of a circuit rely on a specific voltage to work. If it doesn't, then what is wrong with my interpretation of the - terminal sending electrons to the + terminal? Wouldn't that mean their electric charge difference would reduce steadily over time, that being voltage?

8. Sep 5, 2008

9. Sep 5, 2008

### stewartcs

The higher the load (in this case resistance) the greater the power required to drive the current. This extra power (or electrical energy) comes from the chemical reaction in the battery. Therefore it will deplete the battery faster by having a larger load than with a smaller load.

Again, it depends on the load. If the load is constant (i.e. sitting on the shelf waiting to be used or in a flashlight) it will decrease at a steady rate.

CS

10. Sep 5, 2008

### cmos

I disagree with this. True, a larger load will require a larger voltage to drive the same current, but this is not the example we are dealing here with. What we are dealing with is having a constant voltage source (the battery) and having different loads.

The larger load (i.e. larger resistance) will draw less current than the smaller load. So the power supplied will be greater with a smaller load. Therefore, in contrast with what you have claimed, it will deplete the battery faster by having a smaller load than with a larger load.

For example, take the two extremes: infinite resistance (open circuit) and zero resistance (short circuit, e.g. by taking a wire and shorting the battery). In the first instance, the battery will last for several years. In the later, the battery will last for several minutes. Since zero is clearly smaller than infinity, the battery will drain faster by having a smaller load than with a larger load.

Last edited: Sep 5, 2008
11. Sep 5, 2008

### cmos

As I implied above, the voltage of a battery decrease over its lifetime. Yes, a circuit is designed to work at a specific voltage. Again, why do you think you need to replace batteries periodically?

Have you taken a course on introductory chemistry yet? Galvanic cells are usually covered at this level.

12. Sep 5, 2008

### stewartcs

Sorry, of course this is correct, I wrote it backwards!

CS

13. Sep 5, 2008

### stewartcs

I suppose I should clarify this too. It is steady assuming it is an ideal battery. In a real battery the internal resistance would increase as the battery is discharged thus giving it a non-linearity.

CS

14. Sep 5, 2008

### cabraham

What moves electrons is proximity to other charges. Then, current, voltage, power, resistance, etc. are defined in terms of Coulomb's force law, F = k*q1*q2/(r^2), Ohms law, etc. Why 2 charges attract/repel is not known. It is an *axiom*. It is the starting point from which other laws are derived. No one knows its "cause". Attempting to establish "cause and effect" is an endless vicious circle.

I would restrict my sources of info to peer-reviewed text books used in university physics and full electrical engineering curricula. If you use the "shotgun" approach, i.e. ask the whole world and study the responses, you will get a myriad of answers, many of which are incompatible. So who do you trust?

It is said that if I wear but 1 watch, I am reasonably confident I know the time. But if I wear 2 watches, I can't be quite sure.

The classic physics text for starters has been Halliday-Resnick. For e-m fields, it is good to introduce the student to charge and energy concepts. Another good reference is "Electromagnetics" by Kraus and Carver (Ohio State Univ). Your nearby university library should serve as a good source of material. This is reliable info.

Again, your question is valid, and I appreciate your desire to learn. But, these forums have well-informed people who want to help you, and some who are not well-informed who want to help you as well. The problem with those who are wrong is that they usually think they are right. Then back and forth debates and exchanges take place, and eventually the thread gets locked. Questioning a response is ok, but when 2 or more parties are endlessly arguing, it becomes counter-productive. The newbie anxious to learn can't decide who's right.

Just some advice I thought you could use. BR.

Claude

15. Sep 5, 2008

### Redbelly98

Staff Emeritus
A battery's voltage does decrease over time. Circuits designed competently will work over a range of voltages, not just at one value. Even a household wall-outlet voltage will fluctuate, so electrical plug-in devices must also operate over a range of voltages.

16. Sep 5, 2008

### Staff: Mentor

Electronic devices work over a surprisingly large range of voltages. +- 20% is not unusual, and it can be even wider than that. Heck, the nominal voltage of 1.5V NiMH AA batteries is only 1.2V, while AA lithiums are 1.7V!

If you pull batteries out of a device right after it stops working and check the voltage, you may be shocked at how low it can go.

17. Sep 5, 2008

### DocZaius

To try to make myself solve the voltage/ampere conceptual problem that I have, I decided to create an example with two circuits:

Circuit 1: a 10v battery is connected to a closed circuit with a total of 10 ohms of resistance on it. I = V/R and so there is one amp of current on this circuit.

Circuit 2: a 20v battery is connected to a closed circuit with a total of 20 ohms of resistance on it. I = V/R and so there is one amp of current on this circuit.

Then I asked myself: If I were to zoom in at equivalent points on each circuit and look at the behavior of the electrons being handed off from copper atom to copper atom, what would be the difference between each circuit? The amps are the same in each circuit, and the amp is a base unit and therefore cannot be broken down into constituent dimensions. However, I am told that amps represent the amount of charge that passes through per second, so I'll have to go with that interpretation. Anyhow, so they have the same amp. What is the difference? I mean - the implication of them each having the same amp is that for every cross section you look at on each circuit, you'll have as many electrons passing through them. Every electron has the same elementary charge so that can't be the difference...

My only answer was the energy of each electron that passes through the second circuit is higher than the ones that pass through the first circuit. Just as many electrons pass through each circuit, but they pass it at a quicker pace in the second circuit. This is why the 20 ohm resistor in circuit 2 gives off more heat than the 10 ohm resistor in circuit 1. There is more energy running through circuit 2 per second.

The above could be completely wrong but this is what I came up with. Thoughts?

Last edited: Sep 5, 2008
18. Sep 5, 2008

### Staff: Mentor

No, work in a fluid (at constant temperature) is change in pressure times volume, not just volume. Think of the pressure like the voltage.

19. Sep 6, 2008

### Antenna Guy

That sounds pretty close. Look at the power consumed in either circuit:

$$P=I^2 R=\frac{V^2}{R}$$

Regards,

Bill

20. Sep 6, 2008

### DocZaius

The more I think about it, the less I like my hypothesis that the electrons in circuit 2 are faster than the ones in circuit 1, everything else being equal.

If that was the case, the current could not be identical for each circuit at every point measured on each circuit, and yet it is in practice...