# Batteries, Static Voltage

1. Jan 5, 2007

### Nuklear

Let's say you put ina battery that's 9 volts. YOu can vary the voltage on the device. Is it changing the battery voltage or the voltage used throughout the system?

How can you have a set voltage> THe equation for current goes V/R=I. SO if you have a certian resistanve and current the batteriy can't be at a certain voltage. It would depende on the reseistance and current.

Lastly do all wires have the same amount of electrons. Potential energy over COloumbs equalls voltage. FOr a battery maker to know that tany battery has a certain number of volts means there has to be a standard number of COulombs in the wire. Is it the Coloumbs in the wiring or batteryt hat counts to the voltage?

2. Jan 5, 2007

### ZapperZ

Staff Emeritus
You are confusing between the dependent with the independent variable. If I fix the potential difference to be V, and I use a resistor R, then those are the independent variables. The dependent variable in that case will be I, the current. This means that if you use different resistors, you will get different currents! The potential difference doesn't have to change just because you use a different resistor. Try solving this when you put resistors in series. This is essentially what you get.

Zz.

3. Jan 5, 2007

### chroot

Staff Emeritus
The open-circuit (i.e. when nothing is connected to it) potential difference produced by a battery depends only on its chemistry. The actual potential difference across the battery's terminals also depends on what's connected to it. All batteries are imperfect and have some amount of series resistance. When you connect a load to the battery, the resistance of the load forms a voltage divider with the battery's internal series resistance, and the battery voltage drops accordingly. You cannot vary a battery's voltage by any other means than a) putting a large load on it or b) letting it wear out.

I'm not sure what device you're talking about, in which you can vary a voltage, but it's not simply loading the battery. It is more likely using some kind of integrated circuit with a current mirror, and allowing you to vary the very small reference side of the current mirror. Loading a battery is not a good idea, of course, because you wear it out very quickly.

I must confess that I do not understand your last paragraph, but it appears that you are applying an equation (1 V / 1 C = 1 J) incorrectly. All that that equation means is that if you let one coloumb of charge move through a potential difference of one volt, you will have dissipated one joule of energy.

- Warren

4. Jan 5, 2007

### Staff: Mentor

The battery voltage is fixed. How you distribute that voltage across various circuit elements in the device can be altered.

Why do you assume the current is fixed?

What you are interested in is the number of electrons flowing per second through the wire--that's the current. That current just depends on the voltage across the wire and the wire's resistance.

(Edit: Too slow! )

5. Jan 5, 2007

### Nuklear

ALl of you are kind of contradicting or CHroot is. He's sayign the voltage of the battery does drop.

There are devices where the voltage varies. Take radios. Yo can change the volume. SO as the volume increases is the voltage of the battery increasing or is it somewhere in the circuit? IS a battery always putting at it's max output?

The formula we learned for voltage is Potential Energy over Coloums equals voltage. basically it's energy per COloumb. I'm beginning to see it doesn't matter how many Coloumbs are in the wire, just that are flowing by the battery per second...............current. NOw just off a but, do all wires have the same amount of ELectron-Coloumbs inthem?

ANd in Home power, does the current vary whenever you turn on a light, device, or appliance? If there's a set voltage and current coming into your house while you've got the TV,Microwave,lights, and radio on and then you switch on the washing machine do the other appliances loose volts? If not, how do they keep the wattage stable throught the houses wiring?

Last edited: Jan 5, 2007
6. Jan 5, 2007

### Staff: Mentor

Yes, the output voltage across the battery terminals drops due to internal resistance whenever you apply a load to the battery. But the EMF depends only on battery chemistry.

7. Jan 5, 2007

### chroot

Staff Emeritus
Well, your original question was pretty hard to understand. We're all correct, in our own ways. The battery's open-circuit voltage never changes, it's dependent only on its chemistry. The battery's voltage does change under load, but typically only by a very small amount. The battery manufacturers go to great lengths to minimize the internal series resistance. An ideal battery has no internal resistance at all, and its voltage never changes. In most applications, you can completely ignore the very small changes in a battery's voltage due to load.

No, the battery is not changing at all -- it's always producing [essentially] the same emf, or voltage. When you turn the radio up, you're changing the gain on the amplifier, and the amplifier consequently consumes more power, which it sends to the speakers to make them produce louder sounds. Since the radio's supply voltage is constant (the battery does not change its voltage), it draws more current from the battery.

A specific volume of, say, copper has a specific number of atoms in it. Each atom has a specific number of electrons, so you can figure out how many electrons are in a chunk of copper. This is not very important for the study of electronics, though -- we're concerned with the movement of electrons, not the number of electrons initially in the wire.

Consider a garden hose -- you don't really care how much water exists in the hose, and inside the pipes in your house, and in the pipes that bring water from the city water supply into your house. All your really care about is how much water comes out of your hose per unit time, so you can water your lawn.

Indeed, turning on appliances means you're drawing more current from your house's electrical supply. If you go look at your meter, you'll see its disc spinning more rapidly. The meter is not much more than an ammeter (current meter).

If the electrical transmission system were made out of ideal wires (with no resistance), your electrical supply would maintain the same voltage (120VAC, for example), regardless of the load you put on it. The voltage would be rock-solid, all the time.

In the real world, wires and transformers and other devices in the transmission system do have resistance, and do drop some voltage. Consider a very poor length of wire with 1 ohm of resistance -- when you pass one ampere of current through it, it loses 1 volt of emf from one end to the other. It's a small effect, considering one volt out of 120 is rather insignificant, but it's a real effect.

This is why your lights will flicker or dim when your air conditioning unit turns on: the AC motor pulls a very large amount of "in-rush" current as it turns on, and this large current causes your entire house's electrical supply (which has some small, but non-zero resistance) to drop a few volts for a couple of thousands of a second.

- Warren

Last edited: Jan 5, 2007
8. Jan 5, 2007

### Nuklear

Yes but do the lgihts and appliances go back up the their original voltage.

And as for electrons in a wire, are ther efree electrons? In other words are the electrons flowing through a iwire just freelancing with no atoms or molecules to cling to or are they coming fromt he atoms and molecules that make up the wire?

9. Jan 5, 2007

### chroot

Staff Emeritus
Some of the electrons in a metal are essentially free. To a first approximation, they exist almost as a free electron gas.

Let's think about a small chunk of copper, one centimeter on a side.

The number of free electrons in a block of copper a centimeter on a side is about $8.5 \cdot 10^{22}$, or about 13,500 coulombs. This number is very large. If you were to pass that number of electrons through the block of copper in one second, it would constitute a current of a 13,500 amperes, and would vaporize that little cube of copper in an instant.

The number of electrons that actually flow through a wire in normal situations is much, much, much smaller than the number of electrons intrinsically present in that wire.

Similarly, the amount of water flowing out of your garden hose is much, much smaller than the amount of water in the entire city's water supply system.

- Warren

10. Jan 5, 2007

### Nuklear

You stil haven't answered the light question. what if half the city just turned their lights on. WOuld that severly drain the other halfs power supply?

Now in a series circuit the more resistors you add the current goes down or voltage goes up. But in a parrelell I did the equation and if I add a resistors theccurrent goes up or the voltage goes down.

Last edited: Jan 5, 2007
11. Jan 5, 2007

### chroot

Staff Emeritus
Hey, you're welcome. What "light question" are you talking about? This one? Yes but do the lgihts and appliances go back up the their original voltage. If so, I didn't realize it was a question, because it didn't end with a question mark.

The answer is, yes, as soon as the current spike is over, the voltage goes back up. Remember Ohm's law:

V = I * R

If R is the resistance of the household wiring, and I is the current through the house's wiring, then V is the amount of voltage dropped from one end of that wiring to the other. If the current goes down, so does the voltage drop.

I have no idea what you're talking about, really, but please keep in mind that you need to either make the voltage fixed, or the current fixed, and then consider what happens to the other.

Think about these examples. Holding voltage constant, and varying resistance:

- A 10V battery with 1 ohm across it will source 10 amperes of current.
- A 10V battery with two 1 ohm resistors in series will source 5 amperes of current.
- A 10V battery with two 1 ohm resistors in parallel will source 20 amperes of current.

Holding current constant, and varying resistance:

- A 1 ohm load requires 10 volts to push 10 amperes through it.
- A 2 ohm load (two 1 ohm resistors in series) requires 20 volts to push 10 amperes through it.
- A 0.5 ohm load (two 1 ohm resistors in parallel) requires 5 volts to push 10 amperes through it.

- Warren

12. Jan 5, 2007

### chroot

Staff Emeritus
This isn't a question that can be easily answered without knowing more specifics about the entire city in question. Let's imagine a city fed by a distant generator over a single power-transmission line.

If the transmission line between the generator and all the loads is ideal, having no resistance, then the generator would have to work harder to power all those loads, but the voltage on the transmission line would be unchanged.

If the transmission line between the generator and all the loads is not ideal, then it has some resistance, and some voltage is dropped across its length.

Imagine that the city originally pulled 1,000 amperes of current from the generator (which generates 120V), and the transmission line had a resistance of 0.01 ohms. The voltage drop from one end of the wire to the other would be V = IR = 1,000 * 0.01 = 10 V. The end of the transmission line connected to the generator would have 120V on it, but the end of the transmission line in the city would have only 110V on it.

Now, imagine if everyone turned on some more lights at once, and suddenly began pulling 2,000 amperes of current in total. The voltage drop across the transmission line is now V = IR = 2,000 * 0.01 = 20V. The voltage at the generator end is still 120V, but now the end in the city only has 100V on it.

- Warren

13. Jan 6, 2007

### Nuklear

SO if the voltage across the city suddenly drops, does the power power company have a way of dealing with it?

And about parrelell circuits. TO find total resistance it's 1/TR=1/R+1/R+1/R.
SO let's say I have 3 resistors in a circuit. each 4 ohms. The total resistance is 1 1/3 ohms. If voltage is 10V then current is 7 1/2 amps. If I add another 4 ohm resistor total resistance IS 1 ohm. SO if our voltage is 10 our current is 10. Resistance went down, current went up.

14. Jan 6, 2007

### chroot

Staff Emeritus
Well, the situation I described was rather idealized -- power companies use high voltages, not high currents, to transmit power. Large currents require large wires and necessarily involve large amounts of heat dissipated by the transmission line, which is undesirable.

But yes, the power company does have an array of sensors at both ends of its transmission lines which can detect the voltage there, and adjust things interactively.

And yes, you have used Ohm's law correctly in your example.

- Warren