Huge wire

[Emoxihax]

I never got the impression that he thought he was better than anyone

[bobwires]

You're missing some important info, and you have a couple things turned around.

The resistance of a device does not change with voltage. The current changes when the voltage and resistance are fixed.

Ignore R for a second

P=I/E

[Blackedout]

It's threads like these that make me realize how big of a change car audio is in for in the future. The way we run wire is amateur, but given our technological and monetary limitations in car audio, 1/0, 2/0, and in rare cases 4/0 is really the only large wires we can

Use.

Conductors that can withstand 2000 amp currents make me wonder what a 2000 amp short would do to the chassis of a car. Because at that point, the name of the game

Isn't only to fuse for the wire, but to fuse for the car... The current 150 amp fuse rating that we have on our 1/0 wires isn't enough to damage our cars....

We have a lot ahead of us in the next 10 years to overcome, just as we've done in the last 10 years...

[masterbusa2]

holy shizz that a big wire good ground? xD

[Blackedout]

The best thing that could happen for car audio is 48 volt charging systems. More voltage would be great, but air doesn't insulate as well over 50 volts. So we would be hit with another barrier to overcome. At 48 volts, 1/0 wire should handle almost 4x the wattage it does now. How would you feel about two 1/0 runs powering the full extent of 5000-6000 watts (1 power, 1 ground)? Imagine the size of wire needed for everything else in cars... 1/4th the size it is now. What is 1/4th the size of 18

Gauge wire? Not to mention... Demand for copper would decrease... Prices would go down. Just a thought. Technology is a hell of a thing.

[Blackedout]

You guys are getting too technical... You do realize you guys are communicating the same point in two different ways.

From my understanding of ohm's law (which only comes from high school physics), the more volts you apply to a load, the more current goes through it

Power = wattage.

Current = Amperage

However incorrect your statement is, I understand what you are trying to say... all that really matters. The more voltage you apply to a load, the less current that travels through it.

10 watts at 12 v ~ 1 amp

10 watts at 120v ~ .1 amp

The power is the same, the resistance within the wire is the same, the current is the only thing that changed.

[uhoh_45]

500 next to little tsunami 0 gauge

little comparison

[bobwires]

I was wrong to say resistance doesn't change. Not thinking - well I was thinking about other stuff......

The best thing that could happen for car audio is 48 volt charging systems. More voltage would be great, but air doesn't insulate as well over 50 volts. So we would be hit with another barrier to overcome. At 48 volts, 1/0 wire should handle almost 4x the wattage it does now. How would you feel about two 1/0 runs powering the full extent of 5000-6000 watts (1 power, 1 ground)? Imagine the size of wire needed for everything else in cars... 1/4th the size it is now. What is 1/4th the size of 18

Gauge wire? Not to mention... Demand for copper would decrease... Prices would go down. Just a thought. Technology is a hell of a thing.

I agree. I would LOVE a higher voltage standard in cars, or even have parts of the vehicle run AC. Very cool, but more complicated. Imagine if everything but the starter was servicable 3 phase power

Unfortunately, the auto manufacturers haven't gone this route like they planned due to safety issues. 36v batteries require special connectors because they arc more when being connected. I read an article all about this last year. I'll see if i can find it.

You say it yourself, resistance doesn't change. So why do the values for resistance change in YOUR calculations?

I=E/R How is that possible?

I understand that but WHO says power remains constant?

You guys are getting too technical... You do realize you guys are communicating the same point in two different ways.

From my understanding of ohm's law (which only comes from high school physics), the more volts you apply to a load, the more current goes through it

Power = wattage.

Current = Amperage

However incorrect your statement is, I understand what you are trying to say... all that really matters. The more voltage you apply to a load, the less current that travels through it.

10 watts at 12 v ~ 1 amp

10 watts at 120v ~ .1 amp

The power is the same, the resistance within the wire is the same, the current is the only thing that changed.

This is important to understand. You guys have the right idea, but in application there are a few more considerations.

Ohms law and Watts law are different, but they are both laws. They both apply in any electrical situation, but there is a lot more to it.

voltage increase DOES NOT mean there will be more electrons flowing through the wire - voltage, by definition, is potential energy.

Example time. Going back to the water pipe - say you have a 1" pipe with 100psi pump running on it, but it is closed at the end. how much water is flowing? this is like a wire (say #4) with 12.6v on it, but the amp is turned off. No current flow at all. Now swap it out for a 2" pipe (or imagine 1/0 wire) - no change as long as the amp is off. still no current flow.

Now regarding the 'load' - Ultimately you want a certain amount of water in a bucket. Put 2 buckets side by side. You have a 2" line into one, and a 1" line in the other bucket, and both lines have the same amount of pressure (voltage). Open both valves - which will fill up first? The 2" line allows more water molecules (joules, or electrons, however you see it) to pass through per second (watts are joules/second). The 2" bucket will fill first.

Problem - this does not account for ANY resistance at all. It's like saying a 1/0 wire will give your amp more power than a 4 ga. That is just not true - the wire typically only has a small effect on the amp's actual power, unless it fails from being too small to handle the load.

Now turn an amp on full tilt boogie on a test tone. Say you meter 100 amps going through the wire at 12.6v - that makes 1,260 watts (P=ExI). the amp is impeding the charging system with .126 ohms (R=E/I). Now imagine this amp is capable of running at 126 volts (to keep the math easy. The LOAD is determined by the amp. It requires a certain amount of power to run at the level the signal is telling it to. At 126v the amp would only need to pull 10 amps to get 1,260 watts. Power is not calculated in ohms law, but you can get there by E^2/P=R. 126^2/1260=12.6 So the resistance did change with the change in voltage (you are right about this LZTYBRN). I was wrong about the resistance not changing. It does under load for sure.

Meter resistance on a 100w light bulb and it is what it is - say 9.5 ohms. apply 120v to it and you've got 12.63 amps (E/R=I) right? WRONG. That would be >1500watts! It's a 'resistive load'. the resistance when the filament is heated up is way higher than when it's cold. It's actually going to be in the neighborhood of .83amps to get you 100 watts (P/E=I).

Another consideration without work - voice coils not only heat up, they are AC current. resistance changes with phase, inductance in said coil, different frequencies, etc. It gets complicated in a hurry. Read up on 'Reactance', and remember that a 'coil' is just that (inductor)

(pics of big ass lugs)

That's what I'm talking about!

The resistance HAS to change in order to have the same power output at two different voltages.

He was saying that if he plugged his "120VAC 1700W" coffee pot into 12.6 volts, the wattage would be the same and that the amperage would be higher... I don't think this is true at all. According to my calculations, the resistance of the pot should be about 8.5 ohms. 8.5 ohms on 12.6 volts should yield only about 1.5 amps, and 18.7 watts.

The coffee pot requires a certain number of watts. The resistance changes, you are right.

The heating element is a 'resistive load' Go plug a 1700w heating element into an 18w power inverter in your cigarett lighter and see what happens

If you believe "the more voltage you apply, the less current," hook up 2 car batteries in series (24V) and tell me every electronic in your car doesn't get ruined.

Well the car won't fail from too much current! This is a completely different application.

[greatblack]

IS THERE ANY HOT DRUNK CHICKS IN HERE ?????????? PLEASE PM ME IF SO

hahahahaha o LOL'd :lol: :turkey: :drinks:

[bobwires]

The resistance is represented by the diameter of the pipe.

No, that's like saying the resistance in the wire is what restricts current flow. It's such a tiny amount it's not even figured in the most basic equations - frictional losses in physics.

Will your 1000w amp put out 500w if you swap out the 1/0 for 4ga?

Is there a law or rule that says that power dissipation remains constant regardless of the voltage you put it on? I guess that's the main part I'm stuck on... Why do you say that power remains the same?

A 1000w amp is a 1000w amp is a 1000w amp. It doesn't matter if you are talking a 12v DC amp, or 12v AC. The power is the same because that is what it's rated for, and that's how we compare them. The voltage is different, the current is different, and the resistance isn't even important at all. Whatever you calculate for resistance is totally changed when the actual load changes.

a 100w light bulb is a 100w lightbulb is a 100w lightbulb, etc.

I say the power remains the same because that is what is in comparison right now. It is what we usually compare because it's a logical, consistent figure across the board. If you go to by a 500w heater, that's what the package will say. Not '4 amp heater.' What does that mean? Then you'd have to buy a different rating of heater for the same amount of heat in a different application (12vdc, 120vac, etc.) We need a common reference - POWER. If it's plugged into the wall at home it's a 500w heater. Plug it in the car, it's a 500w heater. Power is work over time - it's what we almost always use to determine how useful the device will be, or if it fits our needs in the application.

What about power staying the same in a car, on the same device? When you turn up your 1000w amp in the car, on a 14.4v charging system, let's say the current is 69amps. The voltage will dip a little, because batteries and alternators aren't perfect. What if your batter is really crap, and your stock alternator can't give up another 70amps? well the voltage dips a lot. The Amp, which determines the load, will TRY to produce 1000w. How much current will then flow if the voltage dips down to 11.0v? 90amps. It will take 90amps of current for the Amp to make the same amount of 'sound'. Will it be as loud on a meter? Not quite, because other losses all start to stack up. A little more current causes a little more resistance through every wire and connection, and the amp won't necessarily be able to put out all it's potential.

Did you know a bad alternator can kill a battery and starter? I saw it happen once. The weak charging system doesn't keep the battery full, and the starter cranks at a lower voltage. What happens when the voltage drops to the starter? More current..... all those losses compounding....... resistance goes way up....... wires start to get hot and stiffen over time..... now it's really drawing some current to produce the 1500w, or whatever it NEEDS, to get the job done. The power will remain exactly the same, if the system can handle the strain. in reality power will go down SLIGHTLY due to frictional losses when the voltage goes down.