Please explain to me how using 12 volts decreases the load on wiring. If I have a headlight that is 12 volts and 4.58 amps I get 55 watts. Watts is heat isn't it? So if I drop to 6 volts and increase the current to 9.17 I get 55 watts. I understand the current went down but the voltsage goes up. HELP

 

NO, the amps generate the heat. That's why 6 volt systems have heavier wire. 12 V is thinner guage. They are now working on getting 24 V into cars and trucks and the wiring is super thin. It's not the volts, it's the amps.

Barry

50 F-1

 

Power = Voltage X Current. So if you double the voltage, you reduce the current by one half. This is most evident on your battery cables. A lot of power (Watts) is required to crank your engine. At 12 Volts, the current (Amperes) draw is half what it is at 6 Volts.

In other words, it's not the heat; it's the humidity.

Sorry for that last crack.

 

When the current went down the need for a large diameter wire to carry those amps is decreased. Wire size is determined by the amps a circuit is designed to carry, the insulation on the wire determines the volts it can safely handle. Amps without volts does nothing as it takes a combination to produce power (watts).

 

I think that your confused because either way your using the same amount of energy.

Its just that the higher the volts, the smaller the wire required, thats where savings in construction costs come in.

 

Actually, the higher Voltage is more efficient, though you will not likely notice the savings unless the price at the pump continues on it's rise.

Power loss over transmission lines = Current squared X Resistance.
Since the current is squared (current X current) in this equation, if you can cut it in half, your reduce the power loss by a factor of 4.

Which is why electrical power is sent over high voltage lines and reduce to 110 Volts AC by a transformer as close to your house as possible.

 

Stev8e, your splitting hairs bud and it doesn"t have anything to do with wire size in trucks.

 

I'm not confused.

I stated that what power equals. The left side of the equation didn't change, just the right side (current and voltage). No one in this thread stated that there would be any change in energy (power).

 

I never said "you" were confused. I was replying to christopher2 who asked the question about automotive wiring. Now. Now we all know how smart you are. thanks!

 

You said "your confused". Did you intend to say "you're confused"?

Christopher2 started this thread with his question. Most of us are here to help. It doesn't help him to tell him he is confused. If it were clear to him, he would not be asking a question.

 

Stev8e, It's nice of you to speak for christopher2 who I know asked the question.

Are you sure your not one of my stubborn norwegian cousins?

If you are I'm going to put you on my ignore list. ha! ha!

 

Chris, I used to get this stuff confused all the time. Still do sometimes. Most of what you need to know has already been said - sometimes it just needs to be said in a different way for different people.

The first thing to remember is that the power you are using at your device (headlamps, starter motor, whatever) is NOT the same as the power that is dissipated in the wiring as heat. If you need 55W at your device (headlight), you can use 12V and 4.58A, or 6V and 9.17A - assuming you use the right headlamp bulb. The power lost in the wiring will not be the same, however.

Let's assume you want to wire up the headlamp and you use 10 feet of wire that has a resistance of 0.01 Ohm per foot, for a total wire resistance of 0.1 Ohm. If you use the 6V system, you'll have 9.17A running through the wire and the power dissipated in the wire will be 9.17A x 9.17A x 0.1 Ohm (I x I x R), or 8.4W. If you use the 12V system, you'll have 4.58A running through the wire and the power dissipated in the wire will be 4.58A x 4.58A x 0.1 Ohm (I x I x R), or 2.1W. Note that, since the current is squared in the wire loss calculation, cutting the current in half cuts the wire power dissipation by a factor of four (assuming you use the same size wire in both cases). All the power loss in the wire comes out as heat, and has to be dissipated to the air or material surrounding the wire.

Another way to look at it is that, if you are using a 6V system and you want to draw a certain number of Watts, you have to use wire that has twice the diameter (or four times the cross-sectional area) to get the power loss in the wire down to the same level as a 12V system. That's one reason car manufacturer's are going to the 24V standard as Barry mentioned - to save all the weight and cost of that copper in the larger wiring needed at lower voltages. Of course, as you increase the voltage, there are other limits. I sure as heck don't want to be under the hood of a car designed to be using 120V for the primary electrical system. One touch of the wrong terminal and you could have a very bad day.

 

Thanks for your polite, studied responses, SteV8e & George.

 

Does the switch to 24 volts mean that I can't pull the old "Hey, come here and hold on to this spark plug wire while I spin-test the magneto" on my uneducated friends anymore??

 

not to change the subject, but I ran a 57 ford with 312, and 12:1 pistons, I would use a 6 volt starter, spun that motor up right away, ran it for years and it never burned out, think it was due to the sort period of time it cranked, has anyone else used a 6 volt starter on 12 volt system?