Fight my ignorance about volts and amps.

[Q.E.D]

[QUOTE=Chronos]
Since nobody’s addressed this yet:
[/QUOTE]

Except in post #34.

[CookingWithGas]

[QUOTE=Chronos]
For a given amount of power available, you can increase the current by decreasing the voltage, or vice-versa.
[/QUOTE]
Sorry, I’m lost. Isn’t voltage directly proportional to current? Increase voltage and you increase the current?

OK, for a given amount of power…

[Crafter_Man]

[QUOTE=CookingWithGas]
Sorry, I’m lost. Isn’t voltage directly proportional to current?
[/QUOTE]

Yes, when powering a resistor. But you’re not always powering a resistor - sometimes it’s a non-linear component. And you sometimes have the luxury of controlling the load resistance.

When designing a power transmission system, you have the luxury of controlling most of the parameters. I can, for example, design a system wherein there is a high voltage between the two transmission wires, and relatively low current in each wire. Conversely, I can design a system wherein there is a high current in the wires, and a relatively low voltage between the wires. In either case, the load impedance will be designed for maximum efficiently over the power range.

Clear as mud, right?

Let me try again.

Everyone knows V = IR, right? That’s Ohm’s Law. But I don’t like calling it Ohm’s Law. I prefer to call it the Ideal Resistor Equation. It is an equation that describes what’s going on with an ideal resistor. It also works well (though not perfectly) with practical resistors (as long as you stay within their temperature and power ratings). But it does not work at all with non-linear things. And there are LOTS of those. Examples include diodes, LEDs, batteries, and incandescent light bulbs. Those things are more complicated than resistors, and a simple application of Ohm’s Law will lead you down a wrong path. You also don’t want to use it when looking at the primary of a transformer, since its effective impedance is a function of the load impedance.

[groman]

[QUOTE=CookingWithGas]
Sorry, I’m lost. Isn’t voltage directly proportional to current? Increase voltage and you increase the current?

OK, for a given amount of power…
[/QUOTE]

There’s an excellent online demonstration site that visually shows how certain circuits work:

Two light switches for a single bulb

Full wave rectifier

DC Ohm’s law demonstration

Impedances of same magnitude and different phase

and more

I highly recommend playing with these. Each of them is basically a preloaded circuit in a circuit simulator. You can erase it and draw your own and simulate that (there’s no easy way of saving it, unfortunately).

[Patty_O_Furniture]

The lowest reading I’ve been able to get from sweaty palm to sweaty palm is 145K[symbol]W[/symbol], and that’s by jamming the sharp probes into my skin as hard as my pain threshold will allow. That’s only low enough to draw 80[symbol]m[/symbol]A from 12V, or 1/1500th of the estimated lethal draw.

I then pricked a thumb on each hand and mashed the bloody spot into the probes and the reading only dropped a few more K[symbol]W[/symbol]. It would have to drop all the way down to 100[symbol]W[/symbol] in order to draw the estimated 100mA needed to bother your heart.

Yes, I bled for this message board today.

[groman]

[QUOTE=Patty O’Furniture]
I then pricked a thumb on each hand and mashed the bloody spot into the probes and the reading only dropped a few more K[symbol]W[/symbol]. It would have to drop all the way down to 100[symbol]W[/symbol] in order to draw the estimated 100mA needed to bother your heart.
[/QUOTE]

Sounds like you need some electrolytes

[Rick]

[QUOTE=lobotomyboy63]

@beowulff

Car batteries are a whole other issue. The CCA is a rating which measures how much current the battery can provide to a load in cold weather. But, since the battery is only 12v, that load must be a very low resistance. So, a 500CCA 12v battery would need a load of no more than .024Ω to actually provide 500 Amps. This is why the cables running to the starter are so thick. This also explains why a 12v battery is safe to touch - your body resistance is much, much higher, so only a very small current will flow.

(underlining mine)

I’m always skeptical of claims where money is at stake. In this case, they’re selling me CCAs—I need them for when I am trying to start the car in the winter, home visiting family. The POS battery that was, of course, standard with the car was pretty weak and I wanted to upgrade.

IIRC the battery weighs 30 lbs. Why don’t they advertise it as weighing 5 lbs (fine print: on the moon)? And as I alluded to earlier, my car stereo claims 270 watts of power, but that’s at 4 ohms…my home stereo uses speakers rated at 8 ohms, and at far less wattage it seems a lot louder. I.e. interpreting the numbers is tricky.

So I read the underlined part and my first reaction was, “So did I get ripped off? Does it provide 500 CCA or can it provide 500 CCA under only optimal circumstances?”

If I’ve worked this out correctly in my mind, I guess the ratings are based on industry standards…they assume a particular resistance and keep the other variables constant so that consumers can make a useful comparison. They’re not selling me a “5 lb battery.”

Suppose you hooked up heftier cables to the battery to reduce resistance…would the CCAs go even higher (assuming you haven’t reached a point of hopelessly diminishing returns)? I guess if you did that, you’d run the risk of frying other components of the system that weren’t designed to handle more CCAs.

If the resistance in the cables is higher than it’s supposed to be, I’m not getting the 500 CCAs promised, but that’s not the fault of the battery manufacturer. Right?
[/QUOTE]
It seems to me that both of you misunderstand what Cold Cranking Amps is (are?)
It is a standard that allows you to compare the relative power that can be delivered by a car battery.
In this Wiki article on car batteries there is a good explanation of what the term means

Cold cranking amps (CCA) is the amount of current a battery can provide at 0°F (−18°C). The rating is defined as the amperage a lead-acid battery at that temperature can deliver for 30 seconds and maintain at least 1.2 volts per cell (7.2 volts for a 12-volt battery). It is a more demanding test than those at higher temperatures.

So you decide to go into the car battery business. You start building batteries. You send the to a testing lab, where they take a fully charged battery, chill it to 0F and then see how many amps it will deliver for 30 seconds and still have 7.2Volts left at the end of the test. What ever number they come up with is what you print on your labels.
A few things I would note: You car won’t start if the battery is only providing 7.2Volts, and since chemical reactions slow down as things get colder, this is a very tough test of a battery’s ability to provide power.
Bottom line is this a standard that allows you the consumer* to compare different batteries to determine which is best for your needs.
One last thing, when shopping for batteries, do not confuse CCA with CA which is done at 32F and therefore yields a much higher number.
*CCA is also the basis for battery testing. When using a carbon pile tester, we apply a load equal to 1/2 of the CCA for 15 seconds and read the voltage at the end of the test to determine if the battery is still serviceable.

beowulff Getting back to Ohm’s law for a moment. As a fellow instructor of mine used to explain it:
If you drink too much coffee in class, it is voltage that makes you want to visit the men’s room. Amperage is what you leave in the urinal, and if you squeeze it in the middle, that is resistance.

[HongKongFooey]

[QUOTE=groman]
There’s an excellent online demonstration site that visually shows how certain circuits work:
[/QUOTE]
Thanks for the great link groman, I have a trainee right now and that will be an excellent teaching aid.

Someone already posted a link to this site http://amasci.com/ele-edu.html and I want to second the recommendation, lots of good information there.