Can Ammeter Run Parallel?

[MonzaRacer]

Thats what hes trying to make but the type of system he has wont do it the meter is too heavy internally. If you want to find a volt meter ,gut it and fasten it behind the factroy guage face, ad a pin slighty under the center point and make sure it "looks" like it has a amps guage and is "charging" I talked to a company that does retoreation and they do this every day but its time consuming. Thier big problem is making the fac ammeter "look" like its charging and show a certain charge as its a volt meter.
Good Luck

[68nate]

I don't see why it would be a problem. current= voltage multiplied by resistance. Should roughly divide current by two as indicated in your test. One thing to keep in mind is that a 20 guage wire and 8 guage wire would have about the same RESISTANCE, but it wouldn't take much CURRENT to let the smoke out of the 20 guage wire.

[MonzaRacer]

Wrong the 20 guage wire will have more resisitance. All of the power will go through the 8 guage.
AS amperage goes up the wires will heat, more so for the smaller one BUT if the lagerone can keep up with the power flow nothing will be going through the little one, or not enough to mean anything.
The unit in the car wont respond as its very "heavy " inside if factory unit handled all of the current.

[TheMonkey]

... nothing will be going through the little one, or not enough to mean anything....

when i tested this, a meaningful amount went through the smaller gauge wire through the ammeter. remember, when i split it, just more than half went through the bigger wire. i tested it, it worked the way i want it to.

[baskin]

TheMonkey; your theory is correct, to be more exact, you’ll find a wire resistance chart here http://www.stealth316.com/2-wire-resistance.htm What you’ll need to do to calculate the split is to first find the resistance of each path, then convert that resistance to conductance (1/R) and then find the percentage of conductance for each path. As an example, let’s say you have 10 feet of 10 gauge bypassing the ammeter, giving a resistance of .0102 ohms, and 15 feet of 12 gauge running through the ammeter path, giving a resistance of .0243 ohms. The conductance of the bypass path = 1/.102 = 98.4 S, the conductance of the ammeter path = 41.15 S, and the percentage of current through the ammeter = 100* 41.15/(41.15+98.4) = 29.49%. The resistance given is at 77F and will rise about 6% for every 20 degrees F but, given both paths are at relatively the same temperature, it won't change the final numbers much.

For safety reasons, you'll want to use a fusible link on both paths, the correct gauge for that should be 4 gauge numbers higher than the wire being used, eg; 10 gauge wire would require a 14 gauge fusible link. Finally, remember that the function of the link is to get hot enough to melt the copper on overload, which will also melt solder, so although you can solder the connections to the link, they need to be primarily attached mechanically (crimp connector)

[TheMonkey]

baskin: you rock . that's the information i was looking for. resistance is something i can measure, and that formula seems straight forward.

i was thinking about using a 40 amp circuit breaker before going through the firewall to the ammeter, but i'll check that against the math for the wiring.

cheers,
Scott.

[68nate]

Wrong the 20 guage wire will have more resisitance. All of the power will go through the 8 guage.
AS amperage goes up the wires will heat, more so for the smaller one BUT if the lagerone can keep up with the power flow nothing will be going through the little one, or not enough to mean anything.
The unit in the car wont respond as its very "heavy " inside if factory unit handled all of the current.

Yes, I said "roughly" the same RESISTANCE. Take your Fluke to each and what do you get? About 0.1 on the 8 guage and maybe 0.3 on the 20 guage. Just saying resistance has nothing to do with how much current a wire can handle. In a DC PARALLEL CIRCUIT, the available voltage and the resistance of each circuit controls how much current flows through each circuit.

[baskin]

baskin: you rock . that's the information i was looking for. resistance is something i can measure, and that formula seems straight forward.

i was thinking about using a 40 amp circuit breaker before going through the firewall to the ammeter, but i'll check that against the math for the wiring.

cheers,
Scott.

The lowest resistance you can accurately measure with an ohmmeter is about an ohm, below that, the lack of resolution and contact resistance makes for a pretty inaccurate reading. What you can do is essentially build your own ohmmeter, just go down to radio shack and pick up a couple of 20 ohm, 20W resistors. Connect them in parallel to ground, connect the other ends of them through the wire you want to measure to the positive side of the battery. With a battery voltage of 12.6v applied, each resistor will draw 12.6v/20 ohms = .63A, for a total of 1.26A. Measure the voltage across the wire and divide it by the current in order to get the resistance. This is exactly how an ohmmeter works, but with a far lower current which limits the accuracy in measuring low resistance loads.