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Free Amps...

So I guess I've said enough to either be right or so wrong Hoople will give up on trying to teach me anymore. In any case I've got my electronics book and reading on it when I have time. Fortunately at the current time I really have no need to know any of this so it is not a pressing issue, just curiosity.

Well... I have certainly enjoyed reading all of it and the time all of you have put in to it! I reckon that I am comfortable knowing that I won't be overloading the bike when all is said and done.

I guess my next concern now is to make sure I don't resemble a glowing cigarette lighter, whizzing down the road when I plug that homemade jacket in! :D
 
You people are making this too complicated. whether or not u have an alternator or a generator you will still have a draw. put a wiggy on it and see what the draw of amperage is
 
You people are making this too complicated. whether or not u have an alternator or a generator you will still have a draw. put a wiggy on it and see what the draw of amperage is

The original question was where to put the wiggy :lolrolling
 
Look back at entry point #10. It is the position of the Maxi fuse within the circuit,,nothing more. The Maxi fuse is Before the tap of the Voltage regulator into the wiring harness.
These pictures should clear it up. Entry point #10 was the word version.
 

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Yea I see that. I guess one could test the battery cable and then test the one wire coming out of the regulator and subtract the amps coming out of the regulator from the battery cable amps.

I'm still out to lunch on how the increased voltage output on the line during regulator participation is affecting the line amperage. The parallel line can only have one voltage in all areas. If its going up anywhere on the line, it's going up everywhere on the line.

But this whole affair is still abiguous to me, in more ways than one.

First off you guys must have a different amp probe than me, since mine will only test AC amps with any accuracy. It will read DC current but not accurately. It could be used to test the stator (coming out of it before the regulator). But it won't test DC amps from the regulator or the battery.

My digital tester will only go up to 10amps so I can't just hook it between the battery and the cable, unless I won't to go buy a new tester.

I guess you guys have higher capacity digital amp testers ? (wiggys)
 
If I hadn't met you in person, I would guess you were playing with me.

:lolrolling Yea I get that sometimes even in person. But no I'm not playing, just confused and trying to learn. I suppose many times I'm guilty of playing tricks on my own mind since I tend to become overly concerned with the immediate sometimes confusing the whole picture.

BTW, I did find a Hall-Effect DC amp probe online but it was $229 bucks. Don't need to know my amps that bad yet.

I've always had a mental block on deciphering the separation of Volts and current, all the way back to my night college Electronics I class in 1975. In my Pea brain voltage is dependant on a build up of electrons and current happens when the former is conducted with a low enough resistance to an vaccum of elections (floating around their respective atoms). There can't be one without the other but I get kinda out in the outer boundaries sometimes trying to visualize the ramifications how more current flows WITHOUT more voltage in some scenarios.

In case you haven't notice I am a visual sort of learner. Once I get the concept visualized I can then move into the abstract but not before.

I do realize that by it's chemical nature the battery can only product 2volts per cell. And that 6 cells wired in series is how it steps up the total nominal voltage to 12v give or take the small amount of internal resistance inside the battery.

The internal resistance goes up (or down) depending on the size (area) of the cell platea. Which BTW denotes the amp or current capacity of a battery. The larger (or more square inch of) the plates the less internal resistance inside the battery to the chemical 2 volts per cell, thereby giving more current flow. I = E/R. This same concept is in exact relation to hooking two batteries in parallel, giving more amp capacity at the same chemical volt cell arrangement potential. I can at least visualize that.

Jumping batteries is also a parallel hook up. Of course when one jumps a weak battery with a good battery there is a potential difference in the weaker battery needing jumped so the current is directed at an infinite rate less the internal resistance of the two batteries and the jumper cables. It is my understanding that becuase of this rapid current 'low resistant short' it is easier on the life of the battery to slow charge instead of jumping (if possible of course).

The plot (in my brain) thickens as you move from cell electron generation, into adding a rectified current potential of a larger voltage into the scenario.

The reason I'm confused as to why you say that testing the amps of the battery cable will not be reflective of system amps is two fold.

1. I consider the charging input + the battery input to be the "system current". Why ? Becuase the nature of bike and car systems now days can't operate normally without both. That's just my observation, I have no other source to base that acertation on. So forgive me if that concept seems incredulous.

2. Quite aside from concept (1) and on a separate thread of thought, I do understand ( I thnk) what you are saying as the impact by the increased line voltage provided "during operation" by the regulator input.

I.e. if you physically unhook the regulator input from the system and run totally off the battery (12v nominal), you would have I (sys amps) = E (batt v) / R (sys resistance). So (forgetting for a moment about inductive locked starter rotor ramifications) and lets just say if you have a set total system resistance of X ohms so that this scenario produced 30 amps or -> 30 = 12v / XRt.

Obviously X in this scenario = 0.4 ohms (in paralllel RT is less than the individual Rs). So if XRt remains constant and we hook the regulator supply back into the circuit and line voltage goes up to 14v we now have 35 amps = 14 volts / 0.4 RT.

Anyhow I see you have sent me an email so I guess when we both have time we should continue other theory or instructive concepts privately out of respect for this thread.

But in closing I would like your input here in as much of your opinion of the facet of system amps being described as 'including' charging input or 'not including' charging input. Or in other words why would one want to care what the system amps be without the charging input ?
 
BTW, I did find a Hall-Effect DC amp probe online but it was $229 bucks. Don't need to know my amps that bad yet.

Then make yourself one of these. A simple DC current shunt. DEADLY accurate and can be made on the cheap for $20. It is used in conjunction with an (I may be asking too much here :)) accurate digital voltmeter.
 

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Then make yourself one of these. A simple DC current shunt. DEADLY accurate and can be made on the cheap for $20. It is used in conjunction with an (I may be asking too much here :)) accurate digital voltmeter.

I've heard of doing that a long time ago. You can even buy some that already have the voltmeter built in and is calibrated to show the amps which I surmise some of the more expensive multimeters do. You have to break in the circuit of course.

The one you show there if memory serves me, is constructed so you pre-know the amount of the ratio of ( X mvolts that will occur across it at a pre-known X amps). There by allowing you to calculate the amps by ratio once you measure the voltage across it. Or did I forget that concept totally also ?
 
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