A bit of theory (again...) ?
Ohm's law: U=R * I or Voltage=Resistance * Current , that's it and can't be changed !
Think of it as a garden hose @ home: You got the water pressure (voltage), the hand-gun (resistance) and the flow (current). The more you open the hand-gun (lower resistance): the bigger the flow...
The alternator is voltage-regulated so it outputs always the same voltage in its working range.
An almost empty battery has a relatively low internal resistance, let's say 1,2 Ω and to make it simple 12V for the alternator's voltage.
That would allow 12 / 1,2 = 10 amps to go from the alternator to the battery, thus charging it (hand-gun open).
As it charges, the battery's internal resistance get bigger (hand-gun almost closed), let's say it reaches 12 Ω, the charging current would be reduced to 12 / 12 = 1 amp ...
Got it ?
Also, when current circulates over a resistance, there is a voltage drop (that's why the cable gets warmer or a bulb shines), our garden hose is pretty stiff when hand-gun is closed, but loose stiffness when hand-gun is open...
If TOO much current is drawn, the alternator's would go on its knees and the output voltage would go down. If the pump is too small and the hand-gun fully opened, the flow will decrease once the initial pressure (voltage) drops...
In your case it would be nice to check what current is delivered by the alternator, an Amp clamp will do the job...
ps: I know the purists will say my example is not good since the battery already has a voltage that should enter the equation. I made this purposedly not to add difficulty to understand the basics...
BRs, Bernie
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