Factory ammeter
To illustrate the current relation to voltage drop.
V=I x R
1.5 Volts = 25 amps x Resistance
.06 Ohms = Resistance.
Lets now say the current is 45 amps.
The votlage drop will increase proportionally.
V = 45 amps x .06 Ohms
V = 2.7 Volts
Now what's the regulator going to see and how will it react?
Once the battery is charged (or replaced if defective), the typical current flow will be 5 amps or less.
Now what happens to the voltage drop?
V = 5 amps x .06 ohms
V = .3 Volts
Everything works like normal...
until the lights and other stuff is turned on...
To illustrate the current relation to voltage drop.
V=I x R
1.5 Volts = 25 amps x Resistance
.06 Ohms = Resistance.
Lets now say the current is 45 amps.
The votlage drop will increase proportionally.
V = 45 amps x .06 Ohms
V = 2.7 Volts
Now what's the regulator going to see and how will it react?
Once the battery is charged (or replaced if defective), the typical current flow will be 5 amps or less.
Now what happens to the voltage drop?
V = 5 amps x .06 ohms
V = .3 Volts
Everything works like normal...
until the lights and other stuff is turned on...