/diy/ - Do It Yourself

>>881467
This isn't a situation where a series resistor will work, your explaining is completely wrong, and your math is wrong in a situation where you would use a resistor.


> a device will only pull as much current as its rated for is because it has a current limiting device in it.

No. Current regulators are extremely rare and only really turn up outside of LED drivers.

Every load has a voltage vs current graph. I've attached one as an example. For resistors it's a straight line, for a more complex device like a semiconductor or OP's battery charger it will not be straight but you still get a set current draw for a given voltage. Devices are designed to operate at a set voltage, you give them that voltage and they draw the appropriate current. A 12v 1A lightbulb doesn't decide it wants to draw 50A just because you've replaced its 8 AAs with a car battery.

> high wattage, low resistance element is a great way to limit current

You are only reducing current because you're causing a voltage drop which means the other components in series are operating at a lower voltage and drawing less current. I'll show you how this will effect the battery charger. For simplicity lets assume the charger is like a 28.75 Ω resistor (13.8V/480mA)

Charger is hooked up to its 13.8V power supply and drawing 480mA like its supposed to. Now some asshat comes along and puts a 28.75 Ω resistor in series with it. The total resistance of the circuit is now 57.5Ω. The total current draw of the circuit is now 240mA (13.8V / 57.5Ω). This is the current the charger gets because current is the same through all series elements. The charger is getting HALF of the current it was supposed to get. Furthermore the 28.75 Ω resistor causes a voltage drop of 6.9v (28.75 Ω * 13.8V). Due to Kirchoff's voltage rule this means the charger is getting 13.8V - 6.9v = 6.9v, which means its not going to run at all because devices like that need the right voltage input.