Hi, first of all, I'm not an electrician or whatever lol. I know what ohm means. I ordered some resistors for my low impedance injectors that is rated 1 ohm, 5 watts, 5%. My injectors is rated 5.4 ohm if I remember (bosch 1680cc aqua green top), and I did the maths and 1 ohm resistor is ok ! but, I dont know if 5 watts rated resistors is ok even if I put it after the injectors. someone have the knowledge for that ?

 

It doesn't matter where you put the resistor, the same current will be flowing through it when it is activated.

First, the 5% is the manufacturing tolerance of your resistor. Measure it with an ohm meter and you will find that it is like 1.02 ohms or something.

You need to calculate the current going through your total system to calculate your wattage. V = IR, so in your case, I (current) = 12v/R, where R is the combined resistance of your injector and your resistor.

The wattage calculation is (I^2)*R, where R is the resistance of your resistor (not the total combined resistance).

 

^ +1. If your resistor is connected in series with your injector (doesn't matter if it's in the input or return injector wire), the total resistance will be 6.4 ohms (+/- 5% or so). 12V/6.4 = 1.88 amps through the injectors and your new resistor. Power dissipation in your resistor will be 3.52 watts, which is within the resistor's capability at its rated temperature.

 

voltage (drop) = current * resistance, is show as E = IR

wattage is current * voltage, is show as W = IE, or you can substitute IR for E from above and now you have: W = IIR which is I squared times R.

Then we get into serial and parallel circuits. Best to read a book about DC electronic as a starter.

 

When performing those calculations, it's better to use 14-15V rather than 12V, since the alternator will be running as long as the engine is on. Even if your battery voltage is only 13.0V when the engine is running, it's a good idea to add some safety margin.

So the numbers become 14V / (6.4 ohms) = 2.2A current through the entire circuit, or 15V / (6.4 ohms) = 2.4A. As Chuck mentioned, to solve for the watts of energy that the resistor will be dissipating, the equation is P = (I^2)*R, in your case the numbers become (2.2 * 2.2 * 1) = 4.9 Watts or (2.4 * 2.4 * 1) = 5.8 Watts. Personally I'd look for a resistor with a higher power/wattage rating.


In my opinion, you should make sure your ECU or injector driver box can handle 2.5A of current per channel... if you send excess current through a driver transistor it can become permanently damaged. If the ECU can't handle 2.5A, consider using a higher resistance value (something with higher ohms will limit the total current), but remember you will need to recalculate both current and wattage for the different resistor value. Wattage may increase, since there will be more voltage across the injector.