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Discussion Starter · #1 ·
Can someone with Applied EE or practical electronics background help; my high school physics and civil engineering only go so far.

If I remember, E=IR: so a low-impedance (3 ohm) fuel injector should draw 12 / 3 = 4 amps. A high-impedance (10 ohm) injector should draw 12 / 10 = 1.2 amps. Does a low-impedance injector require more current to fire, and that is why “new” injectors are high-impedance? Or is there something going on in the solenoid resister that I don’t know about? If it is this simple, and I want to run high-impedance injectors, can I simply replace the solenoid?

This of course makes no allowance for voltage variance. I used 12 to make the math a bit simpler.
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