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## Re: Lucas coils

 To: pmcbride@hooked.net Re: Lucas coils jerry@tr2.com (Jerome Kaidor) Thu, 19 Jan 1995 08:08:05 -0800 (PST)
 ```Paul McBride wrote: > > at approx. half of battery voltage. During low speed operation > the points are closed longer, and the resistor heats up to > restrict current flow to the coil. > Less voltage is required to fire the plugs at low speed. > As engine speed increases, the points stay closed for a shorter > duration and the resistor cools off. This allows > more current to flow to the coil, for increased spark output. **** Uh, no. That's not how it works at all. Listen carefully, boys & girls: The spark coil is an inductor. When you put a current through an inductor, it stores energy in the magnetic field surrounding it. When you interrupt that current, the magnetic field collapses, and the inductor delivers the stored energy at its terminals. In the case of a spark coil, the stored magnetic field collapses not only through the primary windings that induced it, but also through the thousands of secondary windings. Zap! When you first hook up a voltage source to an inductor, it looks like an open circuit. This is because a reverse voltage (AKA counter-emf )is produced by the expanding magnetic field. As the magnetic field nears its full expanse, the counter-emf gets less & less, and the coil starts to look like a short circuit. If the coil is not given enough time to charge up, you don't get a spark. Remember, it looks like an open circuit when the points first close. How can we stuff those electrons into it faster? We can stuff electrons faster into most anything by using more voltage. Suppose we applied a 400V source to the coil? Bet we could saturate it real quick! Well, we don't have 400V available. But we do have 12V. And we can choke it down to 6V with a resistor, so the coil doesn't burn up. However, at the moment when we first connect up the coil, and it looks like an open circuit, the RESISTOR DOES NOT DROP ANY VOLTAGE WHATSOEVER, AND WE HAVE A FULL 12V ACROSS THE 6V COIL. Remember, resistors drop voltage by the virtue of current flowing through them ( V=I*R ). No current flowing through, no voltage drop. This makes the coil "start to charge" much faster. Us electronics weenies would say that the resistor converts the car's 12V into a "current source". In other words, a source that wants to deliver a specific value of _current_, unlike the car battery, that really wants to deliver 12V regardless of the current ( or lack of same ) pulled out of it. An ideal current source would be an infinite voltage in series with an infinite resistor. Well, we don't have a source of infinite voltage in the vehicle, so we make do with 12V and a couple of ohms. :-). It's common to have a solenoid terminal that shorts out the resistor during starting. Heck, the battery voltage drops down to 9 or 10 volts anyway then. Especially when you're trying to crank over a big V8. You do lose the benefit of the "current source", but the engine is turning so slowly that it doesn't matter. Bet everybody's *really* confused now :-). - Jerry Kaidor ( jerry@tr2.com ) ```
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