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post #9 of (permalink) Old 07-07-2009, 08:06 AM
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Since you asked for your answer from TR, I stayed out until he posted.

I think the problem here is that you got out the VOM without knowing what to expect.

The formula for the voltage across a coil is;

v = L di/dt

Nothing to be scared of here.

v is the voltage, L is the inductor value, i is current and t is time. The term di/dt is the change in current with respect to time. If there is no change in the current, di/dt is zero and the voltage is zero.

This formula is for an ideal inductor. Because your coil is made of wire, and wire has resistance, you have a combination inductor and resistor.

Using the values you measured, you had a 2 ohm resistance with a difference of 11 volts between the voltage measured at the two ends. Using Ohmís Law, we would calculate a current of 5.5 amps. Thatís not outrageous.

Multiplying the current by the voltage difference across the coil, we get a little over 60 Watts. Yeah, it would get hot just like a light bulb.

You asked about connecting it without the resistor. You ignition circuit, if wired correctly, does exactly that when you try to start the engine. It uses a connection from the starter relay to bypass the resistor when the starter solenoid is energized.

You say you are getting a spark. Is it a strong blue spark or a weak orange or yellow spark? If itís a strong spark, I think you need to look elsewhere for you problem.

If Iím reading this correctly,

Originally Posted by MyJeeps View Post
It ran fine 2.5 years ago, ...
that it hasnít been run in two and a half years, dump the tank and get some new fuel in there.

There are 10 kinds of people in the world.
Those who understand binary and those who don't.
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