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post #9 of (permalink) Old 07-07-2009, 08:06 AM
CJ7Taz
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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,

Quote:
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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