I'm trying to figure out what's wrong with plugging the solonoid back in and zip tieing it up under the tank and out of the way?
Yeah, someone explained that to me before. But you are changing 2 variables, resistance and current (which of course is very applicable here). If all things are kept constant ('cept the resistance) then resistance is proportional to heat, look at a light bulb for example...maybe a bad example...but if you restrict the flow of energy it has to go somewhere, i.e. heat.That does not sound right to me. The higher the resistance, the lower the current draw (Current =Voltage / Resistance, I=E/R). This implies that the higher the resistance, the less heat is going to be generated.
The highest resistance you can get is infinite resistance, which means that nothing is connected to the voltage source, and that means that the least power is drawn (none) and the least heating takes place (none).
If you did not burn out the 10 ohm 1/2 watt resistor, you should be twice as safe with a 20 ohm 1/2 watt resistor. You may even be able to go to 1/4 watt, but since I have waffled about everything else in this post, I certainly am not going to recommend the lower power.
I am not an electrical engineer, I only play one on TV.
Jan
DrBLVD,The higher the resistance, the more heat is going to be generated.
Sir, I stand corrected, Thank you. Good analogy!.....ummmm...nice schematic too, but where is the Flux Capacitor?DrBLVD,
In order to illustrate that the lower the resistance, the more heat is generated, here is a thought experiment:
There is a 1200 watt heater plugged into a 120 VAC outlet.
Power (1200 watts) = voltage (120VAC) x current (amps), therefore
current = power/voltage = 1200/120 = 10 amps.
Ohms law says that resistance of the heater = voltage / current = 120 VAC / 10 amps = 12 ohms.
Now plug in another, identical heater into the outlet right below the one above (the thought-house has 20 amp circuits). That one will also draw 10 amps because it has a resistance of 12 ohms. These two heaters are in parallel, as are all the devices plugged into a given circuit of wall outlets, so their resistances are in parallel. The formula for the equivalent resistance of devices in parallel is:
Equivalent resistance of parallel resistances (ARRL Handbook) = 1/ (1/first resistance) + (1/second resistance)) = 1/ (1/12) +(1/12)) = 1/(2/12) = 1/(1/6) = 6 ohms.
Since the heat increases as you plug in more heaters, and the equivalent resistance of all the heaters together decreases (from 12 to 6 ohms) as you plug in more heaters, higher heat is associated with lower resistance.
The math is the same for a power source of 120 VDC. In fact, if Edison had won the power war instead of Westinghouse and Tesla, it would have been DC.
Jan
I'm not sure if this will work, but it does on OBDII systems in cars and trucks.Ok can any one tell how I can get my FI light to go out after 4 1/2 min.?
The display in your speedo (odometer display) is storing a fault code. The FI light will not go out until the code is downloaded and the fault is corrected. To download the code, there is a white connector with a rubber boot over it down near your ECU behind the right side cover. You dealer has a simple switch that plugs into this connector and that will display the code in your odometer display. Disconnecting the battery will not do it. This link shows how you can make your own switch with a paper clip. Doing this will display the code in your odometer window. If the anomoly that caused the fault is repaired, the code will dissapere and the FI light will not illuminate. If the problem is not fixed, the FI light will again remain on until the code is downloaded and the fault fixed. Mine was code C29, secondary throttle position indicator. Mine had a broken terminal inside. Here is the link, oops, hang on I gotta go find it again.Ok can any one tell how I can get my FI light to go out after 4 1/2 min.?
To answer a PM from Daycruiser.All I can suggest is the resistor mod (50 ohm, 5 watt, heat sink type) once you clear the codes from memory. I've put on about 3 hours of riding and about 30 minutes total of idling and so far no light, no codes.