Hello! I am going to hotwire my Lincoln cont 1970 car.Battrey to coil.?

Question

I was told that I had to put a reistor between the battrey and the coil so I will not burn out my coil.It is now 12 volts what is the reistor that I need to bring it down to?How many voits do I bring it down to?The last Question is,are you having a good day?Thank you

Answer 1

Well, the main reason is for the alternator. the alt puts out about 14-15 volts when operating to charge the battery. However, the battery is merely there to start the vehicle and the alternator if working correctly is there to charge the battery and run the vehicle after engine start. The battery is just a short term backup if alternator fails. I am running my coil of my battery on my old pinto. Just remember to put a switch in the line so the coil doesnt stay charged the entire time otherwise you will kill the battery when you dont have the car running.

Answer 2

We used to use a set of jumper cables and go from the positive of the battery to the positive of the coil. Then take a screw driver and cross the terminals on the starter from selenoid to hot on starter. You have to leave the jumper cable on or it will shut right back off. I never heard of a resistor inbetween! I can't see how you'd burn out the coil when the hot terminal of the battery generally goes to the starter and then a lead from the starter to the hot side of the coil. What's different about that? 12volt is 12volt no matter how you derive at it! This is a make shift deal just to basically get the car to the shop. I wouldn't continue running it that way! Fix the thing right. The alternator isn't going to blow up the battery with 14 volts!

Answer 3

The term also refers to an automobile engine component that lowers the supply voltage to the ignition system after the engine has been started. Because cranking the engine causes a very heavy load on the battery, the system voltage can drop quite low during cranking. To allow the engine to start, the ignition system must be designed to operate on this lower voltage. But once cranking is completed, the normal operating voltage is regained; this voltage would overload the ignition system. To avoid this problem, a ballast resistor is inserted in series with the supply voltage feeding the ignition system. Occasionally, this ballast resistor will fail and the classic symptom of this failure is that the engine runs while being cranked (while the resistor is bypassed) but stalls immediately when cranking ceases (and the resistor is reinserted in the circuit).

9 volts is optimal voltage under load.

ADDED:
I'm always having a great day. I'm still alive!

http://en.wikipedia.org/wiki/ballast_(el...

Answer 4

you are able to merely run a jumper from the battery cable to the small cord the only furthest to the rear.going to the solenoid. to initiate the vehicle capacity up the different cord. 12 volts wont harm it ..notwithstanding it is going to reason the criteria to burn out two times as quickly. no longer if it really is a non everlasting fix ..say per week or so.. in case you recommend to go away it this way ..convinced set up the resistor in case you unplug the ignition change and placed a jumper cord there the production facility resistor will artwork because it really is in the loop