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#1
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62 Catalina with points ignition.
I understand that these old cars start with 12 V going to the + side of the coil to help with easier starting. Once running the ignition switch returns to “run” position and the motor runs on 7-8 volts to coil for longer point life. What is about the resistance wire that makes it reduce 12 volts to 7-8V? I tried hooking it to + side of the battery and reading voltage at the other end of the resistance wire with a DMM but it read 12.7 V. Guess I was getting the battery’s voltage. Please educate me a little. No electrician here so keep it simple, like me!! |
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#2
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Resistors (and resistor wire) only drop voltage if current is flowing. A voltmeter will not draw current (by design).
Was the engine running when you measured the voltage? If not, I would expect battery voltage or close to it. Just a guess.... |
| The Following User Says Thank You to Shiny For This Useful Post: | ||
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#3
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The points were open when you measured the voltage with the engine not running. Turn the engine over very slowly until the points close. Then you should see the lower voltage on coil +. Another way is to ground coil - (which is what the point set does) and then measure coil + to ground.
__________________
My Pontiac is a '57 GMC with its original 347" Pontiac V8 and dual-range Hydra-Matic. |
| The Following User Says Thank You to Bill Hanlon For This Useful Post: | ||
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#4
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Ok, so resistance and voltage drop will not happen without the engine running and/or points opening and closing. . Motor was not running and I simply touched the resistance wire to the battery and checked the voltage and got 12.7 or battery voltage. Duh!!
I thought perhaps the resistance wire was something special and would not flow more than 7 or 8 volts max. Thanks for the “lesson of the day.”. |
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#5
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You have to have an amperage flow to get a resistor to work. No load no resistance. Too large a resistor with too low current flow no resistance. Too small a resistor with too high a current, Poof light bulb or fire.
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