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Discussion Starter · #1 ·
I seem to have a wiring problem. I just bought this 72 Chevelle and while I detailing under the hood, I found a burnt wire going from the junction block on the firewall to the starter. This seem to be a quick fix jumper wire because after looking and removing the starter, I found the original wire going to the starter was burnt into 2 pieces as well. Have any of you guy's ran into this on your Chevelles??? I don't see anything out of the normal along the harness that would make the wires burn up. Thanks for your help..
 

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Almost sounds like the last guy by-passed the fusible link instead of replacing it. That's plain stupid. Put a link in the replacement wire.
 

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i know from experience... definitely replace the link..... i almost burned my car to the ground by not using a fuse. the insulation got cut and created a short circuit...i had to rip thewire off the battery with my bare hands and burned the ****t out of my hand
 

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If a second wire burned up, you know that the wire is overloaded or has a partial or complete short. Can you find the wire on the schematics and do you know where it runs? Has the previous owner tied in additional items on this line? Bottom line, the wire is pulling too many amps, and if replaced will just burn up again (or burn up the fusable link). It could be something as simple as the firewall block (or somewhere else) has a bad connection. A bad connection generates resistance, therefore generating a greater load on the wire. The higher the load, the more current, and small wires have their limitations. On the other hand, I would trace the wire out in detail and look to see what's been tied into the fuse box. I have often found one of the greatest trouble makers being an added stereo system. A small system that pushes only 120 watts can pull over 10 amps.

Good luck.
SS4speed.
 

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Originally posted by SS4speed:
again (or burn up the fusable link). It could be something as simple as the firewall block (or somewhere else) has a bad connection. A bad connection generates resistance, therefore generating a greater load on the wire. The higher the load, the more current, and small wires have their limitations.

Good luck.
SS4speed.
Actually, if the resistance was high, the current would decrease. I would look for a LOW resistance path somewhere on the LOAD end of the burnt wire. If none of the fuses under the dash are blown, look for a problem with something not fused or tied into the supply side of the fuseblock.
 

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Discussion Starter · #7 ·
Thanks for all the replies. The original wire was only burnt through on one side, the fusable link side. I did replace the fusable link, but haven't tried it yet. I drove this car 45 minutes home when I bought it, so I don't know if this was a previous problem or a new one. When I got home a bunch of my buddies was here to see it and nobody smelled a burning wire. I only found it while snooping around. There is nothing aftermarket or extra on the car, and fuse block looks good. I ordered a wire diagram to see what all is on that harness. It does have factory air, but don't see where that would cause a problem. Thanks again for all your guy's advise, wiring is not a good subject of mine.
 

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A shorted alternator could have burnt that fusible link. If it runs OK after you replace it, I wouldn't worry too much.
 

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originally posted by MalibuJerry350..

"Actually, if the resistance was high, the current would decrease. I would look for a LOW resistance path somewhere on the LOAD end of the burnt wire. If none of the fuses under the dash are blown, look for"

True, (E = I X R), but only if you have a purely resistive car, with no Power supplies (Stereo, etc), no motors (Electric fans, etc), and no LC circuits (coil, etc). If this describes your car, and it runs, I really want to see it.

Let me explain my point, if you have a motor or Power Supply (example: Stereo system) that uses or pulls 6 amps (at 12 Vdc). It's rated power in watts very close to 72. Now you have a bad connection (whatever it may be) in the line that drops 4 Vdc. The load now still wants it's 72 watts, but it only has 8 volts to work with. This is true for almost all Power Supplies and Motors (but not all, ofcourse). Therefore, P = I X E, Voltage has dropped, current must increase to 9 amps to maintain the 72 watts. Since all cars have motors and Power Supplies, this was my thinking when I made the above statement.

Regards,
SS4speed.
 

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Usually a bad joint causes high resistance because of the bad joint. Forms a voltage divider. Thought current goes down as resistance goes up...Well, I'll let Jerry type..I'm heading home.
 

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Originally posted by John_Muha:

"Usually a bad joint causes high resistance because of the bad joint. Forms a voltage divider. Thought current goes down as resistance goes up..."

Usually a bad joint does cause high resistance. In the following, I am ignoring carbon arcing or thermal runaway scenarios. A short or partial short forms a voltage divider, not the bad joint, unless it's shorted to something else. Current does go down as resistance rises, E = I X R proves that. As noted above, only in a purely resistive circuit. So if you are talking about the lights, then your correct. If you are talking about your car stereo or a inductive type motor in line with the bad joint, you are incorrect, P = I X E proves that. A power supply or a motor attempts to operate at it's required Power, that's the key here. A motor or Power suppy will draw more current to compensate for the drop in voltage, hence more current flow. The bad joint will cause a voltage drop, therefore droping the voltage to the device beyond it, on that line. If you have ever done any low-line testing on Power Supplies, you will understand.
We are not looking at passive circuits in cars, they are not purely resistive in most cases. Nor are we trying to deal with Real or reactive power here, just how certain devices in cars work, when voltage is dropped to them. No one disagrees with E = I X R, but looking at all circuits from that point of view makes not sense.
 

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I will still stand by my answer. If the connection was faulty in the first place, overheating due to oxidation for example, the resistance would increase, dropping the voltage at the other end. Remember, we are NOt decreasing the resistance on the load side. The "meltdown" would be at the point of resistance (faulty connection, crimp, etc, energy released as heat), not the wire itself. A switching power supply would eventually go into "shutdown" if the input voltage dropped below a point where oscillation could not be maintained. With a DC motor, again, any increase in current would be exhibited by heat through the weakest link....again, the faulty connection OR point of resistance. Place a rheostat in series with a DC motor or switching power supply. Place an ampmeter in series and see what happens when you decrease the voltage. ;) OR, the fusible link could have been too damn close to the headers! :D
 

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Maybe it was a fusible link you were looking at that was burnt? If you short something out momentarily you can burn the fusible link but not cause it to open. The only other thing I can think of is that there's a bad power draw in the wiring. Maybe when you turn on the AC or lights or cigarette ligher or something like that it overloads the power feed (short to ground) and burns the link. Then, the old owner was turning it on and off trying to get it to work and finally gave up.

I'll stand with Jerry and John. For any electrical load that is stock in a 72 Chevelle the current will not go up if there is a resistance added into the wire feeding it.

For the motor. I'd like to see you operate an inductive motor on DC. I always thought they were for AC only but then I'm only an engineer that works in the inductive motor control field ;) The car has a DC motor for the heater/AC. You control the speed of a DC motor by varying it's voltage. Lower the voltage and the speed and current go down. Why do you think the heater motor has a resistor network for speed control?

What you're describing about the voltage falling and the current rising works for AC motors and switching power supplies but not for anything you will find stock in a 72 Chevelle. The only thing I can think of that would be in a car and would exibit this effect would be a power amp with a switching supply in it (any decent amp). I don't know if any manufacturer is using switching supplies in their vehicle these days but they were never used on older vehicles. Linear power supplies don't count since they will draw the same current until they drop-out.

When you want to talk DC there can be no inductive or capacitive loads. That is the nature of DC. Switching loads are not DC loads because they're an active load.

You always have to follow V=IxR since it's a "law" :D

Besides, a resistance of the amount you give in your example would cause everything on the car to act up. When you turned the lights on or tried to start the car the voltage would drop to 0 at the fusebox. If what you prepose was true, the resistance that could be introduced and still allow the car to operate correctly would have negligable effect on the current draw.

Peter
 

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*Psst Jerry*
The problem with the above calculations is that it was converted to Watts. Then Watts were plugged back into a changed circuit. Big no-no. Beyond what George Ohm had to say look towards Thevenins Theorem. That old dead dude thinks everything is a resistor.
"We are not looking at passive circuits in cars, they are not purely resistive in most cases. Nor are we trying to deal with Real or reactive power here, just how certain devices in cars work, when voltage is dropped to them. No one disagrees with E = I X R, but looking at all circuits from that point of view makes not sense."
*EDIT*
Yep, they are resistive in the end. Still a DC circuit. All can be calculated to a simple ohm's law circuit.
 

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Hey Jerry and John,

The only "old dead dude" here is me, Peter just took my head off, very cleanly.

Peter,

Thanks for the education, I've been away from electronic\electrical just too long (over 20 years now). I clearly started to mix AC and DC theory, a Big NO NO, as John stated. If you ever leave your job, you should become a College Professor. Nice job and very well stated, it was crystal clear. Thanks for setting me back on course.

SS4speed.
 
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