Ford recall may be bigger than anticipated
Originally posted by svtbrad
Part III of story.
http://www.detnow.com/wxyz/ys_invest...585993,00.html
Part III of story.
http://www.detnow.com/wxyz/ys_invest...585993,00.html
Originally posted by Crispy98
I wouldn't bet the farm on it. The short in this circuit is high resistance and it's pretty hit or miss if it pops the fuse or not.
I wouldn't bet the farm on it. The short in this circuit is high resistance and it's pretty hit or miss if it pops the fuse or not.
Originally posted by yysenhimer
Explain what a high resistance short circuit is. A short circuit by definition is low resistance.
Explain what a high resistance short circuit is. A short circuit by definition is low resistance.
Bird
From my knowledge of electrical circuits I will try and explain what I “think” they are talking about with a high resistance short and why if the proper fuse is not used where the possibility of fire could develop.
First, if a circuit is designed to be operated at say 5 amps then that is in the design. It has to do with not only the wire size but the insulation package used on the circuit, basically the type and thickness of insulation on a winding (coil of wire), wires, IC’s etc.
I am not sure what is involved with this particular switch but it sounds like there may be a coil involved since “high resistance” circuit is mentioned, or it could be a type of resistor.
Let me try to explain what is meant by a “high resistance” short. Theoretically, there is no such thing as a “high resistance” short. You either have a short or you don’t. Just like an open, a circuit is open or its not.
However, in the real world of electronics and electricity there is no such thing as “theoretical” but rather “real world events”.
Any electrical circuit has heat involved with it. It doesn’t matter how low the current, there is heat involved with it. All circuits have current (amps) and all circuits have resistance. When you put current (amps) through any circuit you will have heat. An electric heater is the most basic example of this as well as your ordinary incandescent light in your home. Touch the light in your home and you feel heat. This is, basically, the result of current passing through the filament inside of the light bulb.
If this particular circuit was designed to be operated at 5 amps then it had that type of insulation package associated with it, to handle the heat it would see in “normal” operation. Usually, insulation packages are designed to handle roughly 2 to 3 times the amount of heat they would see under normal operation.
A “high resistance” short, in theory, is where you could have a short at the output of the circuit but if the resistance of the circuit is high enough (due to a faulty design) the circuit never opens up, or opens up after enough heat is generated to cause a fire.
Looking at this circuit as purely a resistant circuit it may have a total of 2.4 ohms (resistance) in the circuit. If it is operated at 12vdc it would see 5 amps (current) running through it. 5 amps at 12vdc would produce 28.8 watts of heat in the circuit.
Now, if the component is faulty because it has only 2 ohms of resistance it would draw 6 amps if operated at 12vdc. This would produce 72 watts of heat, or over 2 ½ times the heat that it was originally designed to handle. Since the circuit is not protected with a 5 amps fuse it will continue to heat up and not open.
That is possible because the resistance of the circuit is acting like a fuse and until it heats up enough to cause an open it will simple generate heat until it opens and/or causes a fire.
That may not be the best explanation or example but the best I could think of to explain a “high resistance” short. It is why it is important to always fuse circuits to what the designing manufacture recommends. Fuses are not so much meant to protect a circuit as they are more a function of protecting a circuit or device from catching fire from an over current (amps) fault condition as it appears to be with this issue.
First, if a circuit is designed to be operated at say 5 amps then that is in the design. It has to do with not only the wire size but the insulation package used on the circuit, basically the type and thickness of insulation on a winding (coil of wire), wires, IC’s etc.
I am not sure what is involved with this particular switch but it sounds like there may be a coil involved since “high resistance” circuit is mentioned, or it could be a type of resistor.
Let me try to explain what is meant by a “high resistance” short. Theoretically, there is no such thing as a “high resistance” short. You either have a short or you don’t. Just like an open, a circuit is open or its not.
However, in the real world of electronics and electricity there is no such thing as “theoretical” but rather “real world events”.
Any electrical circuit has heat involved with it. It doesn’t matter how low the current, there is heat involved with it. All circuits have current (amps) and all circuits have resistance. When you put current (amps) through any circuit you will have heat. An electric heater is the most basic example of this as well as your ordinary incandescent light in your home. Touch the light in your home and you feel heat. This is, basically, the result of current passing through the filament inside of the light bulb.
If this particular circuit was designed to be operated at 5 amps then it had that type of insulation package associated with it, to handle the heat it would see in “normal” operation. Usually, insulation packages are designed to handle roughly 2 to 3 times the amount of heat they would see under normal operation.
A “high resistance” short, in theory, is where you could have a short at the output of the circuit but if the resistance of the circuit is high enough (due to a faulty design) the circuit never opens up, or opens up after enough heat is generated to cause a fire.
Looking at this circuit as purely a resistant circuit it may have a total of 2.4 ohms (resistance) in the circuit. If it is operated at 12vdc it would see 5 amps (current) running through it. 5 amps at 12vdc would produce 28.8 watts of heat in the circuit.
Now, if the component is faulty because it has only 2 ohms of resistance it would draw 6 amps if operated at 12vdc. This would produce 72 watts of heat, or over 2 ½ times the heat that it was originally designed to handle. Since the circuit is not protected with a 5 amps fuse it will continue to heat up and not open.
That is possible because the resistance of the circuit is acting like a fuse and until it heats up enough to cause an open it will simple generate heat until it opens and/or causes a fire.
That may not be the best explanation or example but the best I could think of to explain a “high resistance” short. It is why it is important to always fuse circuits to what the designing manufacture recommends. Fuses are not so much meant to protect a circuit as they are more a function of protecting a circuit or device from catching fire from an over current (amps) fault condition as it appears to be with this issue.
Originally posted by 01 XLT Sport
From my knowledge of electrical circuits I will try and explain what I “think” they are talking about with a high resistance short and why if the proper fuse is not used where the possibility of fire could develop.
First, if a circuit is designed to be operated at say 5 amps then that is in the design. It has to do with not only the wire size but the insulation package used on the circuit, basically the type and thickness of insulation on a winding (coil of wire), wires, IC’s etc.
I am not sure what is involved with this particular switch but it sounds like there may be a coil involved since “high resistance” circuit is mentioned, or it could be a type of resistor.
Let me try to explain what is meant by a “high resistance” short. Theoretically, there is no such thing as a “high resistance” short. You either have a short or you don’t. Just like an open, a circuit is open or its not.
However, in the real world of electronics and electricity there is no such thing as “theoretical” but rather “real world events”.
Any electrical circuit has heat involved with it. It doesn’t matter how low the current, there is heat involved with it. All circuits have current (amps) and all circuits have resistance. When you put current (amps) through any circuit you will have heat. An electric heater is the most basic example of this as well as your ordinary incandescent light in your home. Touch the light in your home and you feel heat. This is, basically, the result of current passing through the filament inside of the light bulb.
If this particular circuit was designed to be operated at 5 amps then it had that type of insulation package associated with it, to handle the heat it would see in “normal” operation. Usually, insulation packages are designed to handle roughly 2 to 3 times the amount of heat they would see under normal operation.
A “high resistance” short, in theory, is where you could have a short at the output of the circuit but if the resistance of the circuit is high enough (due to a faulty design) the circuit never opens up, or opens up after enough heat is generated to cause a fire.
Looking at this circuit as purely a resistant circuit it may have a total of 2.4 ohms (resistance) in the circuit. If it is operated at 12vdc it would see 5 amps (current) running through it. 5 amps at 12vdc would produce 28.8 watts of heat in the circuit.
Now, if the component is faulty because it has only 2 ohms of resistance it would draw 6 amps if operated at 12vdc. This would produce 72 watts of heat, or over 2 ½ times the heat that it was originally designed to handle. Since the circuit is not protected with a 5 amps fuse it will continue to heat up and not open.
That is possible because the resistance of the circuit is acting like a fuse and until it heats up enough to cause an open it will simple generate heat until it opens and/or causes a fire.
That may not be the best explanation or example but the best I could think of to explain a “high resistance” short. It is why it is important to always fuse circuits to what the designing manufacture recommends. Fuses are not so much meant to protect a circuit as they are more a function of protecting a circuit or device from catching fire from an over current (amps) fault condition as it appears to be with this issue.
From my knowledge of electrical circuits I will try and explain what I “think” they are talking about with a high resistance short and why if the proper fuse is not used where the possibility of fire could develop.
First, if a circuit is designed to be operated at say 5 amps then that is in the design. It has to do with not only the wire size but the insulation package used on the circuit, basically the type and thickness of insulation on a winding (coil of wire), wires, IC’s etc.
I am not sure what is involved with this particular switch but it sounds like there may be a coil involved since “high resistance” circuit is mentioned, or it could be a type of resistor.
Let me try to explain what is meant by a “high resistance” short. Theoretically, there is no such thing as a “high resistance” short. You either have a short or you don’t. Just like an open, a circuit is open or its not.
However, in the real world of electronics and electricity there is no such thing as “theoretical” but rather “real world events”.
Any electrical circuit has heat involved with it. It doesn’t matter how low the current, there is heat involved with it. All circuits have current (amps) and all circuits have resistance. When you put current (amps) through any circuit you will have heat. An electric heater is the most basic example of this as well as your ordinary incandescent light in your home. Touch the light in your home and you feel heat. This is, basically, the result of current passing through the filament inside of the light bulb.
If this particular circuit was designed to be operated at 5 amps then it had that type of insulation package associated with it, to handle the heat it would see in “normal” operation. Usually, insulation packages are designed to handle roughly 2 to 3 times the amount of heat they would see under normal operation.
A “high resistance” short, in theory, is where you could have a short at the output of the circuit but if the resistance of the circuit is high enough (due to a faulty design) the circuit never opens up, or opens up after enough heat is generated to cause a fire.
Looking at this circuit as purely a resistant circuit it may have a total of 2.4 ohms (resistance) in the circuit. If it is operated at 12vdc it would see 5 amps (current) running through it. 5 amps at 12vdc would produce 28.8 watts of heat in the circuit.
Now, if the component is faulty because it has only 2 ohms of resistance it would draw 6 amps if operated at 12vdc. This would produce 72 watts of heat, or over 2 ½ times the heat that it was originally designed to handle. Since the circuit is not protected with a 5 amps fuse it will continue to heat up and not open.
That is possible because the resistance of the circuit is acting like a fuse and until it heats up enough to cause an open it will simple generate heat until it opens and/or causes a fire.
That may not be the best explanation or example but the best I could think of to explain a “high resistance” short. It is why it is important to always fuse circuits to what the designing manufacture recommends. Fuses are not so much meant to protect a circuit as they are more a function of protecting a circuit or device from catching fire from an over current (amps) fault condition as it appears to be with this issue.
Bird
Last edited by Bird; Mar 7, 2005 at 06:34 PM.
Originally posted by Bird
In reading through the problem description, it states that the switch leaks and brake fluid get's up around the contacts of the switch (no coil involved here) and as the contacts are cycled and 'sparked', the fluid eventually burns enough to coat the contacts, thus creating more resistance in the overall circuit across those same contacts. As this resistance increases, it'll draw more current and create more heat until either :a) a fuse is blown or b) the switch burns. Given that they chose to put a large fuse in the circuit that will pass a lot more current that the rating of the switch, it's not surprising that the switch catches fire and is fed by a petroleum product to generate even more heat and flames = bye bye Ford vehicle.
Bird
In reading through the problem description, it states that the switch leaks and brake fluid get's up around the contacts of the switch (no coil involved here) and as the contacts are cycled and 'sparked', the fluid eventually burns enough to coat the contacts, thus creating more resistance in the overall circuit across those same contacts. As this resistance increases, it'll draw more current and create more heat until either :a) a fuse is blown or b) the switch burns. Given that they chose to put a large fuse in the circuit that will pass a lot more current that the rating of the switch, it's not surprising that the switch catches fire and is fed by a petroleum product to generate even more heat and flames = bye bye Ford vehicle.
Bird
Thanks for that information...
Quote-As Charlie Everts puts it, "What needs to be conveyed to the people is that this defective switch has been used in Ford vehicles for 10 years, from 1992 until 2002"
So does this quote mean i am safe with my 2003 f-150 or do i have the same pofsht switch in mine??????
Dan.
So does this quote mean i am safe with my 2003 f-150 or do i have the same pofsht switch in mine??????
Dan.
Originally posted by 01 XLT Sport
I agree, I didn't read anything about brake fluid in the contacts but it makes sense to me and would explain it as well.
Thanks for that information...
I agree, I didn't read anything about brake fluid in the contacts but it makes sense to me and would explain it as well.
Thanks for that information...
Bird


