Keekleberrys
NAXJA Forum User
- Location
- Issaquah
3 efans? Does it not have a mechanical fan?
Oh and just to put this out there, corrosion and high resistence in a circuit normally will cause lower amperage in a circuit and will NOT cause a fuse to blow.
Sounds good but, the facts are the fuses burn through, watts (heat) not amps is what melts them. Instead of say a twenty amp fuse it would be more accurate to call it a 260 watt fuse. Copper heat sinks, the heat travels the entire length of a wire or a bus and radiates some, but the heat can be accumulative and may build over a period of seconds or even minutes before it reaches the threshold needed to melt the fuse. The amperage you are drawing, plus the stored heat, can cause a meltdown on a fuse for a circuit drawing less than max fused amperage. Amperes X Volts equals watts, watts can be either heat or horsepower. The fuse is little more than the weak link and hopefully has the lowest melting point in the system. The heat stored in the wire and what the circuit draw together, are accumulative.
Though you are correct on one point, the 5.5 +/- amp draw of the two O2 sensor heaters together and the heat stored or generated in the wiring or the bus, make burning through a twenty amp (260 watt) fuse unlikely. Though I have seen a whole lot of melted wire and connectors over the years, copper melts at around 2000 F. My guess is the fuse melts at a much lower temperature than that.
I see where you are coming from, however I have yet to see such a thing occur in an automotive environment. Yes, I have seen many melted connectors and wires but never a blown fuse because of them unless they touch ground. I have seen fuses melt the plastic that holds them together before the fuse will actually blow. So, I'm not saying you are wrong, but that is more of a theoretical situation rather than a real world situation.
EDIT: Also, according to ohms law, high resistence in a circuit lowers both the voltage and the amperage, thus drawing fewer watts.
The flaw in your thinking is when you say draws watts, watts (especially in a heater circuit) is heat and it is accumulative. The wire radiates, if the temperature rises faster than it can radiate (or sink) the heat away, the heat continues to build.
It depends on how close the heat source is to the fuse and how much heat is radiated and leeched down the wire. If the wire absorbs enough heat it can get hot enough to burn through a fuse or trip a circuit breaker. In my experience the heat source needs to be within a couple of feet in an automotive system, household systems it can be a lot farther away.
I've seen it numerous times, a classic example is the dash illumination circuit for renix and most HO's. The rheostat (in the headlight switch), in effect a heater or resistor, has a partial melt down, the rheostat is powered by the Tail light fuse, then the power goes back down to the fuse block from the rheostat, through another 7 amp fuse (I think) then back up to the dash lights. The heat generated by that rheostat will blow the 7 amp fuse even though it is after the heat source. It may or may not blow the 15 amp tail light fuse.
Numerous times I've seen fuses burn through because the fuse holder was loose, the heat generated would eventually cause the fuse to fail. Even in your house, a loose connection between the fuse and say a light fixture can trip the fuse, the heat eventually finds it's way back the to fuse or circuit breaker, hopefully before the loose connection starts a fire.
Pretty common failure, really it does happen. If the heat builds faster than the wire can radiate or leech it away, the fuse will fail, it is accumulative, the heat builds, it can take seconds or even an hour.
I started out on auto electric sometime in the early sixties and moved on to bigger stuff. I learned on M-48, through M-60 and then on to M-1 main battle tanks, enough wire in one of those for a fleet of XJ's.
I see where you are coming from, however I have yet to see such a thing occur in an automotive environment. Yes, I have seen many melted connectors and wires but never a blown fuse because of them unless they touch ground. I have seen fuses melt the plastic that holds them together before the fuse will actually blow. So, I'm not saying you are wrong, but that is more of a theoretical situation rather than a real world situation.
EDIT: Also, according to ohms law, high resistence in a circuit lowers both the voltage and the amperage, thus drawing fewer watts.
Better check that again, high resistance = high voltage.
Better check that again, high resistance = high voltage.
I know what the equation says, but how are you going to get more than 14 volts from an automotive charging system? Amperage goes down as resistence goes up with the voltage being unchanged.
I know what the equation says, but how are you going to get more than 14 volts from an automotive charging system? Amperage goes down as resistence goes up with the voltage being unchanged.
I know what the equation says, but how are you going to get more than 14 volts from an automotive charging system? Amperage goes down as resistence goes up with the voltage being unchanged.
You won't get more that 14 volts. A good electrical connection should have 0- .1 volts across it. A bad connection might be .1-.5 volts or more. An open circuit obviously can have no more tha source voltage. It's called a voltage drop test. When there's more or unintentional resistance in a circuit, the voltage across the component or connection increases. You made it sound like volt drop decreases across a bad connection.
You need to think of the circuit as a voltage divider when there is unintentional resistance introduced in series of the circuit. For example and easy math
10v volts across a 100 watt bulb draws 10 amps. The bulb resistance is 1 ohm and wattage across a good connection is 0 volts. The wattage across the connection is 0 since 0 volts x 10 amps=0watts.
If the connection were bad, say .111 ohms there would be 1 volt across the connection and only 9 volts across the bulb. The amp draw would be 9 amps calculated, 10 volts/1.11ohms=9 amps. Since the connection has 1 volt and 9 amps the wattage at the connection is 9 watts and the connector heats up. Since the bulb has 9v and 9 A it is 81 watts and dim.