Dreaded #2 Fuse blows immediatly in TC40D

[PaulInAC]

My experience also supports a weak battery blowing that fuse.

When the voltage drops, the amperage must go up.

Wattage is Voltage X Current. The wattage drawn will remain the same, so if the V drops the C must increase.

I was unable to locate a blade CB for it, so I had to stay with fuses.

But as long as the battery is in good shape, the fuses don't blow.

-paul

 

[jinman]

When the voltage drops, the amperage must go up.

Wattage is Voltage X Current. The wattage drawn will remain the same, so if the V drops the C must increase.

-paul

That's not Ohm's Law, so I guess we can call it Paul's Law? The only way for current (amps) to increase for the same or less voltage is for resistance to be less. Perhaps there is something going on with inductive reactance and/or reverse voltage spikes, but trying to use Ohm's Law to describe a blown fuse due to increased amps with reduced voltage is going to be a tough sell to those of us with lots of electrical/electronic experience.

 

[RickB]

That's not Ohm's Law, so I guess we can call it Paul's Law? The only way for current (amps) to increase for the same or less voltage is for resistance to be less. Perhaps there is something going on with inductive reactance and/or reverse voltage spikes, but trying to use Ohm's Law to describe a blown fuse due to increased amps with reduced voltage is going to be a tough sell to those of us with lots of electrical/electronic experience.

Help me out, Jim. I've never understood that line of thinking, and have read it here more than once. I figure that a given load (lamps, motor, etc) draws say 126 watts, and if the available voltage drops from say 12.6 to 10.4 due to a discharged battery or poor connections, then why would the current draw from that given load not increase from 10 amps to 12.1?

I'm just a mechanic, not an electrical engineer so use plain English.

 

[barbqranch]

The reason the current doesn't generally go up as the volt goes down is the wattage is not constant, the resistance is (approximately). As an example, if you hooked a 12 volt lamp to a 6 volt battery, you wouldn't get the same brightness with twice the current, you would get 1/4 the power and brightness (approximately, because an incandescent lamp doesn't have constant R with temperature, nor is the brightness halve when the power halves). But close enough for this discussion.

However, in a motor, rules are different. The power and current are dependent on the amount of work being done. With a lower voltage, if it can spin, it will take more current to get the same mechanical output. At some point of lower voltage, the extra current will cause it to overheat. Hot enough, no spin and no current.

 

[jinman]

Help me out, Jim. I've never understood that line of thinking, and have read it here more than once. I figure that a given load (lamps, motor, etc) draws say 126 watts, and if the available voltage drops from say 12.6 to 10.4 due to a discharged battery or poor connections, then why would the current draw from that given load not increase from 10 amps to 12.1?

The simplest way I can explain it is to say a 50 hp engine is only 50 hp at a given rpm. If you lower the rpm, it is no longer 50 hp. You can add fuel to bring it back to its rated rpm and hp, but the only time it is 50 hp is when it meets the specified conditions.

When specs are given for watts drawn, it is at a certain set of conditions. There is no magic. It is wrong to say that if voltage drops, current must go up. It is correct to say that if voltage drops, then resistance must also drop if current is to be maintained.

When you talk about our tractors and relays, solenoids, and motors, you have inductive loads to consider. Resistance in an inductive load varies over time. The inductive load in a motor is dynamic based on the motor's operation (speed and load).

The problem with the fuse #2 problem is that it has nothing to do with any high current loads. The darn fuel cutoff solenoid is it's highest single load and it only draws about 1 amp at 12 VDC. What I know is that the fuse seems to blow more often when the battery voltage is low. I also know that it can be fixed with a self-resetting CB. If it were a simple case of Ohms Law, the CB would keep popping over and over. That doesn't happen. I don't know why, but I won't try to explain it by creating a new Ohm's Law either.

 

[Caliche]

Jim you are scaring me man.

Ohms law in a DC circuit is; volts = resistance X current

If voltage goes down and resistance stays the same, current must go down.

When you talk power, watts = volts X amps

When volts go down current and watts go down. The bulb gets dim.....

 

[jinman]

Jim you are scaring me man.

Ohms law in a DC circuit is; volts = resistance X current

If voltage goes down and resistance stays the same, current must go down.

When you talk power, watts = volts X amps

When volts go down current and watts go down. The bulb gets dim.....

I don't know if I'm scaring you because you agree or if you are scared because you misread what I said.:confused2: I agree completely with what you posted and so does barbqranch in his well-stated post. I just said if voltage goes down, resistance will also have to go down to maintain current. It doesn't matter what the rated watts of the load is because those rated watts are for a fixed set of values. If you change one of the values, you change the watts.

All this theory is really enlightening, but what "scares" me the most about this Fuse #2 problem is that nothing in this circuit is supplying the start circuits. It just supplies relay and switch logic power to the seat switch, PTO switch, neutral switch, and cruise control latching in addition to the fuel solenoid. I have never figured out what could suddenly increase the load enough to cause the problem. Perhaps it's a reverse spike from the fuel cutoff solenoid. There is no bypass diode in that circuit and it could conceivably cause a spike, but it would also cause a spike during normal operation. To quote a famous TV commercial, " Wazzup!?":laughing:

 

[Caliche]

Quote:
Originally Posted by PaulInAC
When the voltage drops, the amperage must go up.

Wattage is Voltage X Current. The wattage drawn will remain the same, so if the V drops the C must increase.

-paul

That's not Ohm's Law, so I guess we can call it Paul's Law? The only way for current (amps) to increase for the same or less voltage is for resistance to be less. Perhaps there is something going on with inductive reactance and/or reverse voltage spikes, but trying to use Ohm's Law to describe a blown fuse due to increased amps with reduced voltage is going to be a tough sell to those of us with lots of electrical/electronic experience.
__________________
Jim

I don't know if I'm scaring you because you agree or if you are scared because you misread what I said. I agree completely with what you posted and so does barbqranch in his well-stated post. I just said if voltage goes down, resistance will also have to go down to maintain current. It doesn't matter what the rated watts of the load is because those rated watts are for a fixed set of values. If you change one of the values, you change the watts.


Really Jim, not reading?
Paul just had a problem appling ohm's law. If you had a current regulated power supply Paul would be right. But a battery is not.

 

[Hughman]

One other thing to check... When my TC33 starts blowing fuses a lot, it helps to blow all the dust out of the wire (bullet) connectors under the operator platform (safety circuits for the seat/shifter/etc). I guess dust and dirt seems to add resistance to the starting circuit.

 

[jinman]

All this theory is really enlightening, but what "scares" me the most about this Fuse #2 problem is that nothing in this circuit is supplying the start circuits. It just supplies relay and switch logic power to the seat switch, PTO switch, neutral switch, and cruise control latching in addition to the fuel solenoid.

You know the sure sign of madness is you start talking to yourself, but that's precisely what I am doing.:confused2::laughing:

I have to eat some self-induced crow for the statement above in the quote. As often happens, another TBN member sent me a PM because he is having trouble with the start circuits. As I was describing operation of the circuit, and jumping from page-to-page in the schematics, it suddenly hit me that I had been overlooking one significant component that is powered by fuse #2. It's the starter solenoid!!! Wow! Power from fuse #2 goes through the Neutral Sensing Switch on the transmission and then is applied to the Neutral Start Relay. When the relay is energized by the Ignition switch going to Start, power from fuse #2 goes through contacts in the relay and straight to the starter solenoid coil. I had misread the schematic and thought power from the main fuse powers the solenoid, but that is not the case. It is power from fuse #2 that goes to the starter solenoid and that is the source of the surge current that blows fuse #2. Mystery solved! Fuse #2 power does NOT run the starter, that power comes directly from the battery without any fuse, but fuse #2 DOES supply the power that energizes the starter solenoid.