My New Holland TC- 29 won't start...clicks.only

[brin]

Bob, good for you Did you get a Duralast Gold battery at Autozone? I have them in my Dodge Cummins diesel pickup and they are top notch batteries. Other Duralast batteries are so-so, but the Gold really is a premium battery.

Yep Jim, I got the Duralast Gold battery...they sure are proud of those...you see the price on batteries now...geeez, I haven't bought a battery in 4 yrs. so I was surprised they were so high...I appreciate all the help and advice...looks like I dodged a bullet this time...Whew !

Bob I am glad you got it going. Now you have a multimeter, and a good working tractor, you could make some of those measurements and see how things work when all is well, to better understand how things don't work when all is not well. If you can understand the relationship of Voltage, Current and Resistance (Ohms law) you will always be able to troubleshoot these kinds of problems. I know this will probably make your eyes glaze over, but here it is. E=IR E is "Electromotive force" or Voltage, (in honor of Mr. Volta) I is Current measured in Amperes (in honor of Mr. Ampere) or amps if you prefer and R is resistance measured in Ohms (in honor of Mr. Ohm). You can rearrange the equation using 9th grade algebra any way you want to solve the equation for the unknown variable. Here is an example if the E (or voltage) is known to be 10 volts and you apply it across a 10 ohm resistor, what will the I (or current measured in Amps) be? Just divide thru the equation E=IR thru by R on both sides of the equation, so the R on top and the R on the bottom cancel each other out, and you have on the left E/R=I so so lets plug in the values we know for the E and the R thus: 10/10=I so divide 10 by 10 and that equals 1 so the current is 1 ampere. Now you are thinking what in heck does this have to do with starting my tractor Well as an example say for instance you looked at the huge wires attached to the battery and you measure 12 volts on the battery end and 2 volts on the other end when you turn on a light whose current draw is lets say 1 amps and you know that such a large diameter wire should have way less than 1 ohm of resistance,(you can look this up in a chart) but what is the actual resistance of this wire? Ok lets plug in the values we have a 10 volt voltage drop so E is 10 and the known current draw of the lamp is 1 amp (we can look this up or estimate it) so the I is 1 so now our equation looks like this 10=1R divide through both sides by 1 to get the unknown (the R) on one side of the equation and you have 10=R so the resistance is 10 Ohms.. WAY WAY too much resistance for a starter wire that should be near 0 ohms. So when we try to pull any significant current thru this wire we have a serious voltage drop, in our test case here we have a 10 volt drop. Now if we take the load of the lamp off of the circuit and measure with our digital voltmeter we now measure 12 volts on both ends of the wire.. what the he77? why does that happen? Well the voltage drop appears to be near 0 because the current draw is near 0, (your digital voltmeter draws very very little current to make a measurement) So lets plug in the values to our E=IR equation.. 0=0xR and we know the R from our previous measurement that it is 10 ohms. so 0=0x10 so 0=0 and the equation is balanced. Even a very high resistance wire will measure the same voltage on each end as long as you do not try to draw any real current thru it. Most of you have long ago given up listening to me rattle on or never read this far, or fell asleep so I will stop now. But if you did even understand half of what I said, you are on your way to understanding Ohms Law.

James K0UA

Thanks for the tutorial James....I did not fall asleep, in fact I read it twice and will refer back to it....The problem is if I don't use something like this formula or a multimeter often, I will forget in short order...all things are not like riding a bike....that you never forget....Thank you very much for the time you took to explain that to me...

 

[k0ua]

Yep Jim, I got the Duralast Gold battery...they sure are proud of those...you see the price on batteries now...geeez, I haven't bought a battery in 4 yrs. so I was surprised they were so high...I appreciate all the help and advice...looks like I dodged a bullet this time...Whew !



Thanks for the tutorial James....I did not fall asleep, in fact I read it twice and will refer back to it....The problem is if I don't use something like this formula or a multimeter often, I will forget in short order...all things are not like riding a bike....that you never forget....Thank you very much for the time you took to explain that to me...

Brin I a glad you got some good out of my ramblings. Learning about electronics and the principals behind it can actually be fun. If you like solving mysteries/problems. Which is exactly what a tractor that wont start is, A mystery. There is a reason it won't start, all we have to do is find it. For instance your "bad" battery most likely had a very high internal resistance. A lot of battery's fail in that mode. Not all of course. But a battery with a high internal resistance may measure near normal, especially if just taken off the charger. But with the high internal resistance in the circuit it cannot deliver much current, even though the voltage may be 12.6 to start, as soon as you place a load on it to draw much current the voltage on the terminals drops excessively because the high internal resistance limits the current and the voltage drops. By the next step is to start thinking about the power equation.. Oh boy here it comes. P=EI or stated, the P stands for Power measured in Watts (in honor of Mr. James Watt) equals the E (electromotive force or more commonly called Voltage) times the I or the current measured in Amps. so lets say you have a 12 volt battery (you can measure across it with a voltmeter) and you place a load (lets say a lamp) across it and that load draws 1 amp of current, (you can measure that with your multimeter on the Amps setting, you will have to put the meter leads in series with the load to measure current). So you have 12 volts and 1 Amps. How much power is the lamp consuming? P=EI.. so P=12x1 so you are consuming 12 Watts of Power!. Now keep in mind that you are not putting out 12 watts of light, because incandescent lamps are not very efficient at all! So where is the rest of the Watts going? It is used to produce heat energy, the very thing an incandescent lamp does best. Far more of the Watts consumed go up in heat than are radiated in the form of light. So what is an incandescent lamp? Just a resistor that soaks up Watts and produces a lot of heat and a little light. There is more, but you are probably bored with this game by now. Hope you had some fun, I did.

James K0UA