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current surge stalling engine

crutschow

May 7, 2021
834
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i have this coming from ebay
20k Wirewound Rheostat Variable Resistor
20,000Ω is way too high a resistance.
That will limit the current to a fraction of an amp.
Do you not know Ohm's law, which states that Current = Volts divided by Resistance?

This 50W resistor selected with 0.5Ω should work.
It should limit the maximum current to <10A with the battery attached (less than 5V across the resistance).
 

davit

Aug 16, 2017
39
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Aug 16, 2017
Messages
39
Why won't you just give everyone a break and show what you have there and a diagram of what is connected where.
While you are at it a link to the specs of the alternator.
Thanks.

@Bluejets sorry for any confusion, thought i had gave enough info uploading pics now, you can see in the pick i am using the isolator to connect alt to battery once the revs are very high, appologies for lack of alternator info i bought it new last year from a firm that has since stopped trading ( doh bang goes my warranty ) the little sw in the pic is the energize sw for the alternator,24vdc max 55amp
oppss - pics to large links to dropbox instead thanks
https://www.dropbox.com/s/c0pl4axinglr1e4/DSCN3145.JPG?dl=0
https://www.dropbox.com/s/ujroywaei0deofa/DSCN3142.JPG?dl=0
 
Last edited:

davit

Aug 16, 2017
39
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Aug 16, 2017
Messages
39
20,000Ω is way too high a resistance.
That will limit the current to a fraction of an amp.
Do you not know Ohm's law, which states that Current = Volts divided by Resistance?

This 50W resistor selected with 0.5Ω should work.
It should limit the maximum current to <10A with the battery attached (less than 5V across the resistance).
sorry dont know any laws still learning which is hard at my age and part dementia, really appreciate the link for the resistor though will buy that i ,i just assumed the adjustable one i bought would adjust down to that low amp reading ,
my lack of knowledge is shinning through , we all started somewhere LOL
 

crutschow

May 7, 2021
834
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May 7, 2021
Messages
834
,i just assumed the adjustable one i bought would adjust down to that low amp reading ,
Even if that were a 10W pot (which is not stated). the maximum current the internal wiring could carry would be 22mA, many orders of magnitude below what you need.

If you learn Ohm's law, plus the equations for power (P = V x I, P = V² / R, and P = I² x R) where P is power, R is the resistance, V is voltage across the resistance and I is current through the resistance, then you will be well along to understanding basic circuit operation.
 

Bluejets

Oct 5, 2014
6,862
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Oct 5, 2014
Messages
6,862
What is the heatsinked device on the lower right connected directly across the battery.....?

Seems to be heavy leads connected to some kind of isolator but nothing on the alternator output pins.??

How about a sketch of the wiring you have done..??
Half of the info will get only half an answer at best.

Doesn't look right to me.
 

davit

Aug 16, 2017
39
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Aug 16, 2017
Messages
39
What is the heatsinked device on the lower right connected directly across the battery.....?

Seems to be heavy leads connected to some kind of isolator but nothing on the alternator output pins.??

How about a sketch of the wiring you have done..??
Half of the info will get only half an answer at best.

Doesn't look right to me.

the heatsinked device is a 24v - 12v reducer which works well ( recomened from an earlier post )
the isolater is between the batt and alternator, the electrics all work well its just the current surge can sometimes stall the engine
would be nice if it could be stepped, feel i am going round in circles now ,so will close this post now,
but i would like to thank you all for your reply posts they have been very helpfull and informative , what a wonderfull forum thank you admin,,
 

Bluejets

Oct 5, 2014
6,862
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Oct 5, 2014
Messages
6,862
As I said previously, send a circuit diagram of what you have there.
Looks to me like you have nothing connected to the alternator output studs.
Also as asked for previously, a link to the alternator specs.
Lack of info will result in round and round going nowhere results.
 

davit

Aug 16, 2017
39
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Aug 16, 2017
Messages
39
20,000Ω is way too high a resistance.
That will limit the current to a fraction of an amp.
Do you not know Ohm's law, which states that Current = Volts divided by Resistance?

This 50W resistor selected with 0.5Ω should work.
It should limit the maximum current to <10A with the battery attached (less than 5V across the resistance).
hi m8 (crutschow)
the resistor link you sent ( above) worked well thanks engine did not stall, it reduced charge to 5amp with the details here could you post another link to another resistor please ( still cannot figure ohm,s law )
to give me 10amp, then i am hoping to find a 3 position switch that i could then set up with 3 settings 5amp , 10 amp , full load so i could increase the load in stages to help the engine run, many thanks
 

crutschow

May 7, 2021
834
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May 7, 2021
Messages
834
could you post another link to another resistor please
The link I posted has 0.1Ω resistor.
If 0.5Ω gives 5A then 0.2Ω should give about 10A.
You could use a couple of the 0.1Ω resistors in series to get that.
still cannot figure ohm,s law
Its quite simple.
The voltage drop across a resistor is equal to the resistor current times the resistance, or the current through a resistor equals the voltage drop across it divided by its resistance.

So if 0.5Ω give a 5A charge rate, that means the voltage across the resistor (difference between the generator output and the battery voltage) was 5A x 0.5Ω = 2.5V.
So for double the current, you would need 0.25Ω (0.2Ω should be close enough).
 

davit

Aug 16, 2017
39
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Aug 16, 2017
Messages
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The link I posted has 0.1Ω resistor.
If 0.5Ω gives 5A then 0.2Ω should give about 10A.
You could use a couple of the 0.1Ω resistors in series to get that.
Its quite simple.
The voltage drop across a resistor is equal to the resistor current times the resistance, or the current through a resistor equals the voltage drop across it divided by its resistance.

So if 0.5Ω give a 5A charge rate, that means the voltage across the resistor (difference between the generator output and the battery voltage) was 5A x 0.5Ω = 2.5V.
So for double the current, you would need 0.25Ω (0.2Ω should be close enough).
thanks been trying to use this calculator https://ohmslawcalculator.com/ohms-law-calculator
 
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