how to determine amount of load current to test internal resistance?

[jingpekto]

hi Guys! i have this project to determine the internal resistance of the 12v deep cycle battery, i attached the datasheet.
one of my problem is how to determine the effective amount of current to be drawn by the resistor load where i can already determine its internal resistance?

currently i'm stating that the current to be drawn by the resistor load is 20A, so 12V/20a = 0.6 ohms, resistor rated power should be Power=(20A*20A)(0.6ohms) = 240W !! i think this is too big??

i am thinking if am only putting the load for about 2 to 5 seconds to measure internal resistance,  should i really need to get a 240W rated resistor load or do i only need
smaller wattage of resistor load because i'm only putting the resistor load for only 2 to 5 seconds...  and is 20 amperes of load current correct to effectively determine the internal resistance?

is there a minimum or maximum current to be drawn from the load if your determining the internal resistance of a 12v battery?


hope you give some tips and advise guys! thanks in advance!!

regards,
pekto

[Siwastaja]

DC resistance, especially for such a non-ideal battery as lead-acid, is not a constant, it's a function of (too) many parameters.

The best you can get is to measure the DC resistance near the operating condition that is of interest to you. I.e., what are you going to do with the result?

Usually, step test with two test currents, neither of which is zero, is used. For example, 10A and 50A. R = (U2-U1)/(I2-I1)

Give it a few seconds to be meaningful for real-life seconds-long peaks.

Temperature, state-of-charge and condition of the battery affect the resistance a lot. If you are just testing to analyze the condition, then keep state-of-charge and temperature constant.

[Kilrah]

The average car battery tester will draw about 100A on the battery during a few seconds and bring a big fat piece of wire to red hot in that time.