By the math:
V = battery terminal voltage
V_oc = open circuit battery voltage
R_int = internal resistance of the battery
I = battery current
Accepted battery model equation
V = V_oc - (I * R_int)
At short circuit, by definition, V = 0
Then V_oc -(I * R_int) = 0
or
V_oc = I * R_int
or
I = V_oc / R_int = I_sc = short circuit current
2 batteries in series, total voltage = 2 * V where V is the voltage of one battery. I is battery current which is same for both batteries. Equation is
V_total = 2 * V = 2 * (V_oc - (I * R_int))
Set V_total = 0 for short circuit, then
0 = 2 * (V_oc - (I * R_int))
0 = V_oc - (I * R_int)
or
V_oc = I * R_int
so
I = V_oc / R_int = I_sc
or you could multiply the terms on the right side of V_total equation by 2 giving
V_total = 2 * V_oc - (I *(2 * R_int))
Where V_oc & R_int are values for a single battery and I is found by setting V_total to zero. Then
2 * V_oc - (I *(2 * R_int)) = 0
or
2 * V_oc = I *(2 * R_int)
or
V_oc = I * R_int
or
I = V_oc / R_int = I_sc
So the short circuit battery current is equal to the open circuit battery voltage divided by the battery internal resistance. This is true for one battery, or two batteries in series, or any number of batteries in series.
Example:
Car battery. V_oc = 12.8 volts and R_int = 0.01 ohms
I_sc = 12.6v / 0.01ohms = 1260amps
2 batteries in series V_oc = 25.6volts and R_int = 0.02ohms
Then I_sc = 25.6v / 0.02ohms = 1260amps.
Elements in series add. Voltage adds. Resistance adds. So the division, V / R remains the same.
I = V / R = 2V / 2R = 3V / 3R and so on. Adding batteries in series doesn't change the short circuit current value.
bi
And yet fails basic ohms law. In the circuits pictured below, i have allowed 0.1V across the short, so as easy calculations can be made. I remember way back in one of the TPU threads, it was mentioned that a series of 9 volt batteries could have powered the lightbulbs. Another very well versed member hear determined that the series resistance of all those cells in series could not even come close to providing enough current to power a 100 watt bulb. And some where on this forum, there is a thread where i built a very large magnetizer. I had 24 x 2V, 1300 amp batteries powering this device, and although each single battery could deliver the 1300 amps individually, the max current i could get out of the 24 in series was around 300 amps, and the loss was not in the cables. Once again, another well versed member said the loss was due to the now series connected internal resistance of each battery. Some times things look great on paper, in ideal conditions, but in reality, it just doesn't work out that way. But we can agree to disagree  Brad
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Topic: Where i'm at 1+1=3 (Read 14537 times)
