Actually no.
The two batteries in series will have half the short circuit current than a single battery will have.
Brad
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 = 12.6 volts and R_int = 0.01 ohms
I_sc = 12.6v / 0.01ohms = 1260amps
2 batteries in series V_oc = 25.2volts and R_int = 0.02ohms
Then I_sc = 25.2v / 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
*edit {25.2 was 25.6 in 2 places} changed due to error. 2*12.6=25.2 not 25.6 for two in series. Also, 12.6 was 12.8 in one place due to error.