is this voltage drop calculation correct?

[Scph9002]

pack is 4s10p 13.2v
individual cells have 0.25milli ohms

Therefore internal pack resistance disregarding bus bars etc is (0.25/10) x 4 = 0.1 milli ohms? Right?

Then if I draw 100 amps from the pack the voltage drop is calculated like this: 13.2v x 0.0001ohms x 100 amps = 0.132 voltage drop, voltage at terminals would read 13.068v while i draw 100a?

 

[Deleted member 23531]

4s would give:
4 * 0.25mOhm = 1mOhm

10 of these in parallel would give:
1mOhm / 10 = 0.1mOhm

Then if you draw 100 amps (assuming each cell can tolerate that, and the cell resistance is still valid), the voltage drop would be:
100A * 0.1mOhm = 10mV

So the pack voltage under 100A load would be:
Voc - 10mV

I assume your open circuit voltage (Voc) is 13.2V but it's sort of irrelevant to this calculation.

 

[Scph9002]

4s would give:
4 * 0.25mOhm = 1mOhm

10 of these in parallel would give:
1mOhm / 10 = 0.1mOhm

Then if you draw 100 amps (assuming each cell can tolerate that, and the cell resistance is still valid), the voltage drop would be:
100A * 0.1mOhm = 10mV

So the pack voltage under 100A load would be:
Voc - 10mV

I assume your open circuit voltage (Voc) is 13.2V but it's sort of irrelevant to this calculation.

ah right. thanks alot makes more sense now =D

 

[Deleted member 23531]

No problem!