SageBrush said:I'm obviously being dense about this, thanks for the patience.lorenfb said:SageBrush said:What you say makes sense: if baseline heat generation in the battery is say 2% of the power draw then a 25% increase in resistance would presumably (?) increase heat generation up to 2*1.25 = 2.5% of the power draw. No big deal.
above is battery power loss - from an increase in battery resistance only
SageBrush said:But then I start from
P = I*I*R and I conclude that an increase in R to R1 would lead to a drop in I = sqrt (R/R1)
Then e.g., an increase in R of 25% would lead to a drop in I to sqrt(0.75) = 86.6%
above is total power loss - from a total load resistance increase
You're confusing the two different situations.
If I just consider the battery in isolation, is my use of I*I*R correct ?
That is, does an increase of R to R1 reduce the current to sqrt(R/R1) ?
Yes, for a short circuit test of the battery, i.e. you connect the two terminals of the battery together (no load) and the battery resistance
increased. Typically the resistance would decrease as it heated. Or you used to different external resistors and did that test.