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 resistanceSageBrush wrote:I'm obviously being dense about this, thanks for the patience.lorenfb wrote:above is battery power loss - from an increase in battery resistance onlySageBrush wrote:
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 total power loss - from a total load resistance increaseSageBrush wrote: 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%
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) ?
increased. Typically the resistance would decrease as it heated. Or you used to different external resistors and did that test.