Originally posted by BillO Okay,
[/INDENT]And below is a picture, taken with my K-3, of a little remembrance they gave me after leaving the University of Toronto in honor of the 4 years I spent in the bowels of the McLennan physics labs earning my double major degree in physics and mathematics. So, yeah, not every one on the internet can be assumed to be full of sh!t.
But mines 3 years older than your so I win !!
And oh look mines in Electronics so pehaps explains you complete failure to understand , if only it had been English
From your linked article (did you read it)
The impact of electronic and ionic resistance can be observed using a dual pulse test. This test involves placing a battery on a low background
drain allowing it to first stabilize and then pulsing it with a heavier load for approximately 100 milliseconds.
Using “Ohms Law”, the total effective resistance is subsequently calculated by dividing the change in voltage by the change in current.
Hey guess what 'change in voltage' means .......yep thats right a voltage drop exactly as I said directly realted to the imternal resistance of the battery.
As you article clearly states
Internal resistance can be calculated based on the voltage drop of the battery under a known load.
And finally to answer your highlighted test in your rude and inaccruate reply putting into context your quoted part.
There are two basic components that impact the internal resistance of a battery; they are electronic resistance and ionic resistance. The electronic resistance plus the ionic resistance
will be referred to as the total effective resistance
The voltage drop of a battery under load is a function of
total effective resistance and current drain rate.
Now to put it all in to simple English for you
the Voltage drop of a battery is a function of the current drain and internal resistance (ionic+electrical) as I states originally and as backed up by you linked article.!!
As you say not everyone on the internet is full of sh!t but you sir protest to loudly !
I would sugest in future you read your posted links before attempting to use them to ridicule others, when you clearly did not understand a word they said nor apreciated they were vindicating my original statement.
If you read the section on flash amps you will understand why I related it to mAh in the context here
Identical batteries rated @2000mAh the other 1860mAh
I said
The mAh rating (if true) will reflect how much the battery voltage sags under load.
Read more at:
https://www.pentaxforums.com/forums/172-pentax-k-3/253707-k-3-failure-couldnt...#ixzz34LRyVTxH
note the 'if true' this was to show both mAh rating would be comparable (i.e same load rating)
therfore as v=ir and the rating implies R constant for mAh rating (see flash amps) then V must drop for a lower mAh rating and the difference in mAh rating is directly proportional to the internal resistance of the battery.
Which makes this
Originally posted by BillO With all due respect, this is not true. The actual voltage under load has absolutely nothing to do with the battery's amp-hour rating. Nothing at all.
complete and utter twaddle
.