Hello,because the discussion on reverse voltage with batteries is going on, I will contribute with drawings. Most of the problems which arose here are because its so difficult to explain such things with words only. There is no new argument here, only drawings.
- diagram with old battery (B), lamp (L) current (i). Each battery (also each generator) has an inner resistance (R) to the current. In new batteries R might be negligible. R is drawn separated from the battery cells. The current generates a voltage drop across the resistance. Thus in my special layout, the initial 9 Volts drop to an effective battery voltage of 5 Volt.
- The 2nd diagram consists of 2 new batteries (R of these is about zero) and an old one (with appreciable inner resistance) in series. You easily see, if R is high enough and the current is high enough, the voltage drop UR could be higher then the cell voltage of 9 Volt in this battery. The total voltage of this battery is reversed. The battery will be charged.
Hope this helps, Günter Marshall Dudley wrote:
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David Bearrow wrote: snip ...etc.....
<<inline: circuit1.jpg>>
