XPost: sci.electronics.basics   
   From: tgm2tothe10thpower@replacetextwithnumber.hotmail.com   
      
   DJ Delorie said something like:   
   > "Thomas G. Marshall"   
   >    
   > writes:   
   >> 1. What happens if I wire the following in parallel:   
   >>   
   >>         3V 1mA   
   >>         6V 1mA   
   >>   
   >> Does #1 above average the voltage to 4.5V?   
   >   
   > This type of circuit depends heavily on the resistance of the wiring   
   > and batteries.  What's going to happen is that a LOT of current is   
   > going to want to flow from the 6V battery to the 3V battery, in an   
   > attempt to fast-charge the 3V from the 6V until they're both the same   
   > voltage, or until something fails.   
   >   
   > Aside from burning out the batteries and/or wires (unlikely for a 1mA   
   > capacity), the actual voltage at the junction depends on the relative   
   > resistances of the two half-circuits (wiring, battery's internal   
   > resistance, contact resistance, etc).  Basically, it's a voltage   
   > dividier, as if you had two ideal voltage sources each with a separate   
   > series resistor.   
   >   
   > Guessing (wildly) from the descriptions, the 6V source would have   
   > twice the internal resistance of the 3V source (else it would have a   
   > 2mA max), so a 2:1 divider across the loads results in a 4v node   
   > voltage.   
   >   
   >> Does #2 above average the current to 1.5mA?   
   >   
   > No, current ratings are maximums, not absolutes.  If those are the   
   > real maximums, the result is a 6V 1mA source.  Actual current draw   
   > depends on the load, not the source.   
      
   Ah...ok.  Thanks!   
      
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