DJ Delorie said something like:
"Thomas G. Marshall"
. com
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!