At 10:45 AM 3/8/99 -0500, you wrote:
>If you connect a capacitor in parallel with any battery
>with high internal impedance, it will delivery a higher
>peak of power at the initial connection since the cap
>has an impedance lower than the battery.

|Won't you actually achieve max. power transfer to the cap when the product
|of the voltage across it and the current thru it is maximum? This won't
|occur when you first hook it up because the voltage across the cap is
|initially zero,so the power transfered is initially zero.

The initial transfer of energy to the cap will not be terribly
efficient, and if the cap is substantially discharged its repl-
enishment won't be terribly efficient either.

Suppose, however, that you have a device powered by a "9 volt"
battery which needs to drive a 100-ohm solenoid for 100ms with
at least 60mA.  If you have a 4,700uF cap in parallel with the
battery it will be able to supply that demand while dropping
less than two volts even if the battery is not in good shape and
has an internal resistance of 100 ohms (in which case connecting
a 100ohm load without the cap would cause the voltage to drop by
half, failing to meet the 60mA requirement).

Note that recharging the cap after each pulse will not be 100%
efficient (since the cap's starting voltage will be around two
volts less than the battery voltage) but it will be much more
efficient than would be the battery's efforts at driving the
solenoid directly.