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Re: Capacitance problem



At 09:55 AM 3/30/97, Donald E. Simanek wrote:

... Why does the result (half the energy goes
elsewhere) *not* depend on these particular details? If the only loss were
radiative, you'd get 1/2 for the energy loss. If the only loss were
resistive, you'd get 1/2. If it is a combination of both, you'd get 1/2.
I chose identical capacitors, which is one reason you get 1/2, but that
was only for convenience of discussion. The principle (if there is one)
could be applied to any pair of unequal capacitors, and the resulting
fractional energy loss would be a different value.
...
Dr. Donald E. Simanek

I see the virtue of your complaint - and I scrabble for a handhold on
the correct ledge. Here is one that may just be involved.

When the electrical engineer wants to transfer energy at the maximal rate
he ensures that the source impedance matches the terminal impedance
and accepts that in this case, half the available energy will be dissipated
in the transfer mechanism. He even assigns a name to this principle.

When the electrical engineer wishes to transfer energy efficiently,
he ensures that the load represents a much higher impedance than the source
and interposes fuses to ensure this will remain the case.
An example of the maximum power theorem is the automobile starter: does
the battery voltage not sag to 6 volts?

Sincerely
brian whatcott <inet@intellisys.net>
Altus OK