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Re: Solving the DC coil mystery
Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>
Tesla list wrote:
> Original poster: "Bert Hickman" <bert.hickman-at-aquila-dot-net>
> The proof follows. Assume a constant voltage source (V) and an initially
> uncharged capacitor being charged through resistor R by closing the switch
> at time T=0 and then leaving it closed forever (since it takes forever to
> fully charge the cap).
>...
> Er = (RC/2)*(V^2)/R) = 0.5*C*(V^2)
>
> This tells us that the total energy dissipated in the resistor is the same
> as the total energy that's eventually stored in the capacitor, giving us a
> 50% charging efficiency. And, since R cancels out, the result is
> independent of the value of R! (Not at all intuitive!)
Consider then this other, less known, effect:
Make your power supply with two outputs, so you have two voltages,
V/2 and V (independent transformer windings, rectifier, and storage
capacitors for both).
Connect first the capacitor through a resistor to the supply V/2 and
wait until it's practically fully charged (more than 5 RC seconds).
The energy stored in the capacitor is 0.5*C*(V^2)/4.
The energy dissipated in the resistor is also 0.5*C*(V^2)/4.
Now connect the capacitor (already charged to V/2) to the supply V
through the resistor and wait again until it's charged.
The final energy stored in the capacitor is 0.5*C*(V^2).
The energy dissipated in the resistor in this second charging is
0.5*C*(V^2)/4 again, because the resistor just "sees" the capacitor
voltage rising by V/2.
The total energy dissipated in the resistor in this two-step
charging is just 0.5*C*(V^2)/2, or just one half of the energy
stored in the capacitor.
Add more charging steps, and the power dissipated in the resistor drops
even more. A two-step power supply is not very difficult to make, and
can improve the efficiency of the system proposed by the original
poster.
Antonio Carlos M. de Queiroz