C
Commander Kinsey
Guest
On Sat, 01 Jul 2023 20:41:45 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:
It makes sense for an amp, but not for raw power. For example, there\'s 240V at that socket there, with virtually no resistance. If I connect a resistive heater to it to it to draw power, I get all the power as usable heat pretty much, even though the resistance of the heating element is vastly more than the supply. What if I connected a fucking big element to it, such that it\'s resistance was equal to the supply line? Half the power would be dissipated in the supply line. I\'d drop from 99.9% to 50% efficiency.
Just take a normal one but moreso. Do you really think it\'s much harder to convert 50V to 5V than 10V to 5V?
On Sat, 01 Jul 2023 20:05:21 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:
On Sat, 01 Jul 2023 15:05:10 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:
On Wed, 28 Jun 2023 11:37:04 +0100, alan_m <junk@admac.myzen.co.uk
wrote:
On 28/06/2023 10:09, Commander Kinsey wrote:
Raindrops creating electricity? But.... wouldn\'t you be better sticking
a solar panel there? I assume you can\'t have both? Or could one fold
up when not in use?
https://www.sciencedirect.com/science/article/abs/pii/S2211285521008193
Don\'t you just need a kite and a battery big enough to store the energy
from the lightening strike? Power for life.
There might be an impedance mismatch charging the battery.
I\'ve never heard that term applied to DC and don\'t know what you mean. I only understand it for audio amplifiers.
The Maximum Power Transfer theorem works for AC or DC.
It makes sense for an amp, but not for raw power. For example, there\'s 240V at that socket there, with virtually no resistance. If I connect a resistive heater to it to it to draw power, I get all the power as usable heat pretty much, even though the resistance of the heating element is vastly more than the supply. What if I connected a fucking big element to it, such that it\'s resistance was equal to the supply line? Half the power would be dissipated in the supply line. I\'d drop from 99.9% to 50% efficiency.
There is a battery that consists of a beta emitter coated rod inside a
metal tube. It develops hundreds of kilovolts at low current, and the
problem has always been, aside from the radioactive hazard, how to
convert that down to something useful.
Can\'t be that hard, we convert voltages all the time.
Please sketch up a 5 volt power supply with a 400KV input.
Just take a normal one but moreso. Do you really think it\'s much harder to convert 50V to 5V than 10V to 5V?