If photovoltaic cell only uses less than half of the light that hits them and reflects the rest away, and fiber optics are long, thin strands of very pure glass wrapped in something that keeps the light from getting out, why can't you line the inside a container with photovoltaic cells, mirrors and a low watt light source and get enough electrity to power itself (after a jump start, ofcourse)?
2007-02-07 21:44:08 · 5 answers · asked by wtboy_2000 1 in Science & Mathematics ➔ Engineering
I am not an engineer, but I actually made a very simple circuit not to long ago that was self sustaining. Of course I needed a battery to get the thing going, but I was set out to do what everyone told me was impossible. I still have work to do with it, but I have done it and am waiting for a little extra money to continue. Either way, don't listen to the engineers replying to you. I have done it and no matter how many calculations they can come up with to explain that it cannot be you can rest assure that it can. It took me a while and a lot of waisted money. I have documented everything on video tape. Maybe within the next I will be able finish it up. Good luck in your endeavors.
2007-02-07 23:55:40 · answer #1 · answered by Anonymous · 0⤊ 2⤋
First of all, you can't get more energy from a system than you put in. Secondly, fiber optics don't contain the light because of a covering. They do so because of total reflection. As the light goes through the fiber, it hits the sides at a very shallow angle so all the light is reflected. It's like a rock skipping off a pond.
2007-02-07 23:22:24 · answer #2 · answered by Gene 7 · 0⤊ 0⤋
It doesn't work for the same reason as any perpetual motion device won't work. It is due to efficiency (or loss of energy) of the items we can manufacture. Only if your light source and your photo cell are >/=100% efficient, then this would work.
But even if it was, there would still be no practical use of it, because - I assume - you want to build a lamp that "shines by itself", and so some of the light must "escape" to "shine", and that light doesn't get onto the photo cell.
2007-02-07 21:59:03 · answer #3 · answered by Marianna 6 · 1⤊ 0⤋
In a closed system, the energy is conserved. so, you cannot produce more electricity than that supplied to the light. There is always some loss since no system is 100% efficient. So, without an external supply, the system cannot sustain.
2007-02-07 21:53:21 · answer #4 · answered by Swamy 7 · 0⤊ 0⤋
Have you not yet been introduced to the Second Law of Thermodynamics?
2007-02-07 22:13:42 · answer #5 · answered by Helmut 7 · 1⤊ 0⤋