While learning electronics, I have experimented with different circuits. After trying some different configurations, I built the schematic shown in the picture, and when I joined the wires together (after the capacitors), the lightbulb lits. This phenomenon, as far I know, is not possible. If you join 2 wires, you get a short circuit. But when I connect the wires to the short circuit, the light bulb works.
Anyone can explain why this works?
show in your diagram where exactly you make the short circuit. i beleive the IR value gives enough voltage for teh bulb to work. did you try to put the bulb very close to the shorting points & observed?
It is possible if the short circuit is a long wire and/or you make the connection just by pressing two stripped wires together. Long wire creates inductance (and resistance) and bad connection resistance.
show in your diagram where exactly you make the short circuit. i beleive the IR value gives enough voltage for teh bulb to work. did you try to put the bulb very close to the shorting points & observed?
From left to the right after the capacitors, instead of connecting the positive to one wire to the bulb and then the negative to the other wire of the bulb, I connect the positive and negative together (short circuit) and then I connect each wire of the bulb to the short.
You can see what I mean in the attachment.
It is possible if the short circuit is a long wire and/or you make the connection just by pressing two stripped wires together. Long wire creates inductance (and resistance) and bad connection resistance.
You mean to use a long wire instead the light bulb? explain it more please.
The wire I used is stripped and it's only one wire, I mean that I connect one end to the output of one capacitor and the other end to the output of the other capacitor. In that stripped wire I connect the light bulb, but the wires of the light bulb must be separated I(prevent to touch each wire of the light bulb).
But, why I cannot replicate this effect while using a DC battery without the capacitors and the spark? I mean shorting the battery and connect a 12 V bulb.
I have been testing more the circuit. Now I'm able to say that:
You can touch the short circuit and not get shock.
I can lit the bulb underwater.
But, why I cannot replicate this effect while using a DC battery without the capacitors and the spark? I mean shorting the battery and connect a 12 V bulb.
Not, of course. But it's possible that if I try to replicate the same test using a 10kV AC power supply without the caps and the spark, I will get shocked and the light bulb would only lit only a few.
You are creating a voltage divider with the wires and the short circuit. With 10kV there is enough voltage to light up the lamp, but with the battery the (allready low) voltage drops too much. The internal resistance of the battery is much higher than the resistance of your short.
At the higher voltage, the wire used to short the circuit has a higher resistance, whereas at a lower voltage, it has a lower resistance. The more resistance there is, the more current will flow through the bulb.
By the way, are you seriously using a light bulb at 10kV?
Der Strom
Touching a circuit is not a good way to test if you can get shocked or not. Same thing with white powders.. tasting is not a good way to find out if it's cocaine or not.
At the higher voltage, the wire used to short the circuit has a higher resistance, whereas at a lower voltage, it has a lower resistance. The more resistance there is, the more current will flow through the bulb.
By the way, are you seriously using a light bulb at 10kV?
Der Strom
Touching a circuit is not a good way to test if you can get shocked or not. Same thing with white powders.. tasting is not a good way to find out if it's cocaine or not.
I have took a lot of precautions before touching it. But why you cannot get a shock? when I see high voltage installations they always say: danger high voltage!
I suspect that what you have is two close coupled coils forming a simple air core transformer of sorts and your spark discharge and related one half of the shorted circuit is working like a primary coil and your second half of the shorted circuit is working like the secondary.
By shear luck you just happen to have created two tuned coils that happen to have similar enough resonance frequencies or harmonics of each other to work.
There is a pretty good chance the you could keep the length of wire shorted between the two capacitors and then use the other half of the short circuit on the lamp as an independent loop circuit that would pick up enough induced power to work without being connected to the primary side of the circuit at all.
I have took a lot of precautions before touching it. But why you cannot get a shock? when I see high voltage installations they always say: danger high voltage!
You have to be part of a closed circuit to get a shock. If no current can go through you, you won't get shocked.
Watch this wideo about high voltage power line worker YouTube - High power line workers
The fact that the bulb lights normally suggests that significantly less than 10kV is actually passing through the filament of the bulb. I think you would know it if the full current was passing through the bulb. You are just creating a parallel circuit with two resistors in each branch--the wire acts as a low-resistance resistor, and the bulb acts as another resistor. As long as there's some resistance in the "short" wire, you will get current through the bulb. Since the voltage is so high, a significant amount of voltage also goes through the light bulb, causing it to light.
As for not getting shocked when you touch it, I believe that has already been explained.
I think TCMtech has it right. Add a short where you are putting it now. Then clip off the lamp right at the short and add a short to the lamp wires. Now see if you can light the lamp by getting it close to the first short. Even bigger magic?