Tesla Coil as Power Transmitter

AI Thread Summary
Tesla coils function similarly to transformers, converting low voltage and high current into high voltage and low current while maintaining total power. High voltage is essential for efficient wireless power transmission, as it allows for better electromagnetic wave propagation and impedance matching with antennas. The discussion highlights that a well-matched impedance maximizes power transfer, which is crucial for effective transmission. Additionally, increasing voltage significantly enhances power due to the squared relationship in energy equations, making higher voltage sources preferable for achieving longer spark lengths. Understanding these principles is vital for optimizing Tesla coils as power transmitters.
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I've been trying to hash this problem out with some friends of mine, but we haven't been able to find a satisfactory answer:

Tesla coils are (in an extremely basic sense) like a transformer. There are two coils of wire... the first one having a large current and not-so-large voltage. The second coil has many more turns, so you end up with a large voltage and not-so-large current. Of course, total power (I*V) remains -in a perfect universe- unchanged.

The secondary coil, having high voltage, can make some great sparks. Perfect for science museums and sci-fi movies.

However, back in the day, Tesla intended these coils to act as wireless power transmitters. And indeed they can do just that over short distances. Like an antenna, the secondary coil of the Tesla coil can set up EM waves that propagate.

So here is my question: for power transmission... why do you need high voltage? I understand that you of course need high *power*, but if the power is unchanged going from the first coil (big I, small V) to the second coil (small I, big V), what's the benefit? Is the high voltage able to set up EM waves better than a high current can? Why would that be?
 
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I think the answer you are looking for is that a good antenna has a specific impedance. At the point of being matched for efficient transmitting it appears as a resistance. To push more power into that resistor, you need to apply a higher voltage.
 
So does the secondary coil have the right impedance, while the primary coil doesn't? I'm not sure I quite understand yet.

In my mind, there are two circuits (with antenna in them). LRC circuits, the whole shebang. One of them is configured to have low voltages across the L&C, say, yet high currents. The other is configured to have high voltage across the L&C, yet low currents. Power (I*V) is identical.

And for some reason, the latter transmits better. Is this just a fact, then, that for the above statements to be true... the first instance has a bunch of reactance (which is bad), and the second instance has a bunch of resistance. And resistance, as you say, is what you want. Am I looking at this in the right way?
 
A transformer is commonly thought of as "stepping up" or "stepping down" the voltage. Your first year studies reveal that it can equally be thought of as "stepping down" or "stepping up" the current. Both of these involve the turns ratio 1 : n

More accurately, the transformer can be said to be stepping up or stepping down impedances (or resistances), and by the square of the turns ratio, 1 : n²

So if you have an antenna with an inherent impedance of 300Ω you can drive it with an RF amplifier with an output impedance of 300Ω through a feeder of 300Ω, and everything will be matched. On the other hand, if your amplifier has an output impedance of 50Ω and you have a feeder of 50Ω coax then you could put a transformer right at the antenna to transform 50Ω to 300Ω and again everything will be matched. A suitable turns ratio for the transformer would be 9 : 22

Only with correct matching will you achieve maximum power transfer to the antenna.
 
@TestTubeGames
i am not an electrical student but mechanical engineering student. so sorry in advance if you don't like my answer. i have found this answer of your question...

The amount of energy available to be sent to the primary is 0.5 x C x V^2
C = Farads
V = voltage that the gap fires at.

You can see here that doubling the value of C (provided your power source is robust enough) will give you twice the power. But doubling the voltage that the capacitor is charged up to will give 4 times the power, because the voltage value is squared, that's why if you want spark length its best to go for a higher voltage power source.
more is the power supplied, more is power carried by secondary circuit, and secondary capacitor (torus) is more charged up to.
 
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