Work done by load in charging a capacitor

[PriceMike]

Sooo confused!

 

[truesearch]

I have attached photographs of the demonstration of em radiation from the connecting wires in the capacitor circuit.
The first picture shows the components, the powersupply is a square wave signal generator to charge and discharge the capacitor. The variable resistor is 0-15Ω.
The black wire is one of the current carrying leads in the capacitor circuit and the yellow wire is the 'pickup' coil.
The second photo is the square wave supply and the third photo is what is picked up by the yellow coil from the black wire. I passed the black wire through the coil several times to obtain a trace that could be photographed. There is a trace with the coil just sitting on the straight wire.
When the resistance is decreased the pickup increases (current is greater)
There is 'something' passing from the current carrying wire to the coil and this is energy being lost from the capacitor circuit.
There are essentially 2 means of dissipating the 0.5QV of energy...(1) Joule heating due to resistance (2) electromagnetic (radio) radiation from the wire as a result of the changing current. The 2 together account for the missing energy.

 

[sophiecentaur]

Very interesting and not at all surprising that you can 'detect' some radiated RF. We have established that there is some. But do you actually know (how can you?) how much Power is being radiated by this mechanism and how do you know how much Power is being dissipated by resistive elements in the rest of the circuit? Do you actually know the resistance in the circuit? What about the output impedance of the sig genny at your operating frequency?
Introducing the idea of superconductivity is a complete red herring because there are NO power sources with zero internal resistance - even if there are some superconducting wires involved in the rest of the circuit.
If all the 'extra' power is being coupled out into space then you have achieved a remarkable feat.

 

[jsmith613]

Very interesting and not at all surprising that you can 'detect' some radiated RF. We have established that there is some. But do you actually know (how can you?) how much Power is being radiated by this mechanism and how do you know how much Power is being dissipated by resistive elements in the rest of the circuit? Do you actually know the resistance in the circuit? What about the output impedance of the sig genny at your operating frequency?
Introducing the idea of superconductivity is a complete red herring because there are NO power sources with zero internal resistance - even if there are some superconducting wires involved in the rest of the circuit.
If all the 'extra' power is being coupled out into space then you have achieved a remarkable feat.

I did not realize anyone answered the q...seems like lots of people had fun but I just want to check what I asked first time
anyway to summarise MY initial question:
- work done by power source = QV
- energy stored = 1/2*QV
- missing energy is transferred to wires

right?

 

[BruceW]

I did not realize anyone answered the q...seems like lots of people had fun but I just want to check what I asked first time
anyway to summarise MY initial question:
- work done by power source = QV
- energy stored = 1/2*QV
- missing energy is transferred to wires

right?

Exactly. And if there are other components in the circuit, some of the missing energy will also be transferred to them.

 

[truesearch]

the missing energy is dissipated (lost) from the wires
There are 2 clear means of dissipating energy
1) resistance of wires (joule heating)...could be as near zero as you want to get
2) electromagnetic radiation from the wires...can never be zero if the current is changing.