Physics/What is the secret of a single wire transmission ?
At first, thank you for your recent answers about the information of impedance (Z) and power relations for AC circuits.
In our last discussion we had a problem of one way signal transmission. Lets assume a capacitor with a capacity of 1 micro Farad, the signal generator has an output of 0,1 watt (how we find this ?- we connected a 1000 ohm resistor across the generator and the other end to an oscilloscope, what we see was a curve, the value was 10 volts (peak to peak), so, peak value should be 5 volts, effective value 5 x 0,707 = 3,535 volt.
I think we made a mistake, is the power (P) calculated over the RMS (effective) value ? If so, the power would not be equal to 0,10 watt, but 3,535 X 3,535/1000, so it would be equal to
0,0125 watt, am I right ?
The electrician does now know anymore the P value of his signal generator, for the experimenting we need the P value while P = i² x Z……..
I connected this 1 micro F capacitor to one end of this signal generator, the other end of this generator was connected to a copper sheet with the dimensions of 30 cm x 30 cm for the grounding, you see here an open circuit !!!
The charge is moving between ground and the first electrode of the capacitor.
The other end of the capacitor goes to a resistor of 10 ohm which is connected to an oscilloscope, and the second end of the oscilloscope goes to the ground (earth).
The grounding of the oscilloscope has not any interaction with the copper plate, where the signal generator is attached.
The internal resistance is ignored (less than 1 ohm)
What can we see at the screen of the oscilloscope ?
If the circuit was closed, as known in many textbooks, we should find P = i² x Z, to find Z, we need near R, also the Xc value, at 10 kHz, it would be equal to = 1/(2 x Pi x C x f) = 15,915 ohm, Z would be equal to the square root of (Xc² + R²) :
Z = 18,796 ohm
0,0125 = 18,796 x i²
i would be equal to 25,78 mA.
VC = Xc x i = 0,4103 volt
VR = R x i = 0,2578 volt
We would see a sinus curve (for the resistor-R) peak to peak = 0,2578 x 1,414 x 2 = 0,729 volt, if the circuit was closed, am I right ?
But, what happens, in my open circuit, the charge is flowing from the SG, between the copper plate and the first electrode of the capacitor, back and forth, this creates an elctrostatic induction in the capacitor, so by this way the signal is transported to the second electrode of the capacitor, which is attached to a resistor, the charge is oscillating between the second electrode of the capacitor and the osilloscope, which is grounded.
Can I also see the same value on the oscilloscope screen which is equal to 0,729 volt ?
Steve, how can I find the C value of each electrode in the capacitor, if the mutual capacity is 1 micro F ?
This may change the XC value, I guess, what do you think ?
I am trying to simulate the one way transmission hypothesis of Nikola Tesla, would be happy if you can help me.
Wow, no, this is both too complex and misses things at the same time. You need to add all three voltages correctly, meaning accounting for their phases. So you need the inductance, too. This is a subject for a physics II book, you need to find the inductance. To find that, you can do it with the phase, but that's weeks of classwork.
To simplify this, because complex circuits seem to be a little out of your calculation area for now, consider that the only element which dissipates real power in a circuit is the resistor. So measure the voltage across just the resistor when the source is putting out 10V. Then just use (I^2)R for the power it actually provides. Make sure the resistor connects on one end to the circuit and the other end to ground in your closed circuit.