### Author Topic: The Right Question?  (Read 31471 times)

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##### Re: The Right Question?
« Reply #88 on: December 20, 2015, 21:19:10 pm »
I'm trying to work out if it's possible for displacement current to create e/m waves and I'm getting weird solutions E=μ0(ε0( c dE/dt  + d2E/dt2))  that's a kick *  ODE lol.
The only thing that resembles a capacitor is a waveguide and this only directs em waves it doesn't radiate normaly.

Hey I solved the ODE and the golden ratio just popped up, interesting I feel like yoda now.

E = e^((-1 - Sqrt[5])/2  ct) C[1] + e^((-1 + Sqrt[5])/2  ct) C[2]
« Last Edit: December 20, 2015, 22:49:03 pm by geon »

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« Reply #89 on: December 21, 2015, 01:11:09 am »
I will share someinfo about the SS wire i came up with..

this ideas may need to be confirmed but seems all clear...

In short words the ss wire distributes the resistance over the coil making its Zo to be normally higher and more stable for a broaden range of frequencies.. and as in a tl the energy travels as voltage not current only than there should not be a voltage drop although restrict amps and increase zo

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##### Re: The Right Question?
« Reply #90 on: December 21, 2015, 17:18:46 pm »
There is a secondary imaginary em wave travelling with each e/m wave in the void that can only exist inside a conductor and is the reflected wave , dependent on charge acceleration thus it has a phase delay or Pi/2 because the sin function has maximum slope during the beginning and zero half way.
« Last Edit: December 21, 2015, 17:51:36 pm by geon »

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##### Re: The Right Question?
« Reply #91 on: December 21, 2015, 18:35:01 pm »
Didnt got it Geon..

today i did several tests and determined the dc impedance of my power supply using the labview i determined that under 1 amp it has got a 2,5 ohm impedance, and at 150ma is up to11 ohms... h

it includes the voltage drops of all dioes and igbt...

the impedance change with load impedance but this is a range i tested with 10-20-30-40-50 ohms resistors...

i set to 1 amp with 10 ohms and see the voltage aplied, the current, and open the circuit and measure the input voltage

than maintain it and change the resistors and make the measurements

i found that it depends more on the current output so is reasonable to make graphs for output impedance vs current consume for each different load range...

« Last Edit: December 21, 2015, 20:22:20 pm by sebosfato »

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##### Re: The Right Question?
« Reply #92 on: December 21, 2015, 19:17:03 pm »
...
« Last Edit: December 21, 2015, 20:34:05 pm by geon »

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##### Re: The Right Question?
« Reply #93 on: December 21, 2015, 20:18:04 pm »
YesGeon

this is what i´m talking about... k is the dielectric inst it?

The Electric field is the derivative of the potential... the equation of the potential of a charge in medium of dielectric field means this!!

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##### Re: The Right Question?
« Reply #94 on: December 21, 2015, 20:44:37 pm »
sorry doesnt make sense really go on with what you were doing before I posted

http://www.wolframalpha.com/input/?i=y%3Dk*e^%28%28%28-1%2Bsqrt2%29%29c*t%2Fr%29+-+k*e^%28%28%28-1-sqrt2%29%29*c*t%2Fr%29

I don't know what k is I took initial conditions E(0)=0 and I got k1=-k2=k that's all if k1 is different that k2 then the E field changes from positive to negative and has a 1/sqrt2 value in the root .............
« Last Edit: December 21, 2015, 21:19:27 pm by geon »