The Nature of Electricity

G0MDK

The Nature of ElectricityBy

Chuck Hobson g0mdk

G0MDK

INTRODUCTION

This “Nature of Electricity” presentation has been modified to include remarks normally used by a speaker as he proceeds through the slides. If you have any questions, you can email them to me and I will try to answer them. [email protected] (g zero mdk at tiscali.co.uk)

G0MDK

Elektrizität! Was ist es?

L'électricité! Qu'est-ce que c'est ?

Elettricità! Che cosa è esso?

Electricity! What is it?

Georg Simon Ohm 1787 - 1854

André Marie Ampère 1775 - 1836

Count Alessandro Volta

1745 - 1827

Michael Faraday 1791 - 1867

Joseph Henry 1797 – 1878

Nikola Tesla 1856 - 1943

Charles Augustin Coulomb 1736 - 1806

James Watt 1736-1819

CREDITS

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1745 - 1827






CREDITS

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1745 - 1827






CREDITS

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1745 - 1827






CREDITS

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1745 - 1827






CREDITS

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1745 - 1827






CREDITS

Note they all have units of

electricity named after them.

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THE ELECTRON FOUND IN ALL MATTER

SOME PROPERTIES

• Radius < 10-15 metres

• Rest mass 9.1 × 10-28 grams

• Charge neg. 1.6 × 10-19 Coulombs

HOW SMALL? A thousand trillion electrons side by side measure 0.5m

HOW HEAVY? 1.2 thousand trillion trillion electrons weigh one gram

HOW POTENT? 6.25 million trillion electrons make a 1 Coulomb charge

One Coulomb flowing per second = one Ampere.

One gram of electrons contains 176,000,000 Coulombs of charge

HEART OF ELECTRICITY

G0MDKPUTTING THE ELECTRON TO WORK

Note: Ideal model used, Wires have zero resistance, light illuminates instantly and resistance is a fixed 100 ohms.



Note: If an oscilloscope and photo cell at the battery/SW end is triggered at SW closure, the photo cell & oscilloscope would see the light 6.67µs later.

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A QUESTION

1. 3.3µs was arrived at using a Radar type calculation

2. Velocity x time results in distance travelled (v x t = d)

3. In Radar v = c, the speed of light- 300 million metres per seconds.

4. Radar range (c x t = d) 300 x 106ms-1 x 3.3 x 10-6s = 1000m (one way)

5. In our case we know d (1000m) and c, so we calculate t

or 3.3µs

Now consider current flow again

WHERE DID THE 3.3µs COME FROM?

s3.3x10s0.33x10ms3x10

m10

c

dt 65

18

3

G0MDKCURRENT FLOW

100

100

100

100

G0MDKCURRENT FLOW

100

100

100

100

G0MDKCURRENT FLOW

100

100

100

100

Current measures 1A in fig 3. What about the current in figures 1 & 2?

G0MDKCURRENT FLOW

100

100

100

100

The battery doesn’t see the 100 ohm load in figures 1 & 2 All it sees is the characteristic impedance of the pair of wires In our example, this impedance is assumed to be 400 ohms.

G0MDK

ANOTHER QUESTION

1. No way Jose!

2. Why not?

Reason:

1. Electrons are particles with mass as previously stated

2. As particles approach c their masses increase enormously

3. This is in accordance with Einstein’s “Special Relativity”

4. This has been demonstrated at Cern and SLAC

5. Cern and SLAC use GeV’s to reach near c velocities

6. No particles including the electron have ever been accelerated to c

Can electrons travel through wire at the speed of light ( c ) where c = 300,000,000 metres per second?

G0MDK

ELECTRON VELOCITY-1

ELECTRON VELOCITY IN A VACUUM TUBE

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ELECTRON VELOCITY-1

ELECTRON VELOCITY IN A VACUUM TUBE

Formula from A-Level Physics

Calculation

x2x20001.759x10v 11

= 26.5 million metres per second (final velocity at the anode)

Let’s increase voltage on the Anode of the tube and calc. velocities

2Vm

ev

o

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ELECTRON VELOCITY-1

Anode voltage (kilovolts)

Electron Velocity million metres/s

Velocity to speed of light ratio

8.00 53. 0.177 (12.5%)

16.00 75 0.250 (25%)

64.00 150 0.500 (50%)

128.0 212 0.701 (70%)

256.0 300 1.000 (100%)

512.0 424* 1.410 (141%)

SOMETHING WENT WRONG! Electrons CANNOT exceed c!

Increase in mass with velocity (relativistic mass) was not taken into account

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ELECTRON VELOCITY-2

cv

1

1

When a mass velocity approaches the speed of light its mass increases

This is in accordance with Einstein’s theory “Special Relativity”

That is to say: relativistic mass (mr) = gamma ( ) times mo

Relativistic mass calculations are done using the following formulas:

or m x m #1

#3

#2

#4

where

2r v0.5meV

eV = electron volt is a unit of energy used in particle physics

2Vrm

ev velocity

2erm 2v

V

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ELECTRON VELOCITY-2

cv

1

1





or m x m #1

#3

#2

#4

where

2r v0.5meV


2Vrm

ev velocity

2erm 2v

V

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ELECTRON VELOCITY-2

cv

1

1





or m x m #1

#3

#2

#4

where

2r v0.5meV


2Vrm

ev velocity

2erm 2v

V

Known as the Lorentz Transform

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ELECTRON VELOCITY-2

cv

1

1





or m x m #1

#3

#2

#4

where

2r v0.5meV


2Vrm

ev velocity

2erm 2v

V

Known as the Lorentz Transform

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ELECTRON VELOCITY-2

On the last entry, notice the significant mass increase (x 17,219)

Velocity m/s Gamma

100,000,000 1.225

200,000,000 1.7320

250,000,000 2.4490

290,000,000 5.4770

299,000,000 17.320

299,999,000 547.72

299,999,999 17,219

G0MDK

ELECTRON VELOCITY-2

On the last entry, notice the significant mass increase (x 547723)

Velocity m/s Gamma Voltage

100,000,000.000 1.2250 34.8kV

200,000,000.000. 1.7320 197kV

250,000,000.000 2.4940 435kV

290,000,000.000 5.4770 1.31MV

299,000,000.000 17.320 4.4MV

299,999,000.000 547.72 140MV

299,999,999.000 17,219 4.4GV

299,999,999.999 547723

Voltages shown in the 3rd column required to obtain velocities in the first column.

The point of this exercise is to show that it is virtually impossible to get the electron to move at the speed of light.

G0MDK

A QUESTION REVISITED

ELECTRONS DIDN’T TRAVEL AT c BUT THE SOMETHING DID

1. How about the “nudge” theory (cue ball effect etc.)?

Electrons out of the negative terminal nudge the next one on etc. The end result could be electrons at the light bulb 3.3µs later (?)

There have been many arguments on this issue since the 1920’s A paper on this notion was submitted to the 1997 IEE.

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RECTANGULAR PULSE

WHAT IS HAPPENING DURING INTERVALS: (A – B), (B – C), (C – D)?

(A – B) and (C – D)? Nothing! There is NO voltage or current B – C? 100V and 0.25A

Note: Z of the transmission line pair = 400 Ohms. What would the situation be in 5.66µs?

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RECTANGULAR PULSE

Negative pulse moving back to the Pulse Generator

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SINGLE RECTANGULAR PULSE EXAMINATION

RECTANGULAR PULSE Time Domain:

Viewed on an Oscilloscope

a2

a2tj)(

1d(t)Af(t) e

RECTANGULAR PULSE Frequency Domain: Viewed on a Spectrum Analyser

Pulse reconstruction formula Fast Fourier Transform formula

)1)(

)1

f

fff ( ( sincsin)A(

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PULSE RECONSTRUCTION-1

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RECAP

1. The nudge (cue ball) explanation of conduction unresolved

2. Electrical energy travels ~ speed of light over wires to a load.

3. Likewise, pulses travel ~ the speed of light over wires.

4. Single pulses are made up of wide spectrums of frequencies

5. Pulse (spectrum of frequencies) travel as TEM signals at ~ c

6. In the circuit comprising a 100V battery, switch & light bulb: the leading edge of a pulse occurs at 100V switch on and the trailing edge of a pulse occurs at 100V switch off

7. Very long pulses have same properties as very short pulses

8. AC signals to µ-wave frequencies) travel as TEM modes

9. Note that the wave length of 50Hz = 6 million metres

G0MDK

CONCLUSION

1.Electrical energy travels from source to load over wires as TRANSVERSE ELECTROMAGNETIC WAVES (TEM mode)

2.Current drift* (Amperes) is a consequence of EM Waves NOT THE OTHER WAY AROUND This may be difficult to visualize in a pair of wires, but if you consider EM microwaves travelling down a wave guide, there will be surface currents in the wave guide walls. These are also drift currents. They are also the consequence of EM energy

* A sample calculation of current drift is shown in appendix 1 of this presentation.

G0MDKAppendix 1 ELECTRON DRIFT

The current drift rate through a conductor is in the order of mm/s. The drift rate of 1A through a 1mm diametre copper wire is worked out as follows:

Current density J = amperes per unit area (J = I/A)

so J = 1Amp./(pi x r2) = 1/(3.14 x 0.00052) = 1.6 x 106

J can also be expressed as J = nevd

Transposing: vd = J/(ne)

Copper has an electron density n of 8.47 x 1028 m-3

With e = 1.6 x 10-19 coulombs of charge: ne = 1.4 x 109

Thus: vd = J/(ne) = (1.6 x 106) / (1.4 x 109) = 1.14mm/s

G0MDK

That is the nature of Electricity as I

perceive it.

Thank you for attending Chuck

Hobson BSc(hons) BA

Documents

The Nature of Electricity