

Electric Circuit and Speed of Light
Name: Andy
Status: student
Grade: 912
Country: Canada
Date: N/A
Question:
I have a question about how fast electricity moves. From
another answer I see that electricity travels at close to the speed
of light. If I have a wire that is 1 light year long from the
positive to the negative terminals on a battery. And I put a light
bulb in the middle of the wire (1/2 light year from either end). How
long will it take for the light bulb to turn on? 1 year? 1/2 year?
Replies:
Andy,
Electricity travels at the speed of light, but electrons do not. Within
a wire, the signal between individual electrons, as well as changes in
the electric field that extends around from one end of a battery to the
other, travels at the speed of light. The electrons respond to these
signals. The electrons absorb energy from these signals and then give
the energy to devices at the other end of the wire. Electrons in a wire
bounce all over, slowly drifting from one end of the wire to the other,
through the device, and then slowly through wire that goes back to the
power source.
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
Well, Andy, I would start by assuming that
your battery is very small compared with 1/2 lightyear.
So the two wires start very close together.
I would further assume that they run closetogether,
sidebyside, to the distant destination.
This makes a familiar configuration called a "transmission line".
The inductance of each wire and the capacitance between adjacent wires
has the effect of allowing waves to be launched down the pair
whenever the voltage at one end is changed.
I assume the voltages everywhere start at zero.
Then the switch at the battery is closed, suddenly applying V volts between the wires.
In order to charge the capacitance to V volts, some current must flow into the wire.
But that current cannot reach all parts of the length at once, because of the inductance.
So a risingstep wavefront of voltage travels down the transmission line.
Everywhere nearer than the front, the voltage is V,
and there is a certain current I (See below to figure current I.)
running out along the positive wire and back along the negative wire.
Everywhere farther than the front, the voltage is still zero, and there is no current.
(If the switch started gradually there is some transition zone at the wavefront.
It is reasonable for a switch to start in 10 nanoseconds.
Then, given the speed of light, the width of the ramping zone would be about 10 feet.)
Meanwhile the lightbulb does not light up, does not show a clue, for half a year.
Then, in 1/2 year, the wavefront reaches the lightbulb, and the lightbulb lights up abruptly.
It might light up dimly, or too brightly and quickly burn out,
if the transmission line impedance does not match the bulb's resistance well.
If it lights up dimly it will have to wait successive periods of one year
for the wave's echoes to go back and forth in the wirepair,
abruptly making the current a little righter each time.
If it is a perfect match, the lightbulb simply lights up exactly right,
and the battery never even gets a little echo in the current to tell it when the light bulb got its start.
In practice there's usually a little bit of echo in the line.
I = V/Z, where Z is the "impedance" of the line.
The precise value of impedance is determined by the amount of
inductanceperunitlength and capacitanceperunitlength:
Z = sqrt(L/C)
These in turn depend on the ratio of wireseparation S to wirediameter D:
Z = k log( S/D ) .
I think you can find this formula on Wikipedia, and look up the value of k.
Closer and fatter wires make a lower impedance.
Farther and thinner wires make a higher impedance, measured in Ohms.
Two wires 1cm apart and 0.05cm in diameter tend to be around 300 ohms,
such as the old but familiar "twinlead" TV antenna cable that looks like a ribbon.
The newer round cable is 75 ohms, because the inner wire and outer braid face each other much more closely.
Sticking two strips of metal tape onto each facetoface other makes a transmission line
with impedance somewhere around 1 ohm,
assuming the adhesive layers between the tapes are insulating.
That's the practical range of normal transmissionline impedances.
All this assumes a 1 lightyear length of perfect superconducting wires and perfectly insulating vacuum.
Nobody is ever going to really have all that.
In practice the resistance of the wires and some resistance in parallel or series with the capacitance of the insulator
limit the distinctive transmissionline effects to far less than a lightyear.
Outer space has lots of conductive plasma, too.
If your wires do not run close together, if they perhaps trace out one big circle or square instead,
that is a slightly more advanced problem.
It is somewhat similar to making the transmission line separation S equal to 1/2 lightyear,
which would make the impedance exceptionally large, and very choking to the current,
so the lightbulb would take centuries to gradually light up, one imperceptibly tiny step per year.
In effect, the inductance of a singleturn loop of wire 1/2 lightyear in diameter would dominate.
You could find a simple formula for the inductance L2 of a loop, too.
There is an inductanceresistance timeconstant, t = L2 / R.
It would take a few t's for the lamp to reach full brightness.
Also some of the energy pumped into those wires for so long
would leak out into space as waves of electric+magnetic field, freed from the wire.
Your wireloop would be an effective radio antenna for an extremely low radio frequency like 1 cycle every month or 2.
That would be less than 0.000001 Hz, instead of the 1,000,000 Hz we use in the AM radio band.
Keeping twinwires close together prevents such leakage.
Jim Swenson
Andy,
If you consider the basic structure of a light bulb, if the light bulb
is connected to a source that provides free electrons, in a closed
circuit from the positive and negative terminals of a battery, when an
electron reaches the bulb, and passes through the filament of a bulb,
it's the bombardment of the free electron(s) into the atoms of the
filament that energize the filaments electrons that produce light.
In an ideal case, where the battery has enough potential difference to
produce an electron flow, without any losses in the wire, at a
distance 1 light year away, then the time it would take for (1)
electron to reach the filament should be 1/2 a light year away.
But (1) electron may not be sufficient to produce light as it bombards
the atoms in the filament. It could take a little longer as the flow
of electrons begin to build up, and it may require that a steady state
of electron flow is achieved. Meaning, for the light bulb to remain
lit, it may require that a complete path of the electrons is
accomplished: from the positive terminal, to the bulb's filament, back
to the negative terminal of the battery, and back to the bulb.
So in summary, it depends upon what you require as 'lighting up' the
light bulb; but in an ideal case, the time for (1) electron to reach
the filament would be 1/2 a light year.
Hope that helps.
Alex Viray
Hi Andy,
In the case of a directcurrent voltage used to light a bulb, you want
to look at the speed of the voltage propagating through the wire (not
the speed of actual electrons). The speed of the (voltage) wave
depends on the medium through which it's traveling  the higher the
relative permitivity, the slower the speed of the voltage
propagation. Many cables and wires come with published "velocity
factors" (VF) which relate the speed of voltages propagating through
the cable to the speed of the signal in vacuum (VFs are commonly in
the 0.60.8 range). So in your example, a 1lightyearlong cable with
a VF of 0.8 would require 0.625 light years to reach full intensity. A
similar cable with a VF of 0.1 would require 5 light years. This is a
timedependent system (think of how water flows through a long hose)
 it starts with a trickle before it reaches full flow  in this
example, that means some voltage (less than the max) would reach the
bulb fractionally sooner than that.
The term "velocity factor" describes how much slower the wave travels
in a wire (or any other medium) than it does in a vacuum. The equation
is VF = (K)^(1/2) where VF is the velocity factor and K is the
dielectric constant. In other words, a material with a relative
permittivity of 4 would have a VF of 0.5. This value may be
frequencydependent depending on the material, so just treat this is a
rule of thumb for describing the effect.
Hope this helps,
Burr Zimmerman
You raise a "simple" question that demands an answer. I suspect the
answer is complex, but frankly I do not know the answer!
Let me restate your question in general terms:
Electrons repel one another by Coulombs Law. The speed of repulsion is
limited by the speed of light.
Put the same electrons in a wire. Why do these electrons not repel one
another by Coulombs Law, and the whole wire explode?
We invoke electrons in the conduction band. But isn't this just a mental
invention?
This is not my field, so I cannot give you a satisfactory
explanation, but you are commended for asking the perspective question.
Vince Calder
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Update: June 2012

