10
\$\begingroup\$

Please take a moment to look at the diagram below:

enter image description here

The question is if the lightbulb will momentarily flash when the switch is closed. I think it will but I get the feeling I am wrong.

The reason why I think it will flash is because when the switch is closed, the transmission line wire electric potential should become the same as the electric potential found on the battery terminal and in order for that to happen, electrons will need to flow through the wire until electric potential balance is reached. As electrons flow through the wire, they will need to go through the light bulb filament causing the light to turn on.

By the way, I realize that the light bulb will not light up a room or that it will even light up at all, I am only using a light bulb here to illustrate my question and not mean to represent some sort of real life experiment.

Thanks.

\$\endgroup\$
  • \$\begingroup\$ What would cause the electrons to move? \$\endgroup\$ – JIm Dearden May 25 '15 at 19:49
  • 1
    \$\begingroup\$ Current doesn't flow, charge does. \$\endgroup\$ – fuzzyhair2 May 26 '15 at 12:13
  • 1
    \$\begingroup\$ This isn't much different than an antenna. A switched DC voltage will cause a wave to travel down the line. \$\endgroup\$ – Adam Davis May 26 '15 at 12:39
17
\$\begingroup\$

Yes, there will be a brief pulse of current through the bulb as the portion of the transmission line (i.e., its capacitance) to the right of the bulb charges to the supply voltage.

\$\endgroup\$
  • 1
    \$\begingroup\$ I would mention some factors that affect how much would flow, i.e. if it depends on the mass of the wire to the right of the light-bulb, any initial charge difference, etc. \$\endgroup\$ – user2813274 May 26 '15 at 17:06
  • \$\begingroup\$ @user2813274: Actually the mass of the wire has nothing to do with it, and the OP specified the initial conditions. See this answer for more details, but the key parameters are the resistance of the bulb R, the characteristic impedance of the transmission line Z0, and the length of the line. If R = Z0, there will be a neat rectangular pulse of current whose magnitude is V/2R, and that lasts just long enough for the transient to propagate to the end of the line and back again. \$\endgroup\$ – Dave Tweed Feb 2 '17 at 22:10
15
\$\begingroup\$

There will be a slight current pulse at switch on even if you consider the circuit a lumped element circuit, i.e. without resorting to transmission line theory. Just keep in mind that in a real circuit stray capacitance is always present, therefore you might model the open end of the transmission line as a capacitor (with tiny capacitance, say ~1-10pF). Therefore you have a lumped RC circuit, where R is the filament. Thus when you close the switch you are charging that tiny capacitor through the filament.

Assuming some ballpark figures like \$R=100\Omega, \; C=10pF\$ and \$V_{DC}=10V\$ you get an initial current \$I=V_{DC}/R=100mA\$ exponentially decreasing with a time constant \$\tau=RC=100\Omega \times 10pF = 1ns\$.

Of course the energy transferred to the filament before the current dies out is so tiny that a normal lamp won't be able to emit any detectable light.

\$\endgroup\$
10
\$\begingroup\$

No and yes and it depends on your view.

No if you view it as a schematic symbol representation. This is typically what most engineers view when doing calculations or designing most of their work around - the schematic. In this view, current flows when you have a continuous connection driven by a voltage, but there is no continuous connection here, so no current flows.

Yes if you view it as a transmission line. As @Andyaka and @DaveTweet mention, a change in voltage will propagate through the transmission line and every point in the transmission where there is a voltage change, you will have current flow (displacement current). However it will stabilize itself relatively quickly, until there is no longer any change in voltage.

As a crude analogy, you can think of it as, if you stand still and don't move, are you moving ? If its relative to the earth, then no you are not, but relative to the sun, you are moving - quite fast actually.

\$\endgroup\$
8
\$\begingroup\$

Electricity has to flow or else how could the power source know there wasn't a load at the end. This is all embodied in transmission line theory. The current that flows is based on the input impedance of the transmission line. It's called characteristic impedance by the way. Another interesting side effect is that an infinitely long lossless transmission line will conduct current indefinitely based upon the voltage supplied and the cable's characteristic impedance.

\$\endgroup\$
  • 2
    \$\begingroup\$ If you look closely when plugging in a long extension cord, it may spark a little with nothing plugged into the other end. AC introduces some other considerations (reactive power in a fairly capacitive transmission line...) \$\endgroup\$ – rdtsc May 25 '15 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.