[Physics] In what order would light bulbs in series light up when you close a long circuit

electric-circuitselectricity

For a few days, I was thinking of this question.

Lets assume we have a simple circuit that is 100 meters long. And lets say that we have bulbs A, B and C connected to the circuit's 30th, 60th and 90th meter relatively (from the + side). When we switch the system on, would all the bulbs light up at the same time? Or would A light up first and C last (or the opposite)?

Best Answer

I'm assuming that you're imagining a long, skinny, series circuit with three simple resistive lamps, like this:

    switch            A                      B                      C
  __/ _____________^v^v^v_________________^v^v^v_________________^v^v^v________
 |                                                                             |
 = battery                                                               short |
 |_____________________________________________________________________________|

(Sorry for the terrible ASCII diagram.)

The story we tell children about electric currents --- that energy in electric circuits is carried by moving electric charges --- is somewhere between an oversimplification and a fiction. This is a transmission line problem. The bulbs illuminate in the order $A\to B\to C$, but reflections of the signal in the transmission line complicate the issue.

The speed of a signal in a transmission line is governed by the inductance and capacitance $L,C$ between the conductors, which depend in turn on their geometry and the materials in their vicinity. For a transmission line made from coaxial cables or adjacent parallel wires, typical signal speeds are $c/2$, where $c=30\rm\,cm/ns=1\rm\,foot/nanosecond$ is the vacuum speed of light.

So let's imagine that, instead of closing the switch at $x=0$ and leaving it closed, we close the switch for ten nanoseconds and open it again. (This is not hard to do with switching transistors, and not hard to measure using a good oscilloscope.) We've created a pulse on the transmission line which is about 1.5 meters long, or 5% of the distance between the switch and $A$. The pulse reaches $A$ about $200\rm\,ns$ after the switch is closed and illuminates $A$ for $10\rm\,ns$; it reaches $B$ about $400\rm\,ns$ after the switch is closed, and $C$ at $600\rm\,ns$.

When the pulse reaches the short at the $100\rm\,m$ mark, about $670\rm\,ns$ after the switch was closed, you get a constraint that's missing from the rest of the transmission line: the potential difference between the two conductors at the short must be zero. The electromagnetic field conspires to obey this boundary condition by creating a leftward-moving pulse with the same sign and the opposite polarity: a reflection. Assuming your lamps are bidirectional (unlike, say, LEDs which conduct only in one direction) they'll light up again as the reflected pulse passes them: $C$ at $730\rm\,ns$, $B$ at $930\rm\,ns$, $A$ at $1130\rm\,ns$.

You get an additional reflection from the open switch, where the current must be zero; I'll let you figure out the polarity of the second rightward-moving pulse, but the lamps will light again at $A, 1530\,\mathrm{ns}; B, 1730\,\mathrm{ns}; C, 1930\,\mathrm{ns}$.

(Unless you take care to change your cable geometry at the lamps, you'll also get reflections from the impedance changes every time a pulse passes through $A$, $B$, or $C$; those reflections will interfere with each other in a complicated way.)

How do we extend this analysis to your question, where we close the switch and leave it closed? By extending the duration of the pulse. If the pulse is more than $1330\rm\,ns$ long, reflections approaching the switch see a constant-voltage boundary condition rather than a zero-current condition; adapting the current output to maintain constant voltage is how the battery eventually fills the circuit with steady-state direct current.

Note that if your circuit is not long and skinny but has some other geometry, then transmission-line approximation of constant $L,C$ per unit length doesn't hold and one of your other answers may occur.

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