In what order would light bulbs in series light up when you close a long circuit? 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)?
 A: The information about you flipping the switch will have to propagate so that current can flow. Thus the bulb the closest to the switch will light up first.
Think of it in the water analogy. You have a long channel, with a gate in the middle. One side of the gate is flooded (high voltage), the other is dry (low voltage). If you flip the switch (open the gate), where will the current flow first? Of course the water the closest to the gate.
A: 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.
A: It depends on the Characteristic Impedance $Z$ of the cable, the output resistance $Z_0$ of the voltage source and the total resistance $R$ of the three lights.  We assume the resistance of the wires is negligible.
Note that the Characteristic Impedance is a function of the geometry of the wire and the dielectric insulator (think of a 75 ohm co-ax cable) and is not the resistance of the wire.  A superconducting  75 ohm co-ax with zero resistance wires still has a Characteristic Impedance of 75 ohms.
Why?
Because when you apply a voltage to a cable, the initial current flows through the self inductance of the wire and charges the self capacitance of the wire. Hence the initial current is determined by the ratio of the self inductance/metre to the self capacitance/metre. This function is called the Characteristic Impedance of the cable. 
After things have settled down, the current $I = \frac{V}{R+Z_0}$.  Assume the bulb needs this current to light.  
Before things have settled down a step voltage travels from left to right charging up the self capacitance of the wire and hindered by the self inductance of the wire.  The voltage is forced to zero when it reaches the short circuit and a second step is reflected back to the source. Because $Z_0 = Z$ no further reflection takes place. So, steady state is reached after 2 x time to travel along the wire.  
Light travels about 1 foot in 1 nanosecond, and electricity at about 2/3 this speed, so this will take about 300 nanoseconds for a 100 foot long cable.
On closing the switch a step function of Voltage $V_s = V\times \frac{Z}{Z+Z_0}$ travels from left to right.  The step function current which flows is $I = (V\times Z /(Z+Z_0) / Z.$    
If the Characteristic Impedance $Z$ and output resistance $Z_0$ are much less than resistance $R$, this current is greater than than $I = V /(R+Z_0).$  The near bulb lights first.  We don't have to consider what happens when the wavefront passes the first lamp as the question is answered.  It is a bit more complex when you account for the lamps.
If the Characteristic Impedance Z is much greater than resistance R, this current is much less than than $I = V /(R+Z_0).$  No bulbs light until the current/voltage wavefront is reflected back from the short circuit.  The far bulb lights first.  As $Z_0 = Z$ no reflection takes place when this step arrives at the voltage source and the steady state current $I = V/(R+Z_0)$ is now flowing along the wire. We can ignore the lamp resistance because we have said $Z_0$ is much greater than R.
It is much more complex if $Z$ is of the same order as R as you will have to allow for the volt drop as the current flows through the lamps.  If the output impedance of the voltage source is not equal to $Z$ there will be lots of reflections back and forth.  In fact, by judicious setting of the values you could probably arrange things so that any bulb lit first.  There would also probably be situations where a bulb lit and then went out, possibly several times, before finally remaining alight.
Search out a book on Transmission Line Theory and all will be explained.  Try Pulse and Digital Switching Waveforms by Millman and Taub - it is available as a PDF.  
See Chapter 3 - Pulse Transformers and Delay Lines and Appendix C - Lumped Parameter Delay Lines.
A: Neither of those choices,in fact 
C then A then B would light up with very small time intervals. No human could ever directly notice these delays.   
Energy emerges from the battery both in terms of a supply of electrons from the negative terminal and the deficiency of electrons at the positive terminal.
These will "apply pressure to" and "remove pressure from" the adjoining wire segments in opposite directions. Such application of pressure and reduction in pressure continues around the circuit forming a small charge movement, i.e. a current. The wire and other conditions being uniform, the speed of the electricity being constant (somewhere between 50% to 99% of the speed of light in vacuum), C will feel some current first as it is nearest to the battery, and before the electricity has even begun to flow all the way around the circuit. The next nearest is A. Therefore C and A will receive earlier thermal excitation, and (all other things being equal), will start to radiate earlier than B.
A: Each electron experiences both inertia and the effect of inductance resisting the direction of travel, this causes the electrons transition from one atom to another to lag with respect to the time the switch is closed. Electrons exit the circuit at the same rate as they enter therefore all the globes would light simultaneously but would take a time period to reach full brightness. So I agree with the last answer.
A: They are in series.  They will light at the same time.  Imagine the switch or batteries in different locations.  Since everything is in series it is all the same.  The current through the switch will be the current through the lights which is the same as the wire.  Kirchhoff's current Law. Sum of current at a node = 0.
