Car headlight voltage If I put a higher voltage bulb in my car headlight, what will happen?  I haven't tried it yet, but I'm wondering if it would matter.  The bulb has the same wattage but a different voltage.
 A: A standard experiment in a Physics teaching laboratory is to find the current-voltage characteristics of an incandescent tungsten filament light bulb.
Results from such an experiment show that the resistance $(=\dfrac{\text{voltage across light bulb}}{\text{current through light bulb}}= \dfrac VI)$ of an incandescent light bulb is not constant.
Here is a graph of current against voltage from such an experiment which shows that as the voltage increases the resistance of the bulb increases.  

If the resistance had been constant then the graph should have been a straight line through the origin.
This increase in resistance is due to the fact that the filament is made of tungsten whose resistance increases with the temperature of the filament.
The starting characteristics of a light bulb show this and the large current when the bulb is switch on is one of the reasons for the failure of a light bulb. 
 
Performance characteristics from Beaty and Fink's Standard Handbook for Electrical Engineers (pages 26-8 to 26-9) show that for small variations from the working voltage the expected life time of a bulb is proportional to $\rm voltage ^{13}$ so if you connect a bulb with a working voltage of $12 \rm V$ to an $11 \rm V$ supply it would be expected to last approximately three times longer.  
The Handbook also relates the current to be proportional to $\rm voltage ^{0.54}$ and the power dissipated by the bulb to be proportional to $\rm voltage ^{1.54}$.
Thus connecting your $12 \, V$ bulb to an $11 \rm V$ supply will reduce the power consumption to approximately $\frac 78$ of its working value.  
The consequence of the reduced power dissipation is that the filament will be at a lower temperature and hence not emit as much visible light.
A unit for the measure of visible light emitted from a source is the lumen and the amount of visible light emitted by a bulb is proportional to $\rm voltage ^{3.38}$
Running your bulb at $11\rm V$ will reduce the number of lumens emitted by the bulb to 
$\frac 34$ of its working value, in other words the bulb will appear to be dimmer.  
As well as being dimmer the spectral composition (colour temperature) of the bulb will change.
This graph, copied from the Wikipedia article Color Temperature shows the effect on the visible part of the spectrum when the temperature of the source is changed.  
 
$B_\lambda$ is a measure of the rate at which light is emitted over a range of wavelength from $\lambda$ to $\lambda + d\lambda$ so the area under the graph is proportional to the emitted power and the $\rm M$ is the micro reciprocal degree $(=\frac{10^6}{T(\rm K)})$.  
Running at the lower voltage the colour temperature of the bulb will change in proportion to  $\rm voltage ^{0.42}$ of its original value which in this case will be approximately $96\%$.  
The Handbook has a graph which may be of use?  


I have assumed the it is an incandescent tungsten filament light which the OP was referring to.
What I find extremely interesting is that my reasonable new edition of the Handbook does not mention LED lighting once!
A: A bulb rated at a higher voltage means that the bulb has a higher resistance for the same power. If you run it at a lower voltage, you will get a lower power output so it will simply emit less light.
As an example, imagine a bulb rated at 1V with 1W. The two equations you need are Ohm's Law ($V=IR$) and $P=VI$. So from this, with a bit of algebra, you can get that:
$$
R=V^2/P
$$
and
$$
P=V^2/R
$$
So for this bulb $R=1^2/1=1\Omega$.
Now take another bulb rated at 2V and 1W. This means it gives 1W output when supplied by a 2V battery. So its resistance must be $R=2^2/1=4\Omega$.
If you run this second bulb off a 1V source (which is what you want to do), it's power is given by:
$$
P=V^2/R
$$
So $P = 1^2/4 =$ 0.25W. It gives out a quarter of the power it gave at 2V.
The point is that the thing that is fixed is the bulb resistance. The voltage is externally defined by the power source and the current and power are dependent on the resistance and voltage.

[I assume here we are dealing with incandescent, resistive bulbs. LEDs
  or fluorescents don't work the same way...]

