> Presumably the resistence of the LED is finite and fixed.

No.  Here's a graph of voltage against current for a typical LED:

http://www.cq.cx/pics/int-led-vi.png

Note that for a fixed value resistor this would be a straight line starting 
at the origin.

Say you want to operate the LED at 20 mA (brightness/lifetime etc of LEDs is 
all related to current, not voltage).  That means you need to drive it at 
exactly 2.0 V.  You need to add a resistor in series to connect it to eg a 
5V supply.  So you need 3V across the resistor with 20 mA flowing through 
it, R=V/I=3.0/0.02= 150 ohms.

Of course there are more complex LED driving circuits that actively keep the 
current constant over a wide range of operating conditions, some can even 
drive chains of LEDs in series and flag up when there are failures etc.

> Now, see, I've never been able to comprehend stuff like this. To me, this 
> diagram just looks like when you close the switch, all the current will 
> flow straight from one rail to the other, shorting out the battery and not 
> providing any current at all to the input of the gate.

Yes, *some* current will flow from one rail to the other, which is why it is 
important to use a *high* value resistor.  If you use a 10kOhm resistor 
across 5 Volts, that's I=V/R=0.5mA which is hardly going to cause a problem. 
But the important thing is, that point in the middle that is directly 
connected to the input will be at 5V when the switch is closed.

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