Richard,

You have to see it in a different way.

Let's use your example of a 60W bulb drawing 5A @ 12V.

Using Ohm's law this bulb has a resistance of R=U/I=12/5=2.4Ohm.

Now let's say that in this circuit there is another resistor, a bad joint or partly broken cable for example.

Let's for example consider a very thin and rather long wire. Here the resistance is calculated by R=rho*l/A, rho (the greek symbol) being the specific resistance, depending on the material. For copper it has a value of 0.01678Ohm*mm^2/m, l stands for the length of the wire in m, A is the cross sectional area in mm^2. You have to imagine this as a hosepipe. The thinner and longer, the less water will go through it, given that the water pressure is constant, or like the artery example Bill proposed.

If we now consider a wire with a cross section of 0.2mm^2 and a length of 20m this will give a resistance of approx. 1.7Ohms (R=0.01678*20/0.2)

This resistance of the wire is connected in series to the bulb, so we can model this whole system as two resistors in series connected to a 12V supply. The two resistors add up, being connected in series, so the overall resistance equals to 2.4+1.7=4.1Ohms.

If we now calculate the current with I=U/R=12/4.1 we get 2.93A. You can see that the added resistance caused by the wiring reduces the overall current through the circuit and therefore the power consumption of the bulb, which will be rather dark now.

If we have the 2.93A through our 1.7Ohms wiring, we will get a voltage drop of U=R*I=1.7*2.93=4.981V, meaning that the bulb will only "see" 12V-4.981V=approx. 7V now.

You have to regard the bulb as a constant resistor (it is not really, as it changes it's resistance with temperature, but we will neglect this for our modelling). You can't say the bulb draws 7.5A at 8V, because this would equal a resistance of R=8/7.5=1.07Ohms, while it really has 2.4Ohms, like calculated before.

A 6V/60W bulb has just a quarter of the resistance of a 12V/60W bulb just for this very reason (R=6V/10A=0.6Ohms), so at 12V it would draw I=12/0.6=20A, that's what causes it to burn in a 12V circuit. Following you logic it would simply draw 5A and shine along happily at 60W, which we all know it doesn't.

Hope these little examples helped a bit to foster understanding.

***smart-idiot mode off***

Cheers, Markus