scott wrote:

> A 9V source has an internal resistance, which will initially take all 
> the voltage as the capacitor draws a huge amount of current (even if it 
> doesn't, the wires will have some resistance).  So across the capacitor 
> terminals will be a very low voltage, but a very high current through 
> it.  Once the capacitor is fully charged, there will be no current 
> flowing, and so no voltage drop across the internal resistance, and then 
> the full 9 V across the capacitor.
> 
> So you see, in this simple case, voltage across the capacitor is 
> definitely not proportional to current flowing through it.  Quite the 
> opposite to a resistor.

Right. So the potential difference between the terminals of a battery is 
9 V, unless there happens to be a capacitor connected to them, in which 
case the potential difference is magically 0 V despite the fact that a 
vast current is being generated?

Right. Sure. Makes perfect sense. Electrons just move of their own 
volition anyway; it's not like it requires a *force* to drive them or 
anything...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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