Invisible wrote:
> 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?
...

Remember that ideal components does NOT exist.

You might want to read about resistance of conductors, internal
resistance of voltage sources and equivalent series resistance (ESR)
of capacitors:

http://en.wikipedia.org/wiki/Electrical_resistance#Resistance_of_a_conductor
http://en.wikipedia.org/wiki/Voltage_source
http://en.wikipedia.org/wiki/Internal_resistance
http://en.wikipedia.org/wiki/Equivalent_series_resistance

-- 
Tor Olav
http://subcube.com

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