Assuming the following cicuit:
A 9V cell in series with a capacitor of 0.0001 F and a bulb.
My question is,
As the capacitor charges up, does the potential difference across the bulb drop?
I can't really get my head around it....
The capacitor is fully charged when there is a potential difference of 9V across it... but what about the bulb? Or does the capacitor not charge up to 9V but is charged up to less perhaps 7.5V and the remaining 1.5V is across the bulb?
Just some clarification on how the other components affect the charging of the process with regards to the potential difference would be greatly appreciated.
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Capacitor Charging watch
- Thread Starter
- 29-09-2010 17:54
- 29-09-2010 18:55
The potential difference across the bulb will be given by V=IR for the bulb.
As R is constant, the pd will depend only on the current flowing through it.
When you first connect the cell to the circuit there is no charge (and therefore no pd) on the capacitor.
As a result a current flows from the cell to the capacitor. The capacitor charges up and the pd across it increases. (V=Q/C)
When the capacitor is fully charged, the pd on it is equal to the emf of the cell and no more charge can flow. The current, in other words, falls off (exponentially) to zero.
As for the bulb. The current flowing through it, the charging current, is large to start with and the pd across it is equal to the emf of the cell. As the charging current falls to zero, the pd across it also falls to zero.
At the end there is no current flowing in the circuit so no pd across the bulb. All the pd is across the capacitor.