Hi:
I am having difficulties understanding the nature of voltage / current / charge etc. of capacitors.
Could somebody please let me fire a series of questions at them as I work my way through the basics?

To start with:
Please see the attached diagram:

When all 3 capacitors in this circuit are fully charged, is the voltage across all of them the same 12V. That of the supply voltage?

 

The voltage at the junction between C2 and C3 will be somewhere between 0v and 12v.

It can be calculated what the voltage at that junction is by the ratio of the two capacitor's uF rating to the voltage applied.

Have a look at our E-book: http://www.allaboutcircuits.com/vol_1/chpt_13/4.html

Note that some software simulations don't support two capacitors in series wired as in your schematic, as there is no path to ground with "ideal" capacitors.

 

I'm no capacitor expert, but wouldn't the 6uF capacitor limit the amount of charge that could pass into the 9uF capacitor so they both had the same charge?
Would that mean the midpoint was 6V?

 

They would have the same charge in terms of coulombs, but the voltage for a given charge is inversely proportional to the capacitor value; a 1 Farad capacitor holding one coulomb has 1V across it, but a 1 Microfarad cap would have one Million volts on it.

 

may i please know that in series RC circuit with dc pulse current source , the power calculated for the circuit is I2 * R ??? Am i right about it?

 

So - trying to keep on topic here - The maximum charge a capacitor can hold is not the same as the voltage source. It depends on the capacitance of the capacitor.

The capacitors in series are calculated the same way as resistors in parallel, giving a total capacitance of 3.6μF. This is because of the increase in distance between plates when they are in series. I can then find the value of charge (Q), by Q=CV, which is (3.6x10^-6)(12)=43.2μC. I have to have the capacitor values combined before I use Q=CV because the increase of distance between plates down that branch in the circuit will affect the charge along that line.

Once I know this, the charge through C2 and C3 is the same (Qt=Q1=Q2=Q3), so I can then find the voltage drop across the C2 capacitor using V=Q/C, which gives 43.2F/6.0μF=7.2V
C3 can be calculated in the same way.
But charge is the quantity of electrostatic electricity that can be held in a capacitor, right? So how can this be the same for both capacitors of different values. Can I think of charge like current in a resistor circuit?

In series, capacitors give an increase in diameter between plates, giving a decrease in capacitance, which will be less than any individual capacitor.
In parallel, capacitors give an increase in plate size, giving an increase in capacitance, which will be more than any individual capacitor.

With ohms law - V=IR, what are the equivalents in a capacitor circuit?

 

Well - I know we're not supposed to answer our last posts but I just found this:

If a source of voltage is suddenly applied to an uncharged capacitor (a sudden increase of voltage), the capacitor will draw current from that source, absorbing energy from it, until the capacitor's voltage equals that of the source.

Here:
http://www.allaboutcircuits.com/vol_1/chpt_13/1.html

Which was my original question:

So - trying to make sense of this still:
Charge is the quantitative measure of the number of electrons passing by a point in the circuit - not the amount of electrical energy held on a capacitor. So the charge across capacitors in series is the same, therefore.
The voltage across the capacitors when fully charged is the same as the source voltage.
Is this correct please?

 

I didn't know much about series capacitors before yesterday, but it is making more sense now. Thanks for showing your workings.
I had a quick go in a simulator and it is giving the 7.2V you said across the 6uF capacitor.
The charge is the same in both capacitors because the smaller one is the limiting factor. Once it is "full" the larger one can't get any more charge.

 

Hi Mark:
I haven't heard about any limiting factors to do with capacitors, but I can see how that would make sense. Once that capacitor is charged fully then there would be no more electrical attraction to the plate. But wouldn't the second capacitor plate still have some attraction that pulls electrons from the plate ahead of it? But then they would need to jump the positive plate as well.

Simulator?