Capacitor Basics: How Voltage Suppliers Impact Charging

[smyth]

Hi guys,
can you please explain me what happens in the circuit I've attached? It looks easy but still puzzles me. I mean if one voltage supplier was missing i know that the two capacitors are connected in series, and they will charge. Having two voltage suppliers do the capacitors still charge? how?
Please consider:
case 1: E1=E2
case 2 E1>E2
http://i45.tinypic.com/10h7lap.jpg

 

[NascentOxygen]

A practical circuit should show some series resistance, as well, even if it is small.

KVL applies, so the sum of the voltages around the loop=0. If the capacitors are initially uncharged, explain what happens.

 

[smyth]

it's not intended to be practical, i just want to understand how it woks/eliminate bugs i have in understanding electricity. The capacitors are initially uncharged.
Like i said, if the circuit would have had only one voltage supplier, I'm pretty sure the answer is that there will be a current until the capacitors charge completely , and then the flow stops . The same happens here? the + of a voltage source and the - of another one can charge a capacitor? if i were to measure the voltage across one capacitor how much it will be?

 

[smyth]

applying KVL ... well that would be easy if instead of capacitors there will be resistors. voltage drops across the resistances must equal the sum of the two sources (according to the signs on the circuit). but here there are only capacitors, so i guess (concluding that they charge up) we can consider them, when fully charged, as just another sources with opposite polarization, so that the sum of all is 0, and if the C1=C2 , the voltage across the capacitors would be (E1+E2)/2

 

[Ratch]

symth,

can you please explain me what happens in the circuit I've attached?

Yes.

It looks easy but still puzzles me.

It is easy. Why does it puzzle you?

I mean if one voltage supplier was missing i know that the two capacitors are connected in series, and they will charge.

The capacitors are in series no matter how many voltage sources there are, or what the value of each source is. Since the voltages are connected in series, the total voltage is the sum of each voltage source value.

Capacitors do not charge. The net charge of a capacitor is zero whether there is zero volts across the plate or X number of volts across the plates. That is because for every electron pushed on one plate, another electron is pulled from the other plate for a net charge change of zero. The imbalance of electrons between the plates causes a voltage to form, which stores electrical energy in an electrostatic field. Therefore, a capacitor becomes energized, not charged.

Please consider:
case 1: E1=E2
case 2 E1>E2

The total voltage available to energize the caps is the sum of the voltage sources, no matter what the values are in relation to each other.

i just want to understand how it woks/eliminate bugs i have in understanding electricity.

A noble goal. Another goal would be to to work on your capitalization.

The capacitors are initially uncharged.

OK, no initial charge imbalance? At the end of the transient time, each capacitor in series will have the same charge imbalance.

Like i said, if the circuit would have had only one voltage supplier, I'm pretty sure the answer is that there will be a current until the capacitors charge completely , and then the flow stops . The same happens here? the + of a voltage source and the - of another one can charge a capacitor? if i were to measure the voltage across one capacitor how much it will be?

Yes, the two voltages sources in series act as one voltage source, just like a connecting batteries in series. Both voltage sources are in the same current loop.

The total voltage is Et, and the total capacitance in series is Ct. The voltage across each cap after the transient period will be Et*Ct/C1 and Et*Ct/C2 .

Ratch

 

[smyth]

Thanks a lot Ratch,

and to briefly answer your observations/questions:

It is easy. Why does it puzzle you?

Because i don't handle electronics to well, and there are a lot of misunderstandings in my head (as you pointed out in the reply).

A noble goal. Another goal would be to to work on your capitalization.

:), sorry for that. I'll try to use it right, and also skip unnecessary information.

One more question.
If we eliminate one of the capacitors, leaving the loop interrupted, will the remaining capacitor energize?

 

[NascentOxygen]

it's not intended to be practical, i just want to understand how it woks/eliminate bugs i have in understanding electricity. The capacitors are initially uncharged.
Like i said, if the circuit would have had only one voltage supplier, I'm pretty sure the answer is that there will be a current until the capacitors charge completely , and then the flow stops . The same happens here? the + of a voltage source and the - of another one can charge a capacitor? if i were to measure the voltage across one capacitor how much it will be?

Without resistance, you have to say the capacitors charge instantly by an infinitely large pulse of current, which immediately ceases. After this, the capacitors are left charged such that the sum of their voltages equals the sum of the voltage sources. Since they are in series, equal charge is added to each (identical currents for identical times).

If there is not a complete circuit for current to flow, then no current will flow, and no charge will be added to either capacitor.

 

[Ratch]

smyth,

Of course. Anytime you put voltage across a cap, it will energize and store electrical energy.

Ratch

Edit: Sorry I did not read your word "interrupted" correctly. The circuit has to be complete before the cap can energize.

 

[sophiecentaur]

All you can say, definitely about your circuit is that the PDs across the voltage sources will be E1 and E2. If your capacitors are ideal then the PD across C1 could be anything (they could have been charged to 1kV prior to being connected) but the PDs on C2 would end up just E1 + E2 greater (or less - according to the sign of the 1kV)