Greetings earthlings. If a dielectric in a capacitor doesn't allow current to flow, how does c2 in circuit drawing get electron buildup on its negative plate? Furthermore, why do not both c1 and c2 build up a voltage equal to the source? Wahtinga (thank you for your help)

 

C1 and C2 are in series, thus acting like a higher voltage cap. The capacitance stays the same but the voltage is added together.

The 12 volts is essentially across one large capacitor rather than two independent caps. If you were to measure the voltage from negative to the center of the two caps you should find half the voltage (6V).

Two capacitors in parallel adds up the capacitance but not the voltage.

 

The dielectric does not allow current to flow "thru" the cap, it does not prevent current from flowing in or out of the cap.

 

C1 and C2 are in series, thus acting like a higher voltage cap. The capacitance stays the same but the voltage is added together.

In series, Capacitance stays the same?

 

Capacitance stays the same?

Have I got that wrong?

Thought that two 10µF caps in parallel equaled 20µF but the voltage stayed the same whereas two of them in series adds to their voltage capability without changing the capacitance. Feel free to correct me if I'm wrong. Better to admit ignorance and learn from it than to refuse to learn and remain ignorant.

 

When adding together, i.e capacitors in series, the reciprocal ( 1/C ) of the individual capacitors are all added together,

 

Two 10μF capacitors in series become 5μF.
The equivalent capacitance is calculated the same way as resistances in parallel.

1/Ceq = 1/C1 + 1/C2

In theory, the voltage is shared between the two capacitors. In practice, the inner voltage can be anything depending on the leakage in each capacitor.

 

This is a good video that explains the whole thing.
https://www.khanacademy.org/science.../circuits-with-capacitors/v/capacitors-series.

 

Greetings earthlings. If a dielectric in a capacitor doesn't allow current to flow, how does c2 in circuit drawing get electron buildup on its negative plate? Furthermore, why do not both c1 and c2 build up a voltage equal to the source? Wahtinga (thank you for your help)

The electric field between the capacitor plates store the electrical energy needed to 'push' current around the DC circuit until the separated charge potential at each capacitor are in equilibrium.