Capacitors In Series

Doc Al

Sure.

Bashyboy

Oh, okay, I see. The impression that I was getting from my textbooks was that the electric potential HAD to drop because of the series configuration. But now I know that isn't true. Thank you, everyone.

Doc Al

What do you mean?

The potential difference across each capacitor is what we are talking about. (Or so I thought.)

Doc Al

Just to be clear, here's an example:

Imagine three identical capacitors hooked up in series. The entire string is attached to a 9 volt battery. Once they are fully charged, each capacitor will have a potential difference of 3 volts.

If the three capacitors were not the same, then the voltage across each would not be equal. But the total voltage across all three must still add up to 9 V.

Make sense?

sophiecentaur

This is the problematic word. We may all be talking at cross purposes here (not an uncommon situation and easily rectified by using diagrams and labels). For two identical capacitors, in series, the PD across each capacitor will be the same or the Potential (reference Earth) on the positive terminal of the 'upper' capacitor will be twice the Potential on the positive terminal of the 'lower' capacitor.

stevendaryl

Yes, if the capacitors in series had the same capacitance, then the voltage drop across each capacitor would be identical. (I'm using "voltage drop" rather than "electric potential", because voltage is the terminology that's usually used in circuits. But it's the same thing as electric potential.)

stevendaryl

This might be a confusion of terminology. Let's make this explicit:


In this picture, we have a 9 volt battery connected in series with two capacitors. The voltage drop--that is, the change in voltage--across the first capacitor is [itex]Q C_1[/itex]. The voltage drop across the second capacitor is [itex]Q C_2[/itex]. If the capacitors are identical, then these two voltage drops are equal. But that doesn't mean that the voltages are equal. The voltage at the positive plate of the first capacitor is [itex]V_2[/itex]. The voltage on the negative plate of the first capacitor is [itex]V_3[/itex]. The voltage on the positive plate of the second capacitor is [itex]V_3[/itex]. The voltage on the negative plate of the second capacitor is [itex]V_1[/itex]. If [itex]C_1 = C_2[/itex], then we would have:

[itex]V_3 - V_2 = V_1 - V_3[/itex]

The voltage drops would be the same.

Crazymechanic

I apologize for my off the mark post yesterday I was kinda sleepy and I somehow thought that your talking about capacitors in parallel.

Well anyways ofcourse you don't get the same voltage on the first plate of the second capacitor.This circuit you draw is a voltage divider.A capacitor one.
They have the same ones with resistors too.
This kinda circut was used in the older smps power supplies in the primary mains circuit to switch from 220v to 110v

In theory the voltage drop across each capacitor assuming the capacitance is the same should be equal.Well the voltage you get is depending on where you measure with respect to the ground.
Or do you want to know why is the voltage divided in capacitors in series?

sophiecentaur

If they both start off uncharged and have the same C, when they have the same current passed for the same time, they will have the same pd each.

Bashyboy

All right, I understand, now. Thank you every one.