# Capacitor charged and disconnected then connected to capacitors

1. Jul 21, 2012

### careless25

Initially, a 15uF capacitor is charged by a 100V battery. Once charged, it is disconnected from the battery and connected to two uncharged capacitors(C_2 = 20uF, C_3 = 30uF) in parallel. What is the charge and potential difference on each capacitor.

(The two capacitors mentioned above are in series to each other but parallel to the charged capacitor.)

So, I found out the Q on C_1 as being 1.5mC when fully charged.

Now since they are in parallel the voltage across the two uncharged capacitors is 100V.
I find the equivalent capacitor of those two as 12uF and the charge being 1.2mC.

Now I am confused about how there is 1.5mC on C_1 but just 1.2mC on the two capacitors in series.
Shouldn't the charge across the whole circuit be the same? Or am I mistaken?

2. Jul 21, 2012

### vk6kro

When you add the extra capacitors, the voltage on the combination will drop.

I make the series combination 12 μF and the parallel combination 27 μF when you add the 15 μF.

The charge must be constant,

Initial charge = Q = C * V = 0.000015 Farads * 100 volts = 0.0015 Coulombs, as you said.

So, the new voltage will be Q = C * V

or 0.0015 Coulombs = 0.000027 Farads * V volts

V= 55.56 volts