Very confusing capacitor question

[burns12]

Homework Statement

C1=2[μF] and C2=6[μF] are first connected individually to their own 12[V] battery ; after disconnecting from the batteries, C2's + side is then connected to C1's − side
. . . the battery 2's + side is connected to the battery 1's − side , and the other sides are connected to C3=4[μF] ; after disconnecting from the batteries, C3's + side → C1's + side , and C3's − side → C2's − side .
(a) How much charge is on each capacitor
(b) what Voltage is across each Capacitor
(c) how much Energy does each store ?

Homework Equations

Q=VC, PE=.5CV^2

The Attempt at a Solution

I am at a loss as to how this circuit is supposed to end up looking. I ended up drawing 3 capacitors in series, C1 into C3 into C2. Now, to find the charge I just did Q=VC for each one and got answers, but then the next question is the voltage across them, if I use V=Q/C it's just going to give me the voltage I used in the last problem. So I'm doing one of those wrong. And the question is just very confusing to me.

Thanks for any help

 

[anathema]

!

Hello there,

First, let's clarify the circuit setup. The first step is to connect C1 and C2 separately to their own 12V battery. This means that C1 and C2 are in parallel with each other, each connected to their own battery. After disconnecting from the batteries, C2's positive side is then connected to C1's negative side. This creates a series circuit with C1 and C2.

Next, the positive side of battery 2 is connected to the negative side of battery 1, and the other sides are connected to C3. This creates a parallel circuit with C3 in parallel with the series circuit of C1 and C2.

Now, to answer the questions:

(a) To find the charge on each capacitor, we can use the equation Q=VC. For C1, we have Q1=C1V1, where V1 is the voltage across C1. Similarly, for C2, we have Q2=C2V2, where V2 is the voltage across C2. And for C3, we have Q3=C3V3, where V3 is the voltage across C3.

Since the capacitors are connected in series, the charge on each capacitor will be the same. So Q1=Q2=Q3=Q. We can also use the fact that the total charge in a series circuit is the sum of the individual charges. So Q=Q1+Q2+Q3.

(b) To find the voltage across each capacitor, we can use the equation V=Q/C. Using the same approach as above, we have V1=Q/C1, V2=Q/C2, and V3=Q/C3.

Since the capacitors are connected in parallel, the voltage across each capacitor will be the same. So V1=V2=V3=V. We can also use the fact that the total voltage in a parallel circuit is the same as the voltage across each individual component. So V=V1=V2=V3.

(c) To find the energy stored in each capacitor, we can use the equation PE=.5CV^2. Using the same approach as above, we have PE1=.5C1V^2, PE2=.5C2V^2, and PE3=.5C3V^2.

Since the capacitors are connected in parallel, the energy stored in each