Charge on 4 capacitors in parallel then series

[neshepard]

Homework Statement

C. What is the charge on each of the four capacitors if switch S is closed? I have the charges if the switch is open, C1 and C2 are 8μC and C3 and C4 are 6μC since charges are same on series.

Homework Equations

The Attempt at a Solution

I know to find Ceq and work backwards, and that voltage in parallel is the same. But what happens to the volatage's and charges in both a series and parallel layout?
C1 and C3 are parallel Ceq=4μF
C2 and C4 are parallel Ceq=6μF
Each parallel group is in series.

 

[gneill]

A diagram would go a long way to making the question clear.

 

[neshepard]

Trying to figure how to add on.

 

[gneill]

I know to find Ceq and work backwards, and that voltage in parallel is the same. But what happens to the volatage's and charges in both a series and parallel layout?
C1 and C3 are parallel Ceq=4μF
C2 and C4 are parallel Ceq=6μF
Each parallel group is in series.

Call C1 in parallel with C3 "C13". Similarly "C24" is C2 in parallel with C4.
Now, C13 = 4μF and C24 = 6μF, as you've calculated above.

The first thing to do is determine how the 12V supply voltage will divide over the series combination of C13 and C24.

 

[rwisz]

Ctotal = Ceq1 + Ceq2 = 4μF + 6μF = 10μF

When capacitors are connected in parallel, the voltage across each capacitor is the same and the total charge on each capacitor is the sum of the individual charges. So if the switch is closed, the total charge on the four capacitors will be the same as when the switch is open. However, the charges on each individual capacitor may change depending on the capacitance and voltage of each capacitor. In this case, C1 and C3 are in parallel with each other, so they will have the same charge of 8μC. Similarly, C2 and C4 are in parallel with each other and will have the same charge of 6μC.

When capacitors are connected in series, the total capacitance decreases while the total charge remains the same. In this case, the total capacitance of the circuit is 10μF. This means that the total charge on the four capacitors will still be the same as when they were in parallel, but the individual charges on each capacitor may change. To find the charge on each capacitor, you can use the formula Q=CV, where Q is the charge, C is the capacitance, and V is the voltage. Since the total charge is the same, you can set up the following equations:

C1V1 + C2V2 = C3V3 + C4V4 (since the capacitors are in series)
V1 = V2 (since C1 and C2 are in parallel)
V3 = V4 (since C3 and C4 are in parallel)

Solving for V1, V2, V3, and V4, you can find the individual charges on each capacitor. Keep in mind that the voltage across each parallel group may be different, but the total voltage across the entire circuit will still be the same.