Circuits - using equivalent capacitance

[shawli]

Homework Statement

C1 = 1μC ; C2 = 2μC ; C3 = 3μC ; C4 = 4μC
V = 12.0V

If both switches are closed, what is the charge on C₁?

Homework Equations

q = CV

In parallel - V is constant across capacitors
C_xy = C_x + C_y

In series - q is constant across capacitors
C_xy^-1 = C_x^-1 + C_y^-1

The Attempt at a Solution

I'm REALLY lost with this circuit.
This is what I broke it down to:

(C1 || C2) in series with (C3 || C4)

And then I attempted to work backwards from there, but I can't seem to get the correct answer (which is 8.40μC).
I added rough sketches of the circuit I was trying to break this down into... Not sure if I'm even on the right track!

 

[gneill]

Homework Statement

C1 = 1μC ; C2 = 2μC ; C3 = 3μC ; C4 = 4μC
V = 12.0V

Are C1 through C4 meant to represent capacities (microfarads) or initial charges (microcoulombs)? Units make a difference!

Assuming that they are actually capacities in μF,

. . .


I'm REALLY lost with this circuit.
This is what I broke it down to:

(C1 || C2) in series with (C3 || C4)

That's good. What values did you get for C12 and C34?

Are you familiar with the voltage divider equation for series capacitors?

 

[shawli]

Sorry - the "C" means capacitor, I should have clarified in the question!

Is this the voltage divider equation? http://www.electronics-tutorials.ws/capacitor/cap_7.html I've been trying to read up on this stuff but no, I don't actually know what it is haha.. (have yet to cover it in class)

I got C12 = 3μF and C34 = 7μF (because you simply sum up voltage across parallel capacitors right?).

I then got Ceq = 2.1 μF (taking C12 to be in series with C34).

 

[gneill]

Sorry - the "C" means capacitor, I should have clarified in the question!

Is this the voltage divider equation? http://www.electronics-tutorials.ws/capacitor/cap_7.html

Yes, that page depicts the concept quite well. For the case of just two capacitors in series you can just remember that the total voltage across them divides inversely as the individual capacitances. So if capacitors Ca and Cb are in series and have total voltage E = Va + Vb across them, then
Va = E \frac{Cb}{Ca + Cb} ~~~~~~\text{and}~~~~~~Vb = E \frac{Ca}{Ca + Cb}

I've been trying to read up on this stuff but no, I don't actually know what it is haha.. (have yet to cover it in class)

I got C12 = 3μF and C34 = 7μF (because you simply sum up voltage across parallel capacitors right?).

I then got Ceq = 2.1 μF (taking C12 to be in series with C34).

That's correct. If you can determine the voltage across the capacitor that you're interested in (C1) then you can determine the charge on that capacitor from your formula q = CV. The voltage across C1 must be the same as the voltage across C1||C2, which you can determine with the voltage divider concept.

 

[shawli]

Ah, got it! Thank you :)