Find the individual charge (and potential difference?) on each capacitor

[Color_of_Cyan]

Homework Statement

Four capacitors are connected as shown in the figure below. (Let C = 18.0 µF.)

Find the equivalent capacitance between points a and b

Calculate the charge on each capacitor, taking ΔVab = 17.0 V

Homework Equations

Q = CΔV

C = Q / ΔVΔV1 = ΔV2 = ΔVtot (for parallel capacitors)

Q tot = Q1 + Q2 (charge on parallel capacitors, it says)ΔVtot = ΔV1 + ΔV2 (for series capacitors)

Q = Q1 = Q2 (charge on series capacitors, it says)

The Attempt at a Solution

The only thing I could solve was the equivalent capacitance across, which is 6 x 10^-6

And I broke it up like this (and it's correct):
http://desmond.imageshack.us/Himg213/scaled.php?server=213&filename=prob1m.jpg&res=landing I have NO IDEA at all how to determine the individual charge on each of them, even with the formulas...

I can see ΔV across series is supposed to add up to 17V but no idea how.

Where should I start on that?Edit: ok used Q = CV for the (20 x 10-6 f) capacitor and got (102 x 10-6 C) because charge is supposed to be the same across series

 

[truesearch]

What you have done so far is correct...now find the total capacitance of the 2 in series... (You were able to find the equivalent of C and the 3uF in series !) then you should be able to get the total charge.

 

[Color_of_Cyan]

ok i got them all

you need to just mainly worry about using total charge Q = Q1 = Q2 for series capacitors and V tot = V1 = V2 for parallel capacitors, just to find the individual potentials / charges.

Then you need to work your way down with these and the other two formulas.

 

[truesearch]

thats correct !

 

[asteriatic]

capacitors.

To find the individual charge on each capacitor, you can use the formula Q = CΔV. Since the capacitors are connected in series, the potential difference (ΔV) across all of the capacitors will be the same, which is 17V.

Using this information, you can calculate the charge on the first capacitor (20 x 10^-6 F) by plugging in the values: Q = (20 x 10^-6 F)(17V) = 340 x 10^-6 C.

Since the charge is the same for all capacitors in series, the charge on the second capacitor (18 x 10^-6 F) will also be 340 x 10^-6 C.

For the two capacitors in parallel (10 x 10^-6 F and 6 x 10^-6 F), you can use the formula Q = CΔV to find the individual charges. The potential difference (ΔV) across these capacitors will also be 17V.

So, for the first parallel capacitor (10 x 10^-6 F), the charge would be Q = (10 x 10^-6 F)(17V) = 170 x 10^-6 C.

And for the second parallel capacitor (6 x 10^-6 F), the charge would be Q = (6 x 10^-6 F)(17V) = 102 x 10^-6 C.

Therefore, the individual charges on each capacitor are:

- 340 x 10^-6 C for the first series capacitor (20 x 10^-6 F)
- 340 x 10^-6 C for the second series capacitor (18 x 10^-6 F)
- 170 x 10^-6 C for the first parallel capacitor (10 x 10^-6 F)
- 102 x 10^-6 C for the second parallel capacitor (6 x 10^-6 F)