Find Capacitor network with a known potental difference

[Northbysouth]

Homework Statement

For the capacitor network shown in the Figure , the potential difference across ab is 12.0 V

A) Find the total energy stored in this network.

B) Find the energy stored in the 4.80μF capacitor.

Homework Equations

C = Q/V

The Attempt at a Solution

I'm honestly not sure where to start. I've tried playing around with the units but I haven't managed to find anything that gives me joules.

 

[nasu]

Playing with the units is not the best way to solve the problems.

If you want to calculate energy, you may start by looking up the formula for the energy stored in a capacitor.

 

[Northbysouth]

I managed to get it I found that the equation for the energy stored in a capacitor is:

U = 1/2*CV^2

I found C by breaking it up into the separate capacities. So, for the left most capacities I found their equilibrium capacitor, which I did by adding the reciprocal of the values:

1/8.60 + 1/4.80 = 0.324612

This becomes 3.080597 when you take it's receiprocal.

Then I found the capacitor for the top-most capacitors.

1/6.20 + 1/11.8 = 0.24603

Taking the reciprocal gives me 4.06444

Then I added 4.06444 to 3.50 because these capacitors are parallel, whereas previously the capacitors had been in series. This gave me 7.56044

Then I did:

1/7.56044 + 1/3.080597 = 0.45687

For which I remember to take the receiprocal, thereby giving me 2.1887593 micro-F

Plugging this into the formula gives me:

U= 1/2*2.1887593 micro-F*12^2
U = 157.59 micro-J

Thanks for your help

 

[Northbysouth]

B) Find the energy stored in the 4.80 microFarad capacitor

I tried the following to answer this part:
U = 1/2*4.80 micro-F*12^2
U = 345.6 micro-J

Unfortunately, this didn't work. What am I missing?

 

[nasu]

The potential difference across that capacitor is not 12 V.
You can look at the circuit with the two rightmost capacitor plus one that replaces all the left most ones. So you have a series circuit with three capacitors which should be something familiar.
Now you know something about the charge for series circuits. It is the same for all (and also the same as for the equivalent capacitor). This should be enough to calculate the charge and then the voltage for the desired capacitor.

 

[Northbysouth]

I'm not quite sure I follow. If there are three capacitors then wouldn't it be a parallel circuit?

 

[nasu]

I am talking about the capacitors 8.6 μF (leftmost), 4.6 μF and the equivalent of that loop on the right. You said that the 8.6 μF and 4.6 μF are in series, right? If you replace the loop with the equivalent (I think you got 7.56 μF) then you have the 8.6 μF, 4.6 μF and this 7.56 μF in series. Do you see it?

 

[Northbysouth]

So here's what I tried:

12 V = Q(1/8.60 + 1/4.80 + 1/7.56044)
Q = 5.482558

V = 5.482558/4.80 micro-F
V = 1.142

I've tried this and it didn't work; I think I've messed up the units as I'm not sure what the units of Q would be.

 

[Northbysouth]

I've just looked at it again and realized that this is the voltage, not U.

But pluggin in 1.142 V into U = 1/2*4.80 micro-F*1.142^2
= 3.1299 micro-J

But it's still not right :/

 

[nasu]

The value of the charge is not right.
Check your calculations.
Q=C*V

 

[Northbysouth]

Finally got it:

So in calculating the capacitance I got

1/C = 1/8.60 + 1/4.80 + 1/7.56044

Therefore C = 2.1887 micro-F

Then, using V = Q/C I get

12 V = Q/2.1887 micro-F
Q = 26.2647

Then using V = Q/C I get

V = 26.2647/4.80 micro-F = 5.4718 V

U = 1/2*4.80 micro-F*5.4718^2
U = 71.9 micro-J

Thank you for all your help.