Competition - post a photo you've taken of nature >>

Capacitance questions - bit lost

shorty.loves.angels

Just as I thought I was getting through it.

I can't follow capacitance questions, I've tried keeping rules/ equations in mind but they don't seem to take me to the right answer.

Q1

Three capacitors, each of 120microfarads capacitance are connected together in series. This network is connected to 10kV supply.

a) Calculate combined capacitance.

(1/120 + 1/120 + 1/120)^-1 = 40microfarads

b) Charge stored.

Q=VC 10x10^3 x 40x10^-6 = 0.4C

c) Total energy stored.

Erm. 1/2CV^2?

So 0.5 x 40x10^-6 x (10x10^3)^2 = 2x10^3 J
(Thanks OMMC :bigsmile:

Q2

A 20microfarad capacitor is charged up to 200V and then disconnected from the supply. It is then connected across a 5.0microfarad capacitor.

a) Calculate combined capacitance.

So connected across means in parallel? If so, then C=20+5 = 25microfarads.

b) The charge they store

Q=VC So 200 x 25x10^-6 = 5x10-3 C.

But the answer should be 4, rather than 5. I'm lost.

c) p.d. across combination:

d) Energy dissipated when they are connected together.

I can't really attempt these without understanding the top.

:sigh:

There's also a question in the past paper that I don't understand, hoping help with this will mean I can do that one.

(edited 12 years ago)

Reply 1

Joinedup

You forgot you were dealing with a capacitance being measured in uF. :smile:

Reply 2

Oh my Ms. Coffey

For c) you put Q in the place of C.

For Q2 c) Charge is conserved and the system now rules in a series loop (If im picturing it correctly) so you can find the total capacitance and the total charge to find the total voltage if I remember those rightly.

(edited 12 years ago)

Reply 3

shorty.loves.angels

Original post by Oh my Ms. Coffey

Ugh, I knew I'd confuse C with Q eventually :colonhash:

Can't believe I didn't spot that. I still can't get the right answer for 2b, so c isn't really happening.

Reply 4

Oh my Ms. Coffey

Original post by shorty.loves.angels

Ugh, I knew I'd confuse C with Q eventually :colonhash:

Can't believe I didn't spot that. I still can't get the right answer for 2b, so c isn't really happening.

Okay I understand it.

Basically, its not parallel its in series.

The capacitor is charged, and then disconnected from the source and connected to the other capacitor.

Q is conserved.

Q before is 20microF x 200v = 4mF.

Q is the same after so they must still store 4mF on both of them (I is the same in series, point rule) (no Q is lost).

Q must be the same on both as they are in series. Therefore you can work out the voltage on each as you have Q and C.

(edited 12 years ago)

Reply 5

Stonebridge

Original post by Oh my Ms. Coffey

No.
After they are connected the charge that was originally on the one is now shared between the two. The total charge is constant before and after.
Q is not the same on both afterwards.
The pd across them is the same as they are effectively in parallel.
So to solve it you need the original charge on the one capacitor. Q
The total charge on the capacitors after connecting is Q= C₁V + C₂V where V is the new common voltage.
Use this to find the new voltage.
Use the new common voltage applied to the 2 capacitors individually to find the charge on them.