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Potential difference in parallel - capacitors

  • #1
potential difference in parallel -- capacitors

Two capacitors C1 = 3 µF and C2 = 9 µF are connected in parallel across a 11 V battery. They are carefully disconnected so that they are not discharged and are reconnected to each other with positive plate to negative plate and negative plate to positive plate (with no battery).

(a) Find the potential difference across each capacitor after they are connected.

(b) Find the initial and final energy stored in the capacitors.

related formulas:
C=Q/V

so far...
so this means that they are in parallel, store up charge, and then are disconnected and reconnected in series without battery yes?

since C(total)=12uF and Voltage = 11v (for both capacitors) I figured that the Q(total) = 1.32*10^-4 C

12uF = Q(tot)/11v

when I apply this total charge to the capacitors in a series (inverting everything) I get 1/3uF=V(1)/1.32*10^-4, V(1)= 44v

before I went on to solve for the other capacitor I submitted this but it was wrong. Where did I go wrong?
 

Answers and Replies

  • #2
887
98


When connecting two capacitors in series the capacitance will decrease: Ct = (C1 * C2) / (C1 + C2) and the voltage will add: V1 + V2.
 
  • #3


mmmk so (C1*C2 / C1+ C2) = C(tot) does the same as (1/C1 + 1/C2)^-1, when you say V(1) + V(2) = V(t) is that the same V as when the capacitors were connected in parallel? I'm still entirely confused...
 
  • #4
887
98


StuckInPhysic said:
so (C1*C2 / C1+ C2) = C(tot) does the same as (1/C1 + 1/C2)^-1
No. It's the total capacitance of the two capacitors when they are in series, after you disconnect from the battery and connect them together in series. So "C(total)=12uF" is incorrect.

StuckInPhysic said:
when you say V(1) + V(2) = V(t) is that the same V as when the capacitors were connected in parallel?
Yes. It's the voltages across the capacitors after you disconnect from the battery. Which of course will be V1 = 11 and V2 = 11.
 
  • #5


When I wrote "12uF = Q(tot)/11v" I was applying that to the capacitors in parallel ie what I had described before. When in parallel this is how I approached:

1/3 uF = delta V(1) / Q
1/9 uF = delta V(2) / Q

where Q in both is the same and is the total charge accumulated by the capacitors when in parallel.

I also approached by combining to find delta V(tot):

(C1*C2 / C1 + C2) = (1/C1 + 1/C2)^-1 = C(tot) = delta V(tot) / Q =
delta V(1) + delta V(2) / Q... In doing this I found V(tot) = (1.32*10^-4)/2.25 = 5.87e-5 which... is.. wrong?
 
  • #6
887
98


StuckInPhysic said:
(C1*C2 / C1 + C2) = (1/C1 + 1/C2)^-1 = C(tot) = delta V(tot) / Q = delta V(1) + delta V(2) / Q
Sorry I misread your original post. You are correct.
 
  • #7


but I'm not... maybe small calculation errors somewhere... baaaagh
 

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