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What is the voltage in kV across each capacitor?

  • Thread starter at3rg0
  • Start date
  • #1
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Homework Statement



A 150 pF capacitor and a 600 pF capacitor are both charged to 3.3 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate.

a) What is the voltage in kV across each capacitor?
b) How much energy is lost (in mJ) is lost when the connection is made?

Homework Equations


Q = CV
Connection in parallel: C (total) = C1 + C2

The Attempt at a Solution


Between all the formulas I have gotten anywhere from 9.9 kV to 19.8 kV and just about everything in between. I am just short of tearing my hair out.
 

Answers and Replies

  • #2
learningphysics
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First step... what is the charge on each capacitor before they are connected together?
 
  • #3
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1980microC and 495mC, right?
 
  • #4
190
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If they're connected parallel,final charge on them must be equal.
 
  • #5
learningphysics
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1980microC and 495mC, right?
yes, but they should be nanoC.

Now they are connected positive to negative... so 1980nC is connected to -495nC... we have the two capacitors in parallel with a net charge of 1980 - 495 nanoC = 1485nC...

what is the equivalent capacitance of two capacitors in parallel... hence what is the voltage across the two capacitors?

now you can get the energy in the capacitors connected in parallel... compare to the initial energy.
 
  • #6
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So, in parallel, they'll be added?

So, divide by the sum of the two capacitances? Does that give me the voltage on both?

Okay, so the answer is right.

Each has 1.98kV.

However, how would I account for the Energy lost?
 
Last edited:
  • #7
learningphysics
Homework Helper
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So, in parallel, they'll be added?

So, divide by the sum of the two capacitances? Does that give me the voltage on both?
yes, it gives the voltage on both.
 
  • #8
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Okay, and I used U=.5CV^2 to get the energy. Thanks so much!
 
  • #9
learningphysics
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Okay, and I used U=.5CV^2 to get the energy. Thanks so much!
no prob! did you calculate the energy lost?
 
Last edited:
  • #10
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Yeah, it was easy. I just did:

.5(195E-12)(3.3E3^2) + (.5(600E-12)(3.3E3^2) - Final Energy (I forget the value), and got the solution.
 
  • #11
learningphysics
Homework Helper
4,099
5
Yeah, it was easy. I just did:

.5(195E-12)(3.3E3^2) + (.5(600E-12)(3.3E3^2) - Final Energy (I forget the value), and got the solution.
cool! yup, that's the way to get it.
 

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