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Series capactors

  • #1
1. A battery with an emf of 60 V is connected to the two capacitors shown in the figure . Afterward, the charge on capacitor 2 is 540 uC.

The capacitors are in series, with C1 = 12microFarads and C2 = ?

Since series capacitors have the same charge Q, I used C1 and solved for Q in C = Q/V, obtaining Q = 720uC. I then solved C2 = 720/60 = 12microFarads, which is incorrect.
 

Answers and Replies

  • #2
ideasrule
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The charge across each capacitor isn't 60 V because the 60V is divided across the two capacitors. However, you know what Q is across the first capacitor. It's 540 uC, because if you consider the two capacitors as one unit, it would have the same charge as the two individual capacitors.
 
  • #3
So.. what is C2 then? I got 18 and that was incorrec
 
  • #4
So.. what is C2 then? I got 18 and that was incorrect
 
  • #5
SammyS
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Old post

Hey! The Original Post is over 11 months old!
 
  • #6
I didn't realize that but this is a question I am currently working on
 
  • #7
SammyS
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So.. what is C2 then? I got 18 and that was incorrect
(What units?)

Show how you got that.
 
  • #8
SammyS
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The capacitors are in series. If the charge on C2 = 540 μC, how much charge is on C1 ?

What's the voltage drop across C1 ?
 
  • #9
I am also working on this question, and I don't know where to begin. i know that the charge on both capacitors is equal
 
  • #10
gneill
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I am also working on this question, and I don't know where to begin. i know that the charge on both capacitors is equal
If the charges are equal, and you know the value of C1, what does that tell you about the voltage across C1?
 
  • #11
i am not sure,
but is it that Q = CV
and so i can say C1V1 = C2V2
and i think the voltage drop has to be different for the two capacitors..
 
  • #12
gneill
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i am not sure,
but is it that Q = CV
and so i can say C1V1 = C2V2
and i think the voltage drop has to be different for the two capacitors..
Yes, but the problem statement gives you the charge on one of the capacitors, right? You've also determined that the charges must be equal for both capacitors. The problem statement also gives you the value of one of the capacitors. Apply your Q = CV to find the voltage on that capacitor.
 

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