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Capacitor charging

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    In the following diagram, The voltmeter reading is 5 volts when the switch K is open and C1 is not charged.
    When you close K calculate the following:
    1- Voltmeter reading
    2- Charge of each capacitor
    3- Voltage of each capacitor
    iRpLz9E.png

    2. Relevant equations
    Usual electronics equation

    3. The attempt at a solution
    Well, I was looking through my friend's private tutor papers and I found this question. My solution is different that his so I thought I could bring it here so someone can check it and see if I am missing something.

    C1 is not charged thus it doesn't contribute to the voltage of the voltmeter (assuming actual current flowing or assuming it is negligible) so the voltage reading is purely C2's voltage

    Now instead of assuming Q as transferred charged, you could choose a much simpler way.

    C1 and C3 (it didnt say if C3 is charged so I assumed it is not) are in series then the resultant is parallel with C2
    So
    ##\frac{Q_t}{C_{eq}} = V_{after~opening~ the switch} ##
    you get ## V = 4.25 volts ##
    The Charge for any of C1 and C3 capacitors is the voltage times the equivalent capacitance of C1 and C3 which is in series
    ## V ~ ( \frac{1}{C_1} + \frac{1}{C_3} ) = Q_{3 ~ or ~ 1} ##
    ## Q = 9 uC ##
    the charge for C2 is simply
    ## Q_2 = C_2 V ##
    ## Q_2 = 51 uC ##

    Now the first requirement needs the voltmeter reading which is:
    ##V = V_{C_2} - V_{C_1} ##
    ## V = 4.25 - Q/C_1 ##
    ## V = 4.25 - 2.25 ##
    ## V = 2 ## which he found to be the same
    the rest is straight forward.

    For the charges and voltages, he got it different because he assumed that C2 and C1 are connected in series which is wrong because they definitely don't have the same charge
     
  2. jcsd
  3. Oct 25, 2016 #2

    TSny

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    Your method of solution looks correct to me. Your answers agree with what I got by first solving for the charge q that leaves C2 when the switch is closed. Your solution gets to the answer quicker!
     
  4. Oct 25, 2016 #3
    Thank you TSny :).
    I usually use that method to derive if I could treat those capacitor is a parallel or series way. Sometimes it can get complicated so I just use the Q method :D

    Anyway, Thank you again. Much appreciated
     
  5. Oct 25, 2016 #4

    haruspex

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    Am I misreading something?
    I would have thought that either the diagram is labelled wrongly or the given statement that C1 starts uncharged is a typo; should say C3. A voltmeter works by allowing a trickle current, so the two in series must both be charged.
    On that basis, the two in series have combined capacitance of 3uF, and including the one in parallel we get 7.5uF. The final voltage is therefore (3/7.5)5=2. I think that was your argument, but I did not find it very clear.
     
  6. Oct 25, 2016 #5

    TSny

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    I worried about the effect of the voltmeter, too. I ended up assuming that the voltmeter had no influence. I interpreted it as just a way to say that the total potential difference in going from the left side of C2 to the right side of C1 is 5 V when C1 is uncharged and the switch is open. That's a weird way to just say that initially C2 has 5 V while the other two caps are uncharged.
     
  7. Oct 26, 2016 #6
    Doesn't the ideal voltmeter have infinite resistance? Which means a negligible amount of charges gets transfered from C2 to C1

    I don't know about C1 being charged or no but I clearly remember that the question statement said C1 is not charged but It didnt mention C3


    Even if we assume C1 is charged with certain charge Q and C3 isn't, we can't be sure that ## Q_2 = Q_1 ## so that we can take it as series unless the question states that they perhaps were connected to the same voltage source initially or providing that they have the same charge
     
    Last edited: Oct 26, 2016
  8. Oct 26, 2016 #7

    haruspex

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    No real voltmeter has infinite resistance, just very high, which means the current will be very small. How much charge is transferred depends on how long the voltmeter is connected. That said, I think we have to assume no significant charge is leaking through, otherwise it would eventually be reading 0. The two capacitors would have equal and opposite voltages.
    I don't think it matters. All we need to know is that the initial voltage reading is 5V. How the charge is arranged on those two capacitors is unimportant.
    I suggest the question should be saying that the third capacitor starts with no charge; it's a misprint in either the text or the diagram.
     
  9. Oct 26, 2016 #8
    Yea I get that no real voltmeter has infinite resistance that is why I said ideal. Also it would be a bit difficult to calculate at least for my level.

    Yes, I derived that just few secs ago. Assuming Q charge transfer method or Using the method you gave me above they both yield in the same variables.

    But just a question, We can't find the B and C part of the question because we have no knowledge of Q1 and Q2 right?
     
  10. Oct 26, 2016 #9
    Can anyone confirm the last question in the comment above?
     
  11. Oct 26, 2016 #10

    gneill

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    The question does say that C1 is initially uncharged. So that's not an issue. The only possible impediment to finding all the steady-state charges is that no mention was made of an initial charge for C3. But you would be justified in assuming it to be initially uncharged and proceeding; Just state your assumption as part of your answer. Alternatively, assume some unknown initial charge and give the answers as expressions (symbolically). That would be more work, and probably not what the question author intended.
     
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