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Homework Help: Disconnecting a capacitor

  1. Dec 15, 2017 #1
    1. The problem statement, all variables and given/known data
    This circuit is given:
    upload_2017-12-15_18-53-23.png
    After the current has stabilized calculate:
    1. The current through the voltage source.
    2. The voltage and charge on each capacitor.
    While the circuit is active and the current is stable, the 4μF capacitor is disconnected from the circuit.
    3. What are the new charges on the capacitors?
    2. Relevant equations
    Q=CV

    3. The attempt at a solution
    I did 1 and 2 correctly. However, I thought that after the capacitor is disconnected we will get two capacitors in a series, and therefore be able to use Q1=Q2 (getting 14.4μC). I don't understand why it's wrong to do so here, according to the answers the charges stay the same, why?
     
  2. jcsd
  3. Dec 15, 2017 #2

    jbriggs444

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    Yes, you get two capacitors in a series. Yes, the capacitance of the circuit element consisting of the two capacitors in series is then 14.4 μC. But that just tells you how much potential difference there is between the two ends depending on how much charge is moved between the two terminals. It does not tell you the potential difference across the individual capacitors.

    An unstated assumption is that this circuit started with all capacitors discharged before the 20V power supply was connected. Without this assumption, the calculation in part 2 is baseless. The potential of the node between the capacitors would be indeterminate.
     
  4. Dec 15, 2017 #3
    My teacher told us that if nothing is said we should assume this.
    I don't understand what you are trying to say in your first paragraph. If this is a series the equivalent capacity would be would be 1.2μF. then the equivalent charge would be 1.2*12=14.4μC, and therefore: Q1=Q2=14.4μC. Why is this wrong?
     
  5. Dec 15, 2017 #4

    jbriggs444

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    The two remaining capacitors were both charged before third was removed. The removal of the third capacitor does not remove this charge.

    In part 2 you had an implicit assumption (which you say your teacher says you should make) that the capacitors started completely discharged before power was applied. You can calculate how much current has crossed each capacitor while the circuit was charging, so you know the potential and charge on each capacitor after it has settled down.

    Note: I agree with your teacher that this is an assumption that should be made. I am asking you to be aware that you are making it.​

    In part 3 you have a different explicit assumption -- the capacitors each start with a known non-zero charge.
     
  6. Dec 31, 2017 #5
    Still can't figure it out...
     
  7. Dec 31, 2017 #6

    SammyS

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    Give your answers to parts 1 and 2. It will help to answer your questions about part 3 using the answers to parts 1 and 2 .
     
  8. Dec 31, 2017 #7

    Delta²

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    Another way to view at 3) is that when we disconnect the ##4\mu F## capacitor the equilibrium state is not disturbed because the ##2\mu F## capacitor maintains the same voltage (those two capacitors were connected in parallel).
     
  9. Dec 31, 2017 #8
    But their charges won't be equal after that? They are connected in a series now, why doesn't it change?
     
  10. Dec 31, 2017 #9

    Delta²

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    Current has to flow in that branch to change their charges, if the equilibrium isn't disturbed, current wont flow in that branch.

    If you were starting from the start with those two capacitors in series, completely uncharged at start, then when the equilibrium would have been reached they would have equal charges (because they would have been in series from the start of the charging process).
     
  11. Dec 31, 2017 #10

    Delta²

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    And something else, capacitors being in series doesn't imply that their charge will be the same for all cases. This holds only if they start with the same initial charge at the starting of the charging process

    Consider the case of two capacitors that are in series but the first is starting with charge ##q_1## and the second with charge ##q_2##. Suppose we apply a voltage V , wouldn't it be that ##\frac{q_1+q}{C_1}+\frac{q_2+q}{C_2}=V## for some q, when equilibrium has been reached?
     
  12. Dec 31, 2017 #11
    I see, so how do I know if a certain change will break the equilibrium? Why at this specific case it isn't broken? Because their voltages aren't affected?
     
  13. Dec 31, 2017 #12

    Delta²

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    Some voltage or some current has to change somewhere in the circuit, otherwise the equilibrium isn't broken.

    Just to say, usually something tricky will happen (that is usually we ll have a new equilibrium state if we are speaking of DC circuits) when we remove or add a branch at an active circuit (one that already has current and voltages formed). BUT in this case the tricky part is to understand that there is nothing tricky, the removal of that capacitor doesn't disturb at all the equilibrium state of the circuit.
     
    Last edited: Jan 2, 2018
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