need help with a VERY basic capacitor problem

Discussion in 'Homework Help' started by nirvanaguy, Dec 13, 2009.

  1. nirvanaguy

    Thread Starter New Member

    Oct 3, 2009
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    I'm going over my physics 2 final review and there is one problem that I think I know how to get but my professor tells me its done a completely different way..

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    So.. it says two capacitors of capacitance C1= 1.5 uF and C2= .5 uF are connected to an 8 V battery as shown in the figure above. Initially both capacitors are uncharged

    A) When the switch S is turned to position 1, the plates of the capacitor C1 acquire a potential difference Vo. The switch is then turned to position 2. what are the final charges q1 and q2 on the corresponding capacitors?
    Q1=3 uC, Q2=3 uC (since they're in series they have to have the same charge)
    B) What is the potential difference across C2?
    Vc1=2V, Vc2= 6V

    So.. the thing I did was use Qeq=CeqV to find the equivalent charge (because both C1 and C2 shoudl have the same charge since the'yre connected in series).

    My professor told me that C1 and C2 are in parallel once the switch is thrown to position 2 which I strongly disagree with.. I'm also taking circuits right now and everything we've learned in circuits completely contradicts that simple diagram being in parallel since they dont share any nodes (connecting 3 or more elements)..

    So even after he told me they were in parallel, if all 8 V goes to C1 when the switch is in position 1, How did he get that C1 and C2 each have 6V? I understand if they were in parallel theyd have the same voltage but why would it be reduced to 6 V each?


    Can someone please help me out with this discrepancy asap because my final is tomorrow and I just want to be prepared for the test:-/ Thank you!
     
  2. Paulo540

    Member

    Nov 23, 2009
    188
    0
    It seems by your diagram that when switch is put to pos 2, that you just end up with a closed loop parallel capacitor arrangement, (no more power) so the charges would actually equal out at some point.
     
  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Your professor is correct. The capacitors are in parallel once the switch is thrown to position 2.

    Also, you are correct. The capacitors are also in series once the switch is thrown into position 2. However, in this problem it's better to think of them as in parallel. This is because parallel capacitors have equal voltages, which is a useful fact in this problem. Series capacitors have equal currents which is a useless piece of information in this problem. Note that there is no law that says that series capacitors have equal charge. The only charge law you can use is conservation of charge which says that the initial charge must equal the final charge.

    From a certain point of view, this is a very bad problem because capacitors should not be directly shorted to voltage sources, nor to each other. Capacitors do not like to have the voltage change instantly across their terminals. This results in very large current spikes. In this theoretical problem the currents go to infinity because the voltage source and capacitors are ideal.

    However, if you are just concerned with the final steady state values of charge and voltage, the problem can be solved. The assumption is that there is a small but nonzero resistance that prevents the current from going to infinity.

    With the switch in position 1, the capacitor charges to 8V. This allows you to determine the initial charge on C1 (Q0=V0*C1).

    When the switch is moved to position 2, the charges will flow until equilibrium is reached. However, equilibrium occurs when V1=V2 because the capacitors are in parallel. Well, now you know the total charge between both capacitors will equal Q1+Q2=Q0, and from this you can solve the problem.
     
  4. Paulo540

    Member

    Nov 23, 2009
    188
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    Honest question, but how are the capacitors ever in series?

    From what I see, C2 is completely uninvolved until the switch is at point 2, at which the power supply is removed from the circuit and C1 & C2 are then in parallel.
     
  5. Paulo540

    Member

    Nov 23, 2009
    188
    0
    Im going to super simplify it here, mainly because Im a simple guy, lol. And hopefully im not skewing this thread to oblivion...

    Anyways, Capacitor 1 has .000012 C of 'charge' in it at 8v. and at the same voltage, C2 would be able to contain .000004 C of charge. So, its akin to filling a smaller air tank with one that is 3x larger.

    for every Volt that C1 drops, C2 actually 'picks up' 3, because C1 has 3x more charge (volume) per volt. So, C1 drops to 7v, C2 ends up with 3, C1 drops another volt to 6 and C2 ends up with 3 more.... or 6 Volts. :)


    If anyone sees fault in this, please say so and I'll delete this and go into hiding. But it kinda makes sense to me, lol.

    Paulo
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Well, yes. That's how I view it too. But, technically when two simple two-terminal devices are connected to each other (and nothing else), then they are both "in parallel" and "in series" at the same time. Both devices then each have the same voltage and same current.

    In this problem, it's best to think of them in parallel. But the true situation is that they are in series because the wire resistance and capacitor ESR are the only things which keep the problem realistic. A proper model of this circuit will be an equivalent resistance in series with both capacitors.

    I really think this is an example of a poor problem, but that's just my opinion.
     
  7. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I agree with you. I get the same answer.
     
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