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Capacitor charging and discharging

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    1. In the circuit shown below calculate how long itwill take to charge the capacitor C2 to 4 V and how long it will take then to discharge it to 1V.
    upload_2014-11-5_17-51-34.png
    2. Relevant equations
    upload_2014-11-5_17-51-53.png

    3. The attempt at a solution
    I tried to solve it by taking t=0 when v=0 to get the k which is then ln(VBat). But I am confused if the resistance are connected in series or parallel? Can someone confirm. And for the C in RC do I take the total capacitance or just the C2? Please someone confirm. Many thanks
     
  2. jcsd
  3. Nov 5, 2014 #2

    Matterwave

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    I'm guessing for charging, the switch is connected to a and for discharging the switch is connected to b? Can you specify is that is the case?
     
  4. Nov 5, 2014 #3
    I have no idea sir. This is exactly the information I have. I think its supposed to be some general case. I have weak understanding of this whole circuit concept.

    Is the total resistance of the circuit R3+R2 then parallel with R1 or is it R1+R2+R3?
     
  5. Nov 5, 2014 #4

    Matterwave

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    Well, I think it would be fair to assume the switch connects to the "a" junction only during charging and to the "b" junction only during discharging. But there could be a possibility for both "a" and "b" to be connected at the same time. Without knowing for sure which case we are talking about, we can't do this problem.

    If indeed "a" is connected during charging and "b" is connected for discharging, you have two different circuits, can you draw the two different circuits?
     
  6. Nov 5, 2014 #5
    So if its connected to junction a, we can ignore R3 right? can we ignore anything else for the charging part
     
  7. Nov 5, 2014 #6

    Matterwave

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    Yes, for "a" you can ignore R3, and for "b" you can ignore the voltage source. That's the point of the switch really.

    But the capacitors are connected in parallel, what can you do about that?
     
  8. Nov 5, 2014 #7
    But it asks only for C2 so shouldn't C1 be ignored somehow?
     
  9. Nov 5, 2014 #8

    Matterwave

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    You can't just ignore it...it's part of the circuit. But it is in parallel with C2, so that should tell you something! What do parallel circuit elements have in common?
     
  10. Nov 5, 2014 #9
    Well capacitors got charge common when in parallel.....Not sure how that helps. Explain please
     
  11. Nov 5, 2014 #10

    Matterwave

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    Charge on a capacitor is not equal when in parallel... any circuit elements in parallel has equal voltage across them.

    Do you know how to simplify circuits?
     
  12. Nov 5, 2014 #11
    Oh yeah man sorry you're right
    I suck at simplifying...if I ignore r3 I will have capacitors in parallel. But what about resistors will they be in parallel too?

    And also how does having same voltage help...
     
  13. Nov 5, 2014 #12

    Matterwave

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    The capacitors are in parallel, then the rest of the circuit is a series circuit. It seems that you are missing some very basic knowledge on circuits. I would suggest you go back and study the basics before trying to solve this more complicated problem.
     
  14. Nov 5, 2014 #13
    No its fine. I just don't understand how can I use only c2 when both the capacitors are in parallel. Just because they share same voltage does it mean I can ignore c1?
     
  15. Nov 5, 2014 #14

    Matterwave

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    You can't ignore c1, but you can turn C1 and C2 into an equivalent capacitor.
     
  16. Nov 6, 2014 #15
     
  17. Nov 6, 2014 #16
    Yes we can combine the capacitors in parallel and get a total capacitance. So is the total capacitance to be used with the total resistance in the relevant equation that I uploaded for this problem in order to get the charging time. Please confirm. Thanks
     
  18. Nov 6, 2014 #17
    Do you know what is the value of a capacitor equivalent to C1 and C2 in parallel?
     
  19. Nov 6, 2014 #18
    Yeah its 156 nf.
     
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