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Help understanding RC circuits

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data
    In a problem I have a circuit that I have not set up to know what happens as far as voltages across my battery(4v), two equal resistors and a capacitor.


    2. Relevant equations



    3. The attempt at a solution
    Because i have not seen a circuit like this im thinking that this circuit is not ideal for a capacitor to collect charge. im thinking that the current will only go through resistor A and not but b or the capacitor. So Voltage bat= 4v Voltage resistor A = 4v and resistor b and capacitor = 0V
     

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  2. jcsd
  3. Feb 12, 2012 #2

    gneill

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    Suppose, just for argument's sake, that resistor A was not there. So there is just the battery, resistor B, and the capacitor. Can you describe how the circuit would behave when the battery is first connected?
     
  4. Feb 12, 2012 #3
    well the resistor is actually a light bulb so when the circuit is first connected lets say with a switch the light bulb will light up but at time goes on the light bulb will go out. No more current will be flowing and the capacitor will be charged.
     
  5. Feb 12, 2012 #4

    gneill

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    Right. So current initially flows until the capacitor is charged up. When the capacitor is charged up, what is the voltage across the capacitor? How about the resistor?
     
  6. Feb 12, 2012 #5
    voltage across capacitor is 4v and resistor is 0v
     
  7. Feb 12, 2012 #6

    gneill

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    Excellent. That is correct.

    Now place resistor A back in the circuit. Note that it is a parallel branch -- it is in parallel with the capacitor branch, which we just looked at, and the battery. Being a branch that is across the voltage source, it will behave independently of the other branch; Nothing that resistor B or the capacitor do will effect the voltage that appears across resistor A.
     
  8. Feb 12, 2012 #7
    So when are switch is closed bulb b will light but eventually turn off and the capacitor will then be charge and while this goes on bulb A will be lit continuously?
     
  9. Feb 12, 2012 #8

    gneill

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    Yes, that's right.
     
  10. Feb 12, 2012 #9
    when switch is closed and i wanted to rank my voltages immediately would they be Vbattery=Vbulb A=Vbulb B > V capacitor . V capacitor= 0

    After switch is closed for a longtime current rankings
    I battery = I bulb A > I bulb B . I bulb B = 0 Is this true?

    If the switch is located next to the negative terminal of the battery and the switch was closed for a long time is the voltage across bulb A after the switch is opened again be 2v?
     
  11. Feb 12, 2012 #10

    gneill

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    Yes.
    Yes.
    Yes, if both bulbs have the same resistance.
     
  12. Feb 12, 2012 #11
    Wow thanks so much for taking the time to help me understand these circuits better gneil. I can't thank you enough, but if it makes you feel better i also help younger students with their work when asked :) its very rewarding.
     
  13. Feb 12, 2012 #12

    gneill

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    You're welcome :smile:
     
  14. Feb 12, 2012 #13
    If you dont mind me asking what would happen if bulb B was placed next to the negative terminal and the switch was closed for a long time? would the current through A be equal to the current through B
     
  15. Feb 12, 2012 #14

    gneill

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    I'm not sure that I can picture the circuit that you have in mind... what happens to the capacitor in this scenario? Can you draw the circuit?
     
  16. Feb 12, 2012 #15
    in this circuit when the switch is closed for a long time...
    the capacitor will charge correct?
    Bulb A will continuously be lit correct?
    Im not sure what happens to bulb B, will it light then dim itself a little because of the capacitor or will it light as bright as A ( bulb A and B have the same current) ?
     

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  17. Feb 12, 2012 #16
    Sorry bulb A is the resistor parallel with the capacitor
     
  18. Feb 12, 2012 #17

    gneill

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    After the switch is closed for a long time both bulbs conduct the same current. The capacitor is charged (to half the battery voltage --- the same voltage that appears across bulb A).

    When the switch is first closed the capacitor is uncharged so there is 0V across it. Bulb B then gets the full benefit of the battery potential for that initial instant, while bulb A has zero voltage (and thus no current). As the capacitor charges and its voltage rises, bulb A conducts current. Bulb A brightens while bulb B dims, and they "meet in the middle" with both bulbs running at half brightness and half the battery voltage across each.
     
  19. Feb 12, 2012 #18
    Ok, That was my last question, thanks again!
     
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