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RC circuit and a capacitor

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=aaf6199a7ac964eede844626d152cc49.png
    2. Relevant equations

    q(t) = Cε(1-e^(-t/RC))

    3. The attempt at a solution

    for part a) the instant the switch is closed there is no charge on the capacitor so I am pretty sure I can ignore it for this part and use keirchoffs rules to find the currents through each resistor. For part b) I am less confident. I think I can use the equation q(t) = Cε(1-e^(-t/RC)) to find the charge. if t is a "very long time" like the question says I can say that t goes to infinity so q(t) = Cε(1-e^(-∞)) = Cε(1-0) = Cε = (4 μF)(42 v)
    but I am not sure if this is correct.
     

    Attached Files:

  2. jcsd
  3. Apr 4, 2015 #2

    NascentOxygen

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    Staff: Mentor

    After a long time, the capacitor has charged to the maximum of the voltage available to it, and the current into the capacitor has fallen to zero.

    What is the maximum voltage the capacitor can reach, here? Not 42V, it can never reach 42V in this circuit.
     
  4. Apr 4, 2015 #3

    ehild

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    What do you mean by ignoring the capacitor?
     
  5. Apr 4, 2015 #4
    I thought that I can find the currents as if the capacitor was not part of the circuit for part a) ?
    for part b) the voltage will drop over R1 and drop again over R3 so the max voltage will be the voltage after those two drops? then I can use Q = CV where V is the voltage after those two drops across the resistors and C is the given capacitance ?
     
  6. Apr 4, 2015 #5

    NascentOxygen

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    Yes, in a manner of speaking.

    Yes, but once the capacitor is fully charged there is no further current through R3 so no voltage drop across R3. So the final voltage at the upper capacitor plate can be found by imagining that R3 and the capacitor aren't there (i.e., have been removed).
     
  7. Apr 4, 2015 #6
    so it will have the same voltage that is across R2?
     
  8. Apr 4, 2015 #7

    NascentOxygen

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    Ultimately, yes. The capacitor is all the time headed exponentially towards that voltage, while charging.

    So does that make it easy?
     
  9. Apr 4, 2015 #8

    ehild

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    What would replace it if it was not part of the circuit?
     
  10. Apr 5, 2015 #9
    I meant I thought I could solve for the currents as if it was a kerchoffs rules problem with the same exact circuit except without the capacitor. R3 would just be by itself on the rightmost side
     
  11. Apr 5, 2015 #10

    ehild

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    OK, so you replace the capacitor with a short - a piece of wire.
     
  12. Apr 5, 2015 #11
    yes, the spot where the capacitor used to be would just be wire
     
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