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RL parallel circuit (AP 2006)

  1. May 7, 2015 #1
    1. The problem statement, all variables and given/known data
    The circuit in the link below contains a capacitor of capacitance C, a power supply of emf E, two resistors of resistances R1 and R2, and two switches S1 and S2. After switch S1 has been closed for a long time, switch S2 gets closed at a time t = 0. Sketch graphs of the current I1 in R1 and I2 in R2 versus time.

    The problem is part d in the link below.
    http://apcentral.collegeboard.com/apc/public/repository/_ap06_frq_physicsc_em_51784.pdf

    The rubric is here: http://apcentral.collegeboard.com/apc/public/repository/_ap06_physicsc_em_sg.pdf

    2. Relevant equations
    E = IR

    3. The attempt at a solution
    My initial reaction was to say that before t = 0, there is no current in the circuit because there is maximum voltage across the capacitor. At t = 0, there is a path for current to flow so that the current in R2 and R1 would jump to E/(R1+R2) and stay constant. However, the scoring rubric says otherwise. My guess is that the capacitor discharges, but then the capacitor would lose its potential difference, and current would flow through the capacitor so the current would oscillate. What is the correct explanation?
     
  2. jcsd
  3. May 7, 2015 #2

    Zondrina

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    There is no current in the circuit, but this is because both switches are open, and hence there is no path for the current to flow. There will be no voltage across the capacitor because the charge on the capacitor is zero, i.e ##V_C = \frac{q}{C} = \frac{0}{C} = 0##.

    When switch ##S_1## closes at ##t = 0##, the current flows around the single loop formed. This single loop happens to be an ##RC## low pass filter with a time constant ##\tau_C = R_1C##.

    The capacitor then charges up according to the transient equation they have derived. You can find the current in the circuit at this time by writing ##i(t) = \frac{dq(t)}{dt}## if you are interested.

    The capacitor will not be able to discharge once it is fully charged as there is no safe path for this to occur (##V_C## and ##\epsilon## oppose each other equally). At this time the current in the circuit will decrease to zero.

    Now at a new time ##t = 0##, close switch ##S_2## completing the parallel branch. The capacitor has a new path to discharge through ##R_2## now, creating an exponentially decaying current ##I_2(t)## across it. This causes an exponentially increasing current ##I_1(t)## to start building up through ##R_1## again because there is a new path for current to flow.

    Eventually a long time later, the circuit reaches a steady-state where the currents across the resistors become almost equal.

    You should try deriving the equations for ##I_1(t)## and ##I_2(t)## yourself.
     
  4. May 7, 2015 #3
    Why is there no current immediately after t = 0? Doesn't the emf source also drive a current through R2?
     
  5. May 7, 2015 #4

    Zondrina

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    Are you referring to the current ##I_1## through ##R_1##? If so, refer to the explanation in my prior post. There is eventually no current because the capacitor is fully charged after some time when switch ##S_1## is closed. At the new time ##t = 0## this is still the case because the new path allows a current to flow across ##R_1## again.

    As long as the capacitor voltage doesn't cancel the emf out like it did before ##S_2## was closed, there will be a current.
     
    Last edited: May 7, 2015
  6. May 7, 2015 #5
    I think I understand what happens before new t = 0. My only question is why immediately after new t = 0 the battery doesn't produce a current through R1. The current in R1 immediately after new t = 0 is 0, but can't the battery produce a current through R1 now that a safe path around the circuit (through R2) is available?
     
  7. May 7, 2015 #6

    Zondrina

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    The battery is partially responsible for some of the current after the new ##t = 0##.

    Yes, and read the above statement.

    To gain more insight, we can use circuit analysis. Recall KCL:

    $$\sum I = 0$$

    Lets write a KCL equation at the node between the three components:

    $$I_1 = I_2 + I_C$$

    Right when ##t = 0##, after switch ##S_2## is closed, what do you think the above equation looks like? How can we use the equation to explain a current appears across ##R_1## for ##t > 0##?
     
  8. May 7, 2015 #7
    I am confused. At t = 0, the capacitor will not have discharged yet to Ic = 0 so I1 = I2, which is incorrect.
     
  9. May 7, 2015 #8

    Zondrina

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    How is ##I_C = 0##? That can't be possible right? Considering we know:

    $$I_C = C \frac{d}{dt} v_C(t)$$

    Surely ##v_C(t) \neq 0## when the capacitor is fully charged.
     
  10. May 7, 2015 #9
    Oh, Ok. That makes sense. I see that there is current through the capacitor, but how does this lead to I1 = 0?
     
  11. May 7, 2015 #10

    Zondrina

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    We know ##I_1 = 0## just an infinitesimal amount of time before and after the new ##t = 0##. After the new ##t = 0##, ##I_1## begins to increase again.
     
  12. May 7, 2015 #11
    Then why does the capacitor have current immediately after the switch is closed while the battery has not produced current in R1?
     
  13. May 7, 2015 #12

    SammyS

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    What is the Voltage across the capacitor at new time, t = 0 ?
     
  14. May 7, 2015 #13

    Zondrina

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    It takes a relatively small amount of time for things to happen in the circuit after the switch is closed.

    You read my mind, so I'll quote it instead :P.
     
  15. May 7, 2015 #14
    If the capacitor's initial voltage is E and it discharges, does R2 get the same voltage at that instant?
     
  16. May 7, 2015 #15

    Zondrina

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    The components are in parallel, so they will have the same voltage drop.

    Getting a bit tired, if this is unsolved tomorrow, I'll check back.
     
  17. May 7, 2015 #16

    SammyS

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    The capacitor does not discharge in an instant.
     
  18. May 7, 2015 #17
    OH. That makes so much sense. If R2 has E voltage and the emf is E, there can initially be no current from the emf. But as the capacitor discharges, the current from the emf increases so the circuit eventually reaches a "uniform" current.
     
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