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2 Loop RC Circuit 2

  1. Feb 21, 2016 #1
    1. Situation and Variables:
    A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 30 Ω, R3 = 66 Ω and R4 = 113 Ω. The capacitance is C = 64 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

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    2. Sub-question:
    After the switch has been closed for a very long time, it is then opened. What is Q(topen), the charge on the capacitor at a time topen = 786 μs after the switch was opened?

    3. Relevant Equations:
    Kirchhoff's Loop Rule: [tex] \Sigma V_l = 0[/tex]

    Kirchhoff's Node Rule: [tex] I_i = I_o [/tex]

    [tex] V_C = \frac {q} {C}[/tex]

    For a discharging circuit:
    [tex] q(t) = q_o e^\frac {-t} {\tau}[/tex]

    4. Attempt:
    When the switch is closed initially and time goes to infinity, the capacitor charges up and reaches it's capacitance, which I found the max charge, qo to be about 243μC.
    Then for a discharging circuit, we have the equation I gave above q(t). I believe my issue is finding tau, which is RC. I have been using [tex] R = R_1+R_2 [/tex] but that hasn't been correct. I've tried different combinations of resistors but I'm not quite getting it.
     
    Last edited by a moderator: Feb 21, 2016
  2. jcsd
  3. Feb 21, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Show your working for the capacitor charging voltage.

    How to determine the capacitor's discharging path when the switch is an open circuit?
     
  4. Feb 21, 2016 #3
    Q(∞)?
    IC = 0 ∴ I1 = I2
    Knowing this:
    Loop for right hand side (I):
    [tex]\frac {q} {C} = I_2R_3[/tex]

    Loop for left hand side (II):
    [tex]V_b = I_2(R_1+R_3+R_4)[/tex]
    Therefore,
    [tex]I_2 = \frac {V_b} {R_1+R_3+R_4}[/tex]

    plugging this into (I):
    [tex]\frac {q} {C} = \frac {V_bR_3} {R_1+R_3+R_4}[/tex]

    then solve for q, which is just multiplying both sides by C.
    The exactly answer they have for Q is 242.526315789474μC, which is what I got, and I have this number stored so I used it for my text calculation of find Q(786μs) from the equation [tex]q(t) = q_oe^\frac{-t} {\tau}[/tex]
     
  5. Feb 21, 2016 #4
    Also I figured the current go across the switch, so I assumed it would only traverse the right hand side, which is why I use [tex]R=R_2+R_3[/tex] the equation I originally posted is not what I had written, it was a typo on my part.
     
  6. Feb 21, 2016 #5

    NascentOxygen

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    Staff: Mentor

    That's the discharge path. So you need to be able to write by inspection the exponential time equation for the capacitor voltage as it discharges. I see you almost have.

    Are you able to calculate τ and finish this now?
     
  7. Feb 21, 2016 #6
    Right. So I have this equation:
    [tex]q(t) = q_oe^\frac {-t} {R_2+R_3}[/tex]
    but because R2+R3 is so large in comparison to t=786μs, the e part is approximately 1, and so my calculator outputs nearly the same answer as I had before, which the website counts as incorrect
     
  8. Feb 21, 2016 #7

    NascentOxygen

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    Staff: Mentor

    The time constant is the product of R and C. It needs to have units of seconds, or it wouldn't be a time.
     
  9. Feb 21, 2016 #8
    Wow, that was a really dumb mistake on my part. Thanks for helping me though, I probably would have never realized that.
     
  10. Feb 21, 2016 #9

    NascentOxygen

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    Staff: Mentor

    I thought you had it right in your first post .....
     
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