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Two Loop RC Circuit

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 41 Ω, R3 = 68 Ω and R4 = 154 Ω. The capacitance is C = 49 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.
    h11_RC_time.png
    What is Q(∞), the charge on the capacitor after the switch has been closed for a very long time?
    2. Relevant equations
    Kirchhoff's Laws

    3. The attempt at a solution
    I attempted to find Q(inf) by setting up kirchoff's voltage and loop equations as shown:
    1. I1 = I4
    2. 14 = I2 + I3
    3. V - I1R1 - I3R3 - I4R4 = 0
    4. V - I1R1 - I2R2 - (Q/C) - I4R4 = 0

    I used the equalities in equation 1 to solve equation 3 in terms of I3. I then used I3 from equation 3 and plugged it into equation 2 to get an equation for I2. I then plugged in what I got for I2 into equation 4 as well as plugging in I4 for I1. The final equation I got for the value of Q is: Q = C(V - I4R4 - (I4 - (V - I4(R1 + R4))/R3) * R2 - I4R4). I had already found the value for I4 in a previous problem using V = IR when the switch is closed for a very long time. I4's value is 0.54A.

    Thank you so much in advance!
     
  2. jcsd
  3. Feb 16, 2017 #2

    gneill

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    Staff: Mentor

    Hi marchcha,

    You may be putting too much effort into this one :smile:

    You're looking for the charge on the capacitor at steady state, Q(inf). What do you know about the value of the capacitor current ##I_c## at steady state? What does that tell you about the loop current for that loop?
     
  4. Feb 16, 2017 #3
    The capacitors current will be zero at steady state. I'm a little confused as to what this tells us about the loop current for that loop.
     
  5. Feb 16, 2017 #4

    gneill

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    Staff: Mentor

    Yes, ##I_c = 0## at steady state.

    Is there any way to distinguish ##I_c## from the loop current? Aren't they identical? Just as the loop current for the first loop is identical to the currents flowing through the voltage source, R1, and R4.
     
  6. Feb 16, 2017 #5
    So would I2 also be 0?
     
  7. Feb 16, 2017 #6

    gneill

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    Yes. ##I2 = I_c = 0## at steady state.
     
  8. Feb 16, 2017 #7
    So then Q = C (V - I1R1 - I4R4)?
     
  9. Feb 16, 2017 #8

    gneill

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    Sure. Or Q = C*I3R3, to use your notation. You really don't need to invent so many different currents if there's just one loop current flowing....
     
  10. Feb 16, 2017 #9
    Ok, then how do you find I3?
     
  11. Feb 16, 2017 #10

    gneill

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    Draw in the current flowing through loop 1. It's the only current flowing at steady state.

    upload_2017-2-16_16-50-58.png
     
  12. Feb 16, 2017 #11
    Ok so I3 is .0456A I dont understand how the capacitors charge has anything to do with the first loop though?
     
  13. Feb 16, 2017 #12

    gneill

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    If ##I_c = 0##, what's the current through R2? Hence, what's the potential drop across R2?
     
  14. Feb 16, 2017 #13
    Zero for both.
     
  15. Feb 16, 2017 #14

    gneill

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    So if the potential drop across R2 is zero, what's the difference between the potential at the top of the capacitor and the potential at the top of R3?
     
  16. Feb 16, 2017 #15
    Would it be just V?
     
  17. Feb 16, 2017 #16

    gneill

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    er, no. If the potential at the top of R3 is, say V3, and you "walk" from there across R2 which you say has a potential drop of zero, what's the potential at the other end of R2 (also the top of the capacitor)?
     
  18. Feb 16, 2017 #17
    Ah so they'd be the same and you can say Q/C = I3R3 the equation above. Thanks!
     
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