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Simple RC circuit in series

  1. Jun 25, 2009 #1
    The values of components in a simple RC circuit in series with a key are: C = 1x10^{-6}F, R=2x10^{6} Ohm and epsilon = 10 V. 10s at the moment after the key is closed, calculate:

    a) the charge in capacitor
    b) the current in the resistor
    c) the rate at which energy is stored in capacitor
    d) the rate at which energy is supplied by the battery


    My solution:

    a) Qmax = C * epsilon = 1x10^5 c

    b) I = - dQ/dt
    Rule of Kirchhoff
    I*R = Q/C
    I = Q / R*C = 5x10^{-6} A

    c) How to calculate?
    d) How to calculate?
     
  2. jcsd
  3. Jun 25, 2009 #2

    LowlyPion

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    Maybe consider:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/filter.html#c2
     
  4. Jun 25, 2009 #3
  5. Jun 25, 2009 #4

    LowlyPion

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    Well, what is the "rate of energy build up"?
     
  6. Jun 25, 2009 #5

    berkeman

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    Nevermind (c) and (d), your answers for (a) and (b) are not correct yet.

    LP is reminding you that there is a time constant associated with the RC circuit, and that the charging is exponential with time. You need to write the equation for the capacitor voltage as a function of time, and use that equation to solve for a-d.
     
  7. Jun 25, 2009 #6
    Yes. New:

    a) Q = C * E * (1 - e^{(-t)/R*C}
    Q = 9,932x10^{-6} c


    b) I = E/R * e^{(-t)/R*C}
    I = 3,37x10^{-8} A

    (c) and (d) I not find equation
     
  8. Jun 25, 2009 #7

    berkeman

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    What is the definition of power, in terms of energy? What is the power dissipated in a resistor in terms of I and V? How is the energy stored on a capacitor defined?
     
  9. Jun 25, 2009 #8
    Energy in a capacitor charged:

    1/2 * C * V^2
     
  10. Jun 25, 2009 #9

    ideasrule

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    Since you already found a relationship between charge and time and another between charge flow rate (i.e. current) and time, it's wise to express energy in terms of charge. Then you can use calculus to find the equation for power.
     
  11. Jun 25, 2009 #10
    c)
    U = 1/2 * E^2 * C
    U = 5x10^{-5}J

    d)
    E = (Q*E)/(t)
    E = 1x10^{-5}J


    It is correct ?
     
  12. Jun 25, 2009 #11

    cepheid

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    Part (c) is not correct. What you have calculated is the total energy stored in the capacitor (after it has charged fully). The question is asking for the *rate* at which energy is stored (during the charging process)

    For part (d), I would follow berkeman's advice in his latest post and investigate power.
     
  13. Jun 25, 2009 #12

    ideasrule

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    I meant that you should express, for part c), the energy stored on the capacitor in terms of the charge on the capacitor.
     
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