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Differential Equations and RQC

  1. Oct 25, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A circuit consists of a voltage source, voltage V , a resistor, resistance R, and a capacitor,
    capacitance C, in series.

    (a) Show that the charge, Q, in the capacitor satisfies the equation RQ' + Q/C = V .

    2. Relevant equations

    [tex] R\frac{dQ}{dV} + \frac{Q}{C} = V [/tex]

    3. The attempt at a solution

    I'm not quite sure what to do for this Question. In this Part you have to show how charge satisfies the given equation. Could anyone help show how to start this question, what it actually is asking you to do?

    TFM
     
  2. jcsd
  3. Oct 25, 2008 #2

    alphysicist

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    Hi TFM,

    To start these types of questions you almost always use a Kirchoff voltage equation (or Kirchoff loop equation). What do you get when you write the loop equation for this circuit? Do you see how it reduces to the given equation?
     
  4. Oct 26, 2008 #3

    TFM

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    So using Kirchoff,

    Current in = Current Out

    Sum of Voltage = Total Voltage

    So, if you calculate the voltage, following from my attached diagram:

    [tex] V_{input} = V_{Resister} + V_{Capacitor} [/tex]

    [tex] V_{resister} = IR [/tex]

    [tex] V_{Capacitor} = \frac{Q}{C} [/tex]

    Inserting into Voltage equation above

    [tex] V_{input} = IR + \frac{Q}{C} [/tex]

    but the equation doesn't have IR,

    known equations featuring I:

    Q=It - Time not in question
    V=IR - would take ypu back to V

    Not sure the above equations help, though?

    TFM
     

    Attached Files:

  5. Oct 26, 2008 #4

    TFM

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    I wonder if I have got my derivative correct. The next part of the question is to find an expression for Q(t)

    Does this mean that RQ' + Q/C = V

    should go to:

    [tex] R\frac{dQ}{dt} + \frac{Q}{C} = V [/tex]

    which would be

    Q = It, I = Q/t

    so:

    [tex] V_{input} = IR + \frac{Q}{C} [/tex]

    [tex] V_{input} = \frac{Q}{t}R + \frac{Q}{C} [/tex]

    Although this can't be quite right because i have Q/t, not dQ/dt

    ???

    TFM
     
  6. Oct 26, 2008 #5
    instead of using Q/t for current, you should be using dQ/dt. This gives you the differential equation you're looking for.
     
  7. Oct 26, 2008 #6

    TFM

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    Thanks,

    for the next part,

    Suppose that R, C and V are constant and that Q is initially zero. Find an expression for Q(t). Sketch the solution.

    [tex] V = R\frac{dQ}{dt} + \frac{Q}{C} [/tex]

    R,C,V = Const.
    Q(t=0) = 0

    I have

    [tex] Vdt = R + \frac{Q}{C} dQ [/tex]

    Which I get as

    [tex] Vt = RQ + \frac{Q^2}{2C} + D[/tex]

    where d = integration constant

    so inserting values

    V(0) = R(0) + [tex]\frac{0^2}{2C} + D[/tex]

    D = 0

    does this look right?

    TFM
     
  8. Oct 26, 2008 #7
    I don't think you manipulated your differential equation correctly. Think about this rationally: as the capacitor continues to store energy of separated charge, it will eventually reach a limit to how much energy it can store. This means that Q(t) should start at zero (uncharged capacitor) and asymptotically approach a fixed value. What sort of functions behave this way?

    The sort of functions that behave this way have a differential equation that looks like dU/U = K dt where U is any function and K is a constant. Get your equation into a form like that and integrate both sides.
     
  9. Oct 26, 2008 #8

    TFM

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    Does this look right. So far I have:

    [tex] V = R\frac{dQ}{dt} + \frac{Q}{C} [/tex]

    [tex] \frac{V}{Q} = \frac{R}{Q}\frac{dQ}{dt} + \frac{1}{C} [/tex]

    [tex] \frac{V}{Q} - \frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C} [/tex]

    Factorise out:

    [tex] \frac{1}{Q}(V - R)\frac{dQ}{dt} = \frac{1}{C} [/tex]

    [tex] \frac{1}{Q}\frac{dQ}{dt} = \frac{V - R}{C} [/tex]

    [tex] \frac{1}{Q} dQ = \frac{V - R}{C}dt [/tex]

    integrate gives:

    [tex] lnQ = \frac{V - R}{C}t [/tex]

    does this look better?

    Edit: No it doesn't

    [tex] \frac{1}{Q}(V - R)\frac{dQ}{dt} = \frac{1}{C} [/tex]

    [tex] \frac{1}{Q}\frac{dQ}{dt} = \frac{1}{VC - RC} [/tex]

    [tex] \frac{1}{Q} dQ = \frac{1}{VC - RC}dt [/tex]

    Integrates to:

    [tex] lnQ = \frac{t}{VC - RC} [/tex]

    Does this look right now

    TFM
     
    Last edited: Oct 26, 2008
  10. Oct 26, 2008 #9

    alphysicist

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    This line is not correct. You have factored out a dQ/dt, but the first term did not have that in it.


    Also, remember that after you get the integrals set up, these are definite integrals and so you have to worry about the limits.
     
  11. Oct 26, 2008 #10

    TFM

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    Should it be then:

    [tex] \frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C} [/tex]

    [tex] \frac{1}{Q}(V - R \frac{dQ}{dt}) = \frac{1}{C} [/tex]

    [tex] \frac{1}{Q} = \frac{1}{C(V - R\frac{dQ}{dt})} [/tex]

    [tex] \frac{1}{Q} = \frac{1}{CV - CR\frac{dQ}{dt}} [/tex]

    This doesn't look right though...?

    TFM
     
  12. Oct 26, 2008 #11

    alphysicist

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    I would suggest starting with the beginning equation and getting everything out of the denominator (even though later you'll get things in the denominator again).

    Once you have that, the goal is to completely separate the equation so that q does not appear on one side of the equation, and t does not appear on the other. (This is where you'll have a fraction again.) At that point you can integrate each side separately. Do you get the answer?
     
  13. Oct 26, 2008 #12

    TFM

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    So starting with:

    [tex] \frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C} [/tex]

    Multiply out the Q

    [tex] V - R \frac{dQ}{dQ}=\frac{Q}{C} [/tex]t

    Multiply by C

    [tex] VC - RC \frac{dQ}{dt} = Q [/tex]

    split the derivative:

    [tex] VC - RC dQ = Q dt [/tex]

    put the Q on the other side by dividing through by Q

    [tex] \frac{VC - RC dQ}{Q} = dt [/tex]

    Does this look better?

    TFM
     
  14. Oct 26, 2008 #13

    alphysicist

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    This is correct up to here.

    The first term is missing a factor of dt.

    Once you correct that, then move all terms with dt to one side, all terms with dq to the other, and factor out the differentials until you get:

    (stuff) dt = (other stuff) dq

    At that point you should be able to separate the variables.
     
  15. Oct 26, 2008 #14

    TFM

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    So:

    [tex] VC - RC \frac{dQ}{dt} = Q [/tex]

    [tex] VCdt - RC \fracdQ = Qdt [/tex]

    [tex] -RC dQ = Qdt - VCdt [/tex]

    [tex] -RC dQ = (Q - VC)dt [/tex]

    [tex] \frac{-RC}{Q - VC} dQ = dt [/tex]

    Does this look okay now?
     
  16. Oct 26, 2008 #15

    alphysicist

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    That looks right to me.
     
  17. Oct 26, 2008 #16

    TFM

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    So:

    [tex] \frac{-RC}{Q - VC} dQ = dt [/tex]

    The right side will just go to t + C:

    [tex] \frac{-RC}{Q - VC} dQ = t + C [/tex]

    [tex] \frac{-RC}{Q} + \frac{RC}{VC} dQ [/tex]

    [tex] \frac{-RC}{Q} + \frac{R}{V} dQ [/tex]

    So does this go to:

    [tex] \frac{-RC}{ln(Q)} + \frac{RQ}{V} = t + C[/tex]

    ???

    TFM
     
  18. Oct 26, 2008 #17

    alphysicist

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    I would set limits on the t integral:

    [tex]
    \int_0^t dt
    [/tex]

    That way the constant is evaluated.

    Or if you want to be a bit more general, you could use

    [tex]\int_{t_0}^{t} dt[/tex]

    but I don't think that is probably necessary here.

    This is okay, but your next line is not the thing to do. Just go ahead and integrate the left side.


     
  19. Oct 26, 2008 #18

    TFM

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    Okay

    [tex] \frac{-RC}{Q - VC} dQ = \int^t_0 dt [/tex]

    So

    [tex] \frac{-RC}{Q - VC} dQ = t [/tex]

    But is the integral for the left:

    [tex] \frac{-RC}{ln(Q - VC)} = t [/tex]

    ???
     
  20. Oct 26, 2008 #19

    alphysicist

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    No, the ln function goes in the numerator. Also, remember that you have to evaluate the limits of the integral.
     
  21. Oct 26, 2008 #20

    TFM

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    Okay, so:

    [tex] \int^Q_0 \frac{-RC}{Q - VC} dQ = t [/tex]

    [tex] \frac{ln(Q - VC)}{-RC} = t [/tex]

    ???

    TFM
     
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