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Self Inductance Question

  1. Nov 13, 2008 #1

    TFM

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

    Give a definition of (self) inductance. Suppose a battery, which supplies a constant EMF ϵ_0 is connected to a circuit of resistance R and inductance L at t = 0. Find an expression for the current as a function of time.

    2. Relevant equations

    V = IR

    [tex] V = -L\frac{dI}{dt} [/tex

    3. The attempt at a solution

    I am assuming that this is to be treated as a Kirchoff Loop, thus the total voltage = 0

    Voltage providers:

    Inductor
    Battery

    Users:

    Resistor

    Thus I have the equation:

    [tex] \epsilon - L\frac{dI}{dt} - IR = 0 [/tex]

    and thus:

    [tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

    treating like a differential equation:

    [tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

    [tex] \epsilon dt - L dI = IR dt [/tex]

    rearrange:

    [tex] \frac{L}{IR} dI = -dt + \epsilon dt [/tex]

    Gives:

    [tex] \frac{1}{L}ln(IR) dI = -t + \epsilon t [/tex]

    multiply by L

    [tex] ln(IR) = -Lt + \epsilon t [/tex

    take exponentials:

    [tex] IR = e^{-Lt} + e^{\epsilon t} [/tex]

    Does this look right so far?

    TFM
     

    Attached Files:

  2. jcsd
  3. Nov 14, 2008 #2

    alphysicist

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    Homework Helper

    Hi TFM,

    This equation does not follow from the previous one.
     
  4. Nov 14, 2008 #3

    TFM

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    No it doesn't

    so:

    [tex] \epsilon dt - L dI = IR dt [/tex]

    see, I have to rearrange to get I onto the left side.

    does this look better:

    [tex] \epsilon dt - L dI = IR dt [/tex]

    [tex] - L dI = IR dt - \epsilon dt [/tex]

    [tex] - L dI = (IR - \epsilon) dt [/tex]

    [tex] - \frac{L}{IR - \epsilon} dI = dt [/tex]

    so and so:

    [tex] - \frac{1}{L}ln(IR - \epsilon) = t [/tex]

    Does this look better?

    TFM
     
  5. Nov 14, 2008 #4

    alphysicist

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    I think this part looks okay.

    This isn't quite right. The integration is not right (if you take the derivative of the left side with respect to I you don't get the left side of the previous equation). Also remember that these are definite integrals, so you have to evaluate the limits.
     
  6. Nov 14, 2008 #5

    TFM

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    Would the limits be for I between 0 and I and for t between 0 and t?

    I am not sure how to integrate this, because I thought when you integrated:

    [tex] \frac{1}{x} dx [/tex]

    you got:

    [tex] ln(x) [/tex]

    and for:

    [tex] \frac{b}{x} [/tex]

    where b is a constant, you got:

    [tex] \frac{1}{b}lnx [/tex]

    ???

    TFM
     
  7. Nov 14, 2008 #6

    alphysicist

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    No, the integral

    [tex]
    \int \frac{b}{x} dx \to b \ln x
    [/tex]
    (plus a constant) because in that case the b can come out of the integral (it is not affected by the integration process at all). However, you do have to take into account that the R is multiplying the I.

    [tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]
     
  8. Nov 14, 2008 #7

    TFM

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    Okay, so:

    [tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]

    [tex] -L(ln(IR - \epsilon) - (0R - \epsilon)) = t - 0 [/tex]

    [tex] -L(ln(IR - \epsilon - - \epsilon)) = t [/tex]

    [tex] -L(ln(IR)) = t [/tex]

    ???

    TFM
     
  9. Nov 14, 2008 #8

    alphysicist

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    The R has to be handled like this:

    [tex]
    \int \frac{1}{IR-\epsilon}\ dI \to \frac{\ln(IR-\epsilon)}{R}
    [/tex]
    (plus the constant).

    Also when you go to apply limits,

    [tex]
    \ln (x)\Big|_{x_0}^{x_f} \neq \ln (x_f - x_0)
    [/tex]

    Instead, it is:


    [tex]
    \ln (x)\Big|_{x_0}^{x_f} = \ln (x_f) - \ln(x_0)
    [/tex]
     
  10. Nov 14, 2008 #9

    TFM

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    Okay, so:

    [tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]

    This will integrate to:

    [tex] -L\left[\frac{ln(IR - \epsilon)}{R}\right]^I_0 = t [/tex]

    [tex] -L \left(\frac{IR - \epsilon}{R} - \frac{0R - \epsilon}{R}\right) = t [/tex]

    [tex] -L \left(\frac{IR - \epsilon}{R} - \frac{- \epsilon}{R}\right) = t [/tex]

    does this look better?

    ???

    TFM
     
  11. Nov 14, 2008 #10

    alphysicist

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    You seemed to have dropped your [itex]\ln[/itex]'s when you put in the limits.
     
  12. Nov 14, 2008 #11

    TFM

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    Sorry, copied through typo :redface:

    It should be:

    [tex] -L\left[\frac{ln(IR - \epsilon)}{R}\right]^I_0 = t [/tex]

    [tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(0R - \epsilon)}{R}\right) = t [/tex]

    [tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t [/tex]

    ???

    TFM
     
  13. Nov 15, 2008 #12

    TFM

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    Does this look correct

    ???

    TFM
     
  14. Nov 15, 2008 #13

    alphysicist

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    What does that give for I?
     
  15. Nov 15, 2008 #14

    TFM

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    Well, if we rearrange it:

    [tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t [/tex]

    [tex] \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = -\frac{t}{L} [/tex]

    Multiply by R:

    [tex] ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L} [/tex]

    take exponentials:

    [tex] IR - \epsilon + \epsilon = e^{-\frac{Rt}{L}} [/tex]

    [tex] IR = e^{-\frac{Rt}{L}} [/tex]

    [tex] I = \frac{^{-\frac{Rt}{L}}}{R} [/tex]

    Does this look correct?

    TFM
     
  16. Nov 15, 2008 #15

    alphysicist

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    This line is not correct, because:

    [tex]
    \exp \left\{ \ln x + \ln y\right\} \neq x + y
    [/tex]

    Before you take the exponential of both sides, you just want a single natural log on the left . That is, you want the left side to be just:

    ln(something)
     
  17. Nov 15, 2008 #16

    TFM

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    so:

    [tex] ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L} [/tex]

    isn't [tex] ln(a) - ln(b) = ln(\frac{a}{b}) [/tex]

    ???

    If so:

    [tex] ln(IR - \epsilon) - ln(-\epsilon) \equiv ln\left(\frac{IR - \epsilon}{-\epsilon} \right) [/tex]

    giving:

    [tex] ln\left(\frac{IR - \epsilon}{-\epsilon} \right) = \frac{-Rt}{L} [/tex]

    taking exponentials:

    [tex] \frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}} [/tex]

    Does this look better?

    TFM
     
  18. Nov 15, 2008 #17

    alphysicist

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  19. Nov 16, 2008 #18

    TFM

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    Excellent. So:

    [tex] \frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}} [/tex]

    [tex] IR - \epsilon = \epsilon e^{\frac{-Rt}{L}} [/tex]

    [tex] IR = \epsilon - \epsilon e^{\frac{-Rt}{L}} [/tex]

    factorise out:

    [tex] IR = \epsilon \left(1 - e^{\frac{-Rt}{L}} \right) [/tex]

    divide by R:

    [tex] I = \frac{\epsilon}{R} \left(1 - e^{\frac{-Rt}{L}} \right) [/tex]

    Does this look correct?

    TFM
     
  20. Nov 16, 2008 #19

    alphysicist

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    That looks right to me.
     
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