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Capacitor and Dependent Voltage Source

  1. Nov 7, 2017 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Select the correct expression for ##i_o(t)## for ##t≥0^+##.
    Figure_P07.30.jpg
    2. Relevant equations
    ##V(t) = V(∞)+[v(0)-v(∞)]e^{\frac{-(t-t_0}{\tau}}##
    ##i(t) = C\frac{dV(t)}{dt}##
    ##\tau = RC##

    3. The attempt at a solution

    I'm a bit stuck on this problem. Specifically, I'm not sure how to deal with the dependent source.

    At ##t=0##, the voltage of the capacitor should be 15 volts since the capacitor is directly in series with the voltage source with no branches leading off anywhere.

    After ##t=0## the current through the capacitor would be ##C\frac{dV(t)}{dt}##, but I'm sure the dependent source changes things somehow. I'm just not sure how.
     
  2. jcsd
  3. Nov 7, 2017 #2

    berkeman

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    Correct.
    Try writing the KVL loop equation for that right-hand loop after the switch changes, and then factor in the initial condition that you have.
     
  4. Nov 7, 2017 #3

    Drakkith

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    For my KVL loop I get ##5i_0 + V -15i_0 = 0##
    ## V-10i_0 = 0##
    ##V=10i_0##

    After this I'm not sure what to do. I thought my capacitor voltage equation would be ##V(t) = 15e^{-3333t}## but using that doesn't lead me to a correct current expression.
     
    Last edited by a moderator: Nov 7, 2017
  5. Nov 7, 2017 #4

    berkeman

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    To get the voltage across the capacitor, you need to use the differential equation version. There may be a better way to approach it, but I'd start out my try adding up the voltage drops across the components in that loop and setting the sum equal to zero. what is the voltage drop across the capacitor as a function of the current?
     
  6. Nov 7, 2017 #5

    Drakkith

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    Didn't I just do that?

    In general, or for this problem?
     
  7. Nov 7, 2017 #6

    cnh1995

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    In general, as a function of time.

    Or, set up a DE in terms of charge on the capacitor.
     
  8. Nov 7, 2017 #7

    Drakkith

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    It should be ##V(t) = \frac{1}{c}\int i(t)dt##
     
  9. Nov 7, 2017 #8

    cnh1995

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    Yes, so what is i(t)?
     
  10. Nov 7, 2017 #9

    cnh1995

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    You have arrived at the correct expression for V.
    What is i0(t) in terms of charge on the capacitor? What is voltage V in terms charge on the capacitor?
     
  11. Nov 7, 2017 #10

    NascentOxygen

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    I'm confused, too. You show a voltage arrow across the capacitor and the arrow appears to be labelled ##i_0##. Is that what you see in your textbook?
     
  12. Nov 7, 2017 #11

    Drakkith

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    It should be just what I put in my original post: ##i(t) = c\frac{dv}{dt}##
    ##i(t) = \frac{dQ(t)}{dt}##
    ##V = \frac{Q}{C}##

    Yes, this picture is directly from the book. ##i_0## is the current through the capacitor.
     
  13. Nov 7, 2017 #12

    cnh1995

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    Your equations for i(t) and V(t) in terms of charge are correct.
    Now substitute these values in the pink equation in #9 and solve the DE (for Q(t)).
     
  14. Nov 8, 2017 #13

    cnh1995

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    @Drakkith, I just realized that I wasn't careful with the polarities.
    To avoid any sign errors, I would reverse the the direction of Io (since the capacitor is discharging), call it I1 (So I1= -Io) and reverse the dependent source polarity (it will now be 5I1).

    Now you can see the dependent source is not really a source but is actually a 5 ohm resistor in disguise.
     
    Last edited: Nov 8, 2017
  15. Nov 8, 2017 #14

    Drakkith

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    Solving for the DE I get: ##i_0(t) = 50,000Ae^{50,000Q(t)}##
    ##Q(0) = CV(0) = 2*10^{-6}(15) = 3*10^{-5}##
    So ##i_0(0) = 50,000Ae^{1.5}##

    To find A, I need ##i_0(0)##.
    Doing a KVL loop: ##15-5i_0+15i_0=0##
    ##i_0 = -1.5A##

    Plugging that in:
    ##-1.5=50,000Ae^{1.5}##
    That would make ##A = -6.69*10^{-6}##
    and:
    ##i(t) = -0.335e^{50,000Q(t)}##

    Well, that's closer to one of the possible answers. The exponential matches some of them, but the constant doesn't.

    To solve my DE, I did:
    ##V-10i_0 = 0##
    ##\frac{Q}{C}-10\frac{dQ}{dt} = 0##
    ##-10\frac{dQ}{dt}+\frac{Q}{C} = 0##
    ##\frac{dQ}{dt}-\frac{Q}{10C}=0##
    ##\frac{dQ}{dt}-50,000Q=0##
    My integrating factor is: ##exp(\int -50,000 dQ)=e^{-50,000Q}##
    Multiplying the equation by the IF:
    ##e^{-50,000Q}\frac{dQ}{dt} -50,000e^{-50,000Q}Q = 0##
    That's just:
    ##\frac{d({e^{-50,000Q}Q})}{dt} = 0##
    Integrating and moving the constant over:
    ##e^{-50,000Q}Q = A##
    ##Q = \frac{A}{e^{-50,000Q}}##
    ##Q = Ae^{50,000Q}##

    Since ##i_0 = \frac{dQ}{dt}##
    ##i_0 = 50,000Ae^{50,000Q}##
     
  16. Nov 8, 2017 #15

    cnh1995

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    Is the given answer I(t)=0.75e-25000t?
     
  17. Nov 8, 2017 #16

    Drakkith

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    Apparently it's ##-0.75e^{25,000t}##
     
  18. Nov 8, 2017 #17

    cnh1995

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    Yes, because I'd reversed Io in #13.

    Your equation for i(t) in #14 is dimensionally incorrect (and you don't need that much math either).

    Use the steps in #13. The fact that the dependent voltage source is actually just a 5 ohm resistor makes the solution a lot simpler.
     
  19. Nov 8, 2017 #18

    Drakkith

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    Do you mean treat the dependent source as a resistor and solve the problem?

    Hmm. Am I supposed to be able to recognize that this is a resistor in order to solve this?
     
  20. Nov 8, 2017 #19

    cnh1995

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    Yes.
    Even if you don't recognize that, you'll realize it when you write the KVL equation. But since the capacitor is discharging, it would be better if you assumed a current I1 (anticlockwise) such that I1= -Io. Remember that you'll have to reverse the dependent source's polarity as well.

    Now you'll have current I1 flowing throgh the dependent source from its positive terminal to negative terminal and this voltage "drop" is 5I1. Doesn't this suggest that the dependent source is acting as a 5 ohm resistor here?
     
  21. Nov 8, 2017 #20

    Drakkith

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    Alright, treating the dependent source as a resistor and using ##i_1 = -i_0 ##:
    ##V(t)=15e^{-25000t}##
    ##i_1(t)=-c\frac{dV}{dt}=-2*10^{-6}(-25000)(15)e^{-25000t}=0.75e^{-25000t}##

    I suppose so. I'm not sure I would have recognized that had you not mentioned it. Correct me if I'm wrong, but it looks like this is only true as long as the polarity of the source is such that the current enters the positive terminal, correct? If the source voltage was proportional to ##-5i_0## then we wouldn't be able to treat this as a resistor.
     
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