Help on First order Diff Eq problem

  1. 1. The problem statement, all variables and given/known data,dJ2iv

    2. Relevant equations

    Current in loop 1: i_1 going counter-clockwise
    Current in loop 2: i(t) going counter-clockwise

    Before opening switch: we know that 2 loops exist, left and right. also the current is constant because it says the circuit was at rest for a long time.

    After we open the switch, we know that the left loop is shorted out, and we somehow use (a) and (b) and plug it into our equations and solve?

    We also know that i(0-) = i(0+)

    3. The attempt at a solution

    Before opening the switch:

    Loop 1 (left) : -v_initial - (i_1 - i(t)) R = 0 ---> v_initial = -R i_1 + R i(t)

    Loop 2 (right): i(t) = - R i(t) + r i_1 - 1/c integral i(t)dt

    After switch is open:

    i(0+) = 1
    v(2microseconds) = 1/e


    I don't understand how to proceed from here? What exactly do I have to do.. I'm stuck. Please let me know as soon as possible. Thanks.
  2. jcsd
  3. rude man

    rude man 6,086
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    Why do you have two current loops?
  4. Before switch is open, it's been closed for a long time and that means there are 2 loops; one on the left and one on the right. And is current not going into the left loop?
  5. rude man

    rude man 6,086
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    How can there be a loop when the switch is open?
  6. It says switch is closed beforehand, i'm writing the equations for when it's closed..

    I don't know that's why I am asking. What am I supposed to do here?
  7. rude man

    rude man 6,086
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    You're only interested in what happens after the switch is opened. The switch having been previously closed (for infinite time, in fact) determines your initial condition(s) when the switch is opened.

    So, what is/are the initial condition(s) just before, i.e. when, the switch is opened?
  8. We know that current is a constant value, and that current before opening switch must equal current after opening the switch.
  9. rude man

    rude man 6,086
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    Right. So how about writing an expression for that current? What role if any does C play in determining it?
  10. Because current through a capacitor is C dv/dt , we know that current is constant before switch is open which means current through capacitance is 0 before AND after opening switch?

    Thus, we can just say that ( -i(t) / 1 - i(t) R = 0 ) using KVL <-- but isnt voltage through capacitor 1/C integral i dt? So shouldnt we include that?

    So we get: i(t) = i(t)*R .. so if we divide by i(t) we know R = 1 ??

    But if we included the Capacitor equation in KVL, then it should be:

    i(t) = - R i(t) + r i_1 - 1/c integral i(t)dt

    not sure which is right..

    How do we use part (a) and (b) that they gave us?
  11. If I actually use KVL on the right loop I get this:

    -1/c integral i(t)dt - v(t) - i(t)R = 0

  12. rude man

    rude man 6,086
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    Right, but you need your equation to have one dependent variable, not two. What can you write in place of v(t)?

    Get back to you in about 8-9 hrs.
  13. -i(t).

    So our equation becomes:

    -1/c integral i(t) dt - i(t) - i(t) R = 0

    So do we jjust plug in i(0+) = 1A?

    so -1/c t - 1 - R = 0 ?

    How do I use the voltage equation? (v 2us) = 1/e ?
  14. rude man

    rude man 6,086
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    First, I ask you to label the 1 ohm resistor as R1. That way you can keep track of your units term-by-term. That's a powerful way of catching math errors.

    So now your equation is
    -(1/C)∫i(t)dt - R1i(t) -Ri(t) = 0.

    This is an integral equation. What do you think should be done with this situation? It's entirely a math problem at this point.
    Last edited: Feb 13, 2012
  15. Are you familiar with the Laplace transform? This tool enables you to transform your differential equation into an algebraic one, simplifying the math.
    You can then use tables to convert from the Laplace domain back to time domain to get your answer.

    If not, you can still solve the differential equation. Remove the integral by differentiating the equation and solve for I. Remember then to apply the initial conditions to determine the I when the switch is first opened up.

  16. Okay so now we solve for i(t) using first-order differentials.

    We end up getting this:

    i(t) = Ke^ (-t/ (C(R+1))

    Now, we can use CONDITION a) that i(0+) = 1A to find K, which we find out K = 1.

    So our equation is now:

    i(t) = e^ (-t/(C(R+1))

    Now we can use our second condition v(2us) = 1/e to get a RC relation.

    v(t) = i(t)*1 <-- due to the voltage drop across the resistor

    so v(2us) = i(2us) = 1/e

    Plugging it in:

    e^-(2*10^-6)/C(R+1) = 1/e

    Take natural log of both sides, we end up getting:

    C = ( 2*10^-6 ) / ( R + 1 )

    But now.. I still have 2 variables with 1 equation. How do I solve for C and R ??
  17. rude man

    rude man 6,086
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    EDIT: oops! You're right!

    OK, you know that i(0+) = 1 amp. So what does that tell you about R1?

    Another way of putting it - what causes i(t) to flow once the switch is open and the 2V source is out of the picture?
    Last edited: Feb 13, 2012
  18. I thought R_1 is just 1 Ohm here?

    so that V(t) = i(t)
  19. rude man

    rude man 6,086
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    You're right, sorry. I meant R.
  20. Voltage is somehow still in the Resistor?
  21. rude man

    rude man 6,086
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    No, a resistor can't store voltage. What componernt can?
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