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Homework Help: RL Partial Circuit

  1. Apr 24, 2008 #1
    Partial circuit shown in attachment. A current pulse is fed to the partial circuit shown in Figure P32.25. the current begins at zero, then becomes 10.0A between t = 0 and t = 200 microseconds, and then is zero once again. Determine the current in the inductor as a function of time. I1 is the current before the junction, I2 is the current through the resistor, I3 is the current through the inductor, and R is the 100 Ohm resistance.

    The current is 0A when t < 0s. For t between 0 and 200 microseconds, I used Kirchoff's Method and came up with:
    L*dI3/dt = (I1 - I3)*R...from here on I came up with a differential equation and got:
    I3 = 10(1 - e^(-10000t), which is the answer I'm supposed to get, although when coming up this, I had to assume I1 was unaffected by the inductance and remained as 10A through the time interval.

    Where I am stuck at is finding the current when t > 200 microseconds. The answer I'm supposed to get is I3 = (63.9A)*e^(-10000t), in which 63.9A is the initial current right after the current pulse is gone. I can get the format of the equation, but I just dont know how to find that the initial current is 63.9A...help?

    Attached Files:

  2. jcsd
  3. Apr 24, 2008 #2
    Not sure, but can you show us what you did to find the 'format of the equation'?

    Are you sure you specified the integral limits correctly when solving the DE?
  4. Apr 24, 2008 #3
    After the current pulse is gone, there should be an emf induced by the inductor that would provide for a current. I took the inductor-resistor loop and used Kirchoff's method to come up with:
    -L*dI/dt - IR = 0...using that, I came up with I = Io*e^(-10000t), in which Io is the initial current.
  5. Apr 24, 2008 #4
    63.9A is not the current after the current pulse is gone. you get the current after the
    pulse is gone by substituting t = 200 microseconds in (63.9A)*e^(-10000t).

    if you evaluate the first equation you had for 0 <= t <= 200 microseconds, you will
    find that that will give the same value for I3 at t = 200 microseconds.
  6. Apr 25, 2008 #5
    wait...can you expand on this (I'm not exactly getting it)?
  7. Apr 25, 2008 #6
    if you have [tex] I = I_0 e ^ {\frac{- t}{R C}} [/tex] than I_0 is only the initial current, if you mean the current at t=0 with that. Here the initial current after the switch closes is

    [tex] I = I_0 e ^ {\frac{- 0.0002}{R C}} [/tex]
  8. Apr 25, 2008 #7
    wait...doesn't RC refer to resistor-capacitor system? There are no capacitors in this partial circuit.
  9. Apr 25, 2008 #8
    sorry. replace [itex] \frac {-t}{R C}[/itex] by [itex] \frac { - R t} {L} [/itex] in the previous post.
  10. Apr 26, 2008 #9
    oh, ok....I think I get what you're saying; there is a functional continuity, in which at t = 200 microseconds for the first one and t = 0s for the second one come down to the same value, and I sort of realize my mistake in modeling the situation after the pulse in that I was assuming the resistor carried the same current the inductor did, although there is a junction where the current could split. So then, how did they come up with 63.9 or the equation in general?
  11. Apr 26, 2008 #10
    How they came up with the equation in general is easy, you just repeat what you did for the first question (kirchhoff).

    This yields: [tex]-iR -L \frac{di}{dt} = 0[/tex]
    Solving this yields: [tex]- \frac{R}{L} dt = \frac{di}{i}[/tex] and taking integrals:
    [tex]\int_0^t - \frac{R}{L}dt = \int_{I_0}^i \frac{di}{i}[/tex].
    Then after rearranging you get:
    [tex]i = I_0 \exp (- \frac{Rt}{L} )[/tex]

    How you get the value of [tex]I_0[/tex] however I can't see so quickly... I'll think about it.
  12. Apr 26, 2008 #11
    yeah...that value doesn't seem to come to me.
  13. Apr 26, 2008 #12
    I did some thinking (just short) but I still don't know... anyone else maybe?

    What I did think was weird though, if the current is actually jumping from 10A to 0A in an infinitely small amount of time (like the question implies), shouldn't the emf induced in L be infinite? Since the emf is [itex]L \frac{di}{dt}[/itex] and the rate of change in [itex]i[/itex] is infinite? Argh I dunno :p
  14. Apr 26, 2008 #13
    heh...thats what I was sort of thinking. I typed the question and did the diagram exactly as they are. I'm not sure whats going on with this question.
  15. Apr 26, 2008 #14
    The current through the inductor can not jump. It will be the same just before and just after 200 microseconds. that is how you find I_0.
  16. Apr 26, 2008 #15
    So what you're saying is if I set the before and after equations equal, I could solve for I_0...but when I do that, the limit as to how much I_0 can go up to is 10A, isn't it?
  17. Apr 26, 2008 #16
    The current through the inductor is limited to 10 A. that does not mean that I_0 is limited in the expression below. (I is smaller than 10 A if t>= 0.0002)

    i = I_0 \exp (- \frac{Rt}{L} )
  18. Apr 26, 2008 #17
    oh...I see what happened. I've always been taking t = 0s for that expression...a mistake I missed. Why is it .0002s?
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