switches, inductors and impulse

Discussion in 'Homework Help' started by tAllann, Nov 3, 2013.

  1. tAllann

    Thread Starter New Member

    Oct 26, 2013
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    [​IMG]

    Sorry for the low quality image. The switches are at the top connecting the wires which connect the wires connecting the resistors and inductors. They open at t = 0. The values of the inductors pictured are 3mH on the left, and 6mH on the right.

    This problem is kind of confusing to me. The switches are open for t < 0 and close at t = 0. It seams that in both configurations, the resistance of the resistors is not an essential component for solving the problem which is to find V(t). But in this case, V(t) seams to be 0, except for the short time when the switch opens and there is a sudden impulse.

    So I went about solving the problem by finding the initial current through the inductors and then solving for the current as a function of time after for t >= 0 when the switches open. But there is no resistance, so I just get I(t) = Ioe^(-tR/L) = Io u(t), where u(t) is the unit step function.

    So to get the voltage, which it seams is equal to the voltage across either inductors over time. I use v(t) = L di(t)/dt.

    And here is where it gets funny, because normally this would just be 0, but we are after the impulse, and so we take the derivative also of the step function giving us the impulse function which is 0 everywhere except t = 0.

    So the answer I get is 3mH * 24 imp(t) A, which is 8X10^(-3) imp(t) volts. But I'm not sure I am doing this correctly. The book is very confusing, and the answer in the back of the book just says V(t) = 24 imp(t).

    If anyone can help clarify what is happening in this circuit and verify that the impulse is 24mVolts not just 24Volts, that would be much appreciated. I'm most confused about taking the derivative of the x*u(t), and especially about the nature of the currents I1(t) and I2(t), both positive and heading towards each other, and constant for all of t > 0.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Let's make sure we are on the same page from the beginning:

    First you said, "The switches are at the top connecting the wires which connect the wires connecting the resistors and inductors. They open at t = 0."

    Then you said, "The switches are open for t < 0s and close at t = 0."

    Which is it? Do they open at t=0s or close at t=0s?

    Assuming they open at t=0s, there is no way to determine how current is flowing in each inductor just before the switches open. We only know that the sum of the currents is 12A. That being the case, if there is a firm answer then is must be independent of the sharing ratio.
     
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  3. tAllann

    Thread Starter New Member

    Oct 26, 2013
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    Yes sorry, the switches open at t = 0.

    I had also considered adding the inductors in parallel, in which case they become a single 2 mH inductor with 12 amps flowing through it, and then if you do the same technique of taking the derivative with unit step function, then you still get 24 mV imp(t), which means that momentarily (or I guess only at t = 0) you have a potential difference of 24mV, and then immediately after, no potential voltage.

    I have no idea if I am doing this correctly. I was thinking about finding the stored energy at DC steady state, and then seeing if that can tell me something.
     
  4. WBahn

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    Mar 31, 2012
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    If you combine the two inductors, then where will the current flow immediately after the switch is opened?
     
  5. tAllann

    Thread Starter New Member

    Oct 26, 2013
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    That is a good question because in that case there is no loop, so there should be no current.

    What I think is strange about the way I have gone about this problem is that when you have no resistor in the loop with the inductor, and so goes R/L goes to 0, then you get that the current after time 0 is a constant 12 Amps, and this is independent of the value of L. But when you want the voltage impulse, then you get different values depending on the values of L.
     
  6. WBahn

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    Mar 31, 2012
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    Basically what you have is that the two inductors have incompatible currents once the switch is opened. Both inductors have a certain amount of energy in them and the current (and thus the energy) has to change. Thus, one inductor will act as a source in order to deliver energy to the other inductor and this will continue until the currents in the two are compatible. Since there is no resistance, this happens instantaneously via an impulse voltage strong enough to instantly transfer the needed amount of energy from one to the other.
     
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  7. tAllann

    Thread Starter New Member

    Oct 26, 2013
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    One of the inductors has 96mJ of energy stored and other has 48mJ. Assuming that they need to be equal in order for the current to be stable, then each needs 72mJ = (96+48) / 2. So the 96mJ inductor will need to transfer (96-72)mJ = 24mJ to the other inductor.

    Am I thinking about this correctly?
     
  8. WBahn

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    Mar 31, 2012
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    How are you determining the initial energy in each? How do you know how much current if flowing in each inductor when the switches are closed for a long time?
     
  9. tAllann

    Thread Starter New Member

    Oct 26, 2013
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    When the switches are closed for a long time, there has been 12 Amps flowing through the two inductors in parallel. So I used current division to get that 4 Amps had been flowing through the 3mH inductor and 8 Amps had been flowing through the 6mH inductor.

    I used the equation E = .5 L Io^2 to get the energy stored in each. E1 = .5*6mH*(4A)^2 = 48mJ ...
     
  10. WBahn

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    Mar 31, 2012
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    But what is the current division equation you are using based on? It's based on RESISTANCE, not inductance.

    Imagine putting a small series resistor along with each inductor. Say 0.01Ω for one of them and 0.001Ω for the other. Take your pick which has which. Now you can determine the steady state current in each inductor and it is completely determined by the resistors, not the inductance.

    So all you know is that one inductor has I1 and the other inductor has (12A-I1).

    An argument can be made (but a weak one) that the current will share so as to result in minimum total energy stored in the pair of inductors. But that is a pretty week argument.
     
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