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Dependent source parallel an inductor with a sinusoidal source

  1. Aug 6, 2012 #1
    an independent source with a forcing function modelled by cos(500t) is in series with an 100 ohm resistor. A dependent (current controlled) source is placed across the remaining resistor terminal and ground. A 0.3mH inductor is in parallel with the dependent source. The value of the dependent source is 0.2 multiplied by the inductor current.

    now i have tried many attempts.I've tried applying KVL around the outer loop giving me a standard first order linear DE Li' + 80i = cos(500t). I've tried KCL, giving me a integro-differential equation. This was pretty much the same DE after i differentiated through.


    If i try to remove the inductor that means the dependent source goes, so i can't use thevenin's theorem. There has to be a simplified way. please help
     
  2. jcsd
  3. Aug 6, 2012 #2

    CWatters

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    Think about the dependent current source. It has the same voltage across it as the inductor. It's current behaves like the current through an inductor so what can you replace it with? Then simplify the circuit.
     
  4. Aug 6, 2012 #3

    gneill

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    What is it you need to solve for? Some particular voltage or current?
     
  5. Aug 6, 2012 #4
    Hey gneill.I'm solving for the inductor current IL. The current source is 0.2IL and i should have mentioned before the dependent current flows in the opposite direction to the inductor current.

    Hello CWatters. Are you suggesting that I can model the dependent source as an inductor parallel to the other inductor?
     
  6. Aug 6, 2012 #5

    berkeman

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    The voltage source has a fixed frequency. You don't need to write the full differential equations. Are you familiar with using Phasors and complex impedance?
     
  7. Aug 6, 2012 #6
    yeah, we are learning phasors and impedance soon. i've just started to solve basic inductive circuits by replacing the source with a complex forcing function. I think the person who wrote the question wants us to go through the hard way of solving it, so that when we do learn phasors, we can compare the two methods.
     
  8. Aug 6, 2012 #7
    here is the cct diagram with the DE. I didn't show the DE solution because it would have been too much work to upload
     

    Attached Files:

  9. Aug 6, 2012 #8

    berkeman

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    So to move forward with this, assume a solution for the inductor current iL(t), differentiate it, and substitute iL(t) and i'L(t) back into your last equation. What would be a good guess at the form of the solution iL(t)?
     
  10. Aug 7, 2012 #9
    had to sign in with a different account. So the form will be iL = Acos(500t) + Bsin(500t).

    I substituted this back in and solved for iL to get A = (12.5 x 10^-3)
    B = (23.43 x 10^-6).

    But the answer in the book is 5.88cos(500t - 61.9°). {which isn't equivalent to my answer}

    This is why i have a problem in the first place. Hence I'm questioning the logic I used in deducing the DE and also in solving it
     
  11. Aug 7, 2012 #10

    gneill

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    Your result looks fine to me (although I believe that the B term should be negative; you might want to recheck your math). It maybe that there's a mistake in the book's answer key. It's been known to happen.
     
  12. Aug 7, 2012 #11

    NascentOxygen

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    At 500 rad/sec that 0.3mH inductor has an impedance of 0.15Ω. So that 100Ω resistor will dominate the scene, and current angle will be very close to 0.

    ▸ Try solving with the resistor set to 100mΩ.
     
  13. Aug 7, 2012 #12
    Ah yes, i did notice that after. thank you.
     
  14. Aug 7, 2012 #13
    Yes I found that with resistor set to 80mΩ that the answer is perfect. although that value for a resistor seems unlikely somehow.

    I think they may have made a mistake
     
  15. Aug 7, 2012 #14
    thank you CWatters, gneill, berkeman and NascentOxygen
     
  16. Aug 8, 2012 #15

    gneill

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    Not necessarily; it might for example represent the resistance of the wire leads from the power source to the load if the circuit was meant to represent some 'real life' device.
     
  17. Aug 9, 2012 #16
    That is true. Thankyou for the help guys
     
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