Vcvs in rc circuit

Discussion in 'Homework Help' started by jjtjp, Apr 27, 2014.

  1. jjtjp

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    Mar 3, 2014
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    short title because can't post??

    I'm just not sure how to go about solving this. I tried labeling a reference node above the resistor through which we want current. The voltage through this is  7V_{c} is it not? Which results in the current being
    7V_{c} \times \frac{1}{R} = 7e^{\frac{-t}{2}}
    Well this makes sense to me, at least, but when I model it in LTSpice the current decays much more quickly. I tried to reverse solve and I got a \tau=1.845 but even plugging this in for 2 in my equation doesn't get the decay to happen fast enough so I'm thinking the formula/method of solving I've chosen is completely incorrect. Sorry for all the detail but I just wanted you to know I've worked on it for a while.
    You can't see from the pic but V_{c}(0) = 10 V and we are solving for i_{x}
     
  2. shteii01

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    Feb 19, 2010
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    Sounds good. Good luck.
     
  3. jjtjp

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    Mar 3, 2014
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    I'm not sure if you're saying good luck because you can't help me or because I failed to ask an explicit question. Either way, I'll ask one:

    Can someone point me in the right direction on how you would solve this? Thanks!
     
  4. jjtjp

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    Mar 3, 2014
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    Maybe the pic is unclear, but it is a Voltage controlled voltage source dependent upon the voltage across the capacitor which is initially at 10V. So the Initial current Ix is 7 amps. But I'm not sure what is wrong with my formula after that.
     
  5. shteii01

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    Look up Unbounded Response in your textbook. That is where I have RC circuits with dependent sources.
     
  6. jjtjp

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    Mar 3, 2014
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    Thanks for the reply but I read and understand that section. But in this case, if you'll notice, the solution is not unbounded but converges to 0.
     
  7. shteii01

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    Ok. Back to my original comment.
    What are we solving for?
     
  8. jjtjp

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    Mar 3, 2014
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    ix as a function of time
     
  9. shteii01

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    I am getting differential equation. It has to be solved by Separation of Variables Technique. This will give you V(t). ix then becomes:
    ix(t)=[V(t)+6*V(t)]/10
     
  10. jjtjp

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    Mar 3, 2014
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    That's what I got, too. But SPICE software gives me a different answer. Who knows..
     
  11. shteii01

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    Did you do the actual integration using the initial condition to find the actual expression for V(t)?
     
  12. WBahn

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    Mar 31, 2012
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    I got an initial current of 7A with a time constant of 0.5 seconds, not 2 seconds.

    I suspect that, had you tracked your units, you would have discovered that you used a reciprocal when you shouldn't have.
     
  13. jjtjp

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    Mar 3, 2014
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    Can you tell me how you got that as a Time constant? That definitely gets closer to the expected behavior based on SPICE but I am unclear on how else you would get a time constant other than R_{eq}C= 0.4\times5 = 2 s

    The only other way I see the cap dissipating is 100% goes through the 10Ω that's in parallel because the potential is always higher at the voltage source. But this would give an even larger time constant (4s) which is going the wrong way from where it should be.

    How would the differential equation look, shteii01? Other than a natural or step response with a first order linear diff eq I'm really at a loss.
     
  14. WBahn

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    I solved it using transform methods, which are probably a bit beyond where you have gotten so far.

    I don't know how you are getting an equivalent resistance of 5 ohms. The two 10 ohm resistors are NOT in parallel!

    When you have a dependent source, you need to find the equivalent circuit using a test source and taking the ratio of Vtest/Itest. So replace your capacitor with Vtest (a DC source will do).
     
  15. shteii01

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    WBahn probably did the Thevenin Equivalent circuit. That is the first way I did it since that is how they do it in my textbook. What I got is Thevenin Resistor of 1.25 Ohm. tau then: tau=Rth*C=1.25*.4=0.5

    Overall, you get: dV/dt=-2V
    At this point you have to use Separation of Variables Technique to find V(t). You, for whatever reason, don't seem to want to do it.
     
  16. shteii01

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    Here is an example of how to solve differential equation of the type we got:
    http://www-math.mit.edu/~djk/18_01/chapter15/example04.html

    In that example they have:
    dv/dt=-v+32
    initial condition: v(0)=0

    You have:
    dV/dt=-2V
    initial condition: V(0)=10

    So just work through the example and then apply the steps to your differential equation.
     
  17. WBahn

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    Note that if you know it is a first order response and you know the initial and final values, then you don't need to solve the differential equation at all since you know the result will be of the form:

    y(t) = y(t=∞) + [y(t=0)l - y(t=∞)] exp(t/τ)

    None-the-less, you should be able to solve the differential equation and if you have difficulties doing so then that is all the more reason to slug through it.
     
  18. jjtjp

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    Mar 3, 2014
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    Thanks for all the responses. I've been out but now that I'm home I'm going to try it again. For the record, I don't have any problems solving differential equations (tho I am a bit out of practice b/c that class was a year ago now). I was just having a hard time coming up with a diff eq that was different from the standard form which yielded me the wrong answer. I was never taught the test source method for obtaining norton/thevenin eq so I am going to try that and I'll report back. Thanks again.
     
  19. shteii01

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    The test source method is used when a dependent source is present in the circuit. The chapter in my textbook that deals with deriving thevenin equivalent circuit has a section on how to derive thevenin equivalent circuit when dependent source is present, the test source method is introduced and used there.
     
  20. jjtjp

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    Mar 3, 2014
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    Okay great! So I got RC of 5, too with an Req of 1.25. I wish my teacher had taught that. It's not her fault. There are just a ton of morons in that class and they slow the class down so much that I think we just moved on. We were probably supposed to read it and I missed that note. Anyway, it is an easy concept once you're aware of it. So anytime there is a dependent source you remove the capacitor/inductor and replace it with a test voltage with a test current and solve that way? Just want to get everything correct in my head.
     
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