RLC, when in series ? when in parallel ?

Discussion in 'Homework Help' started by Hitman6267, May 29, 2010.

  1. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    Capture2.PNG

    In the solution of this problem they wrote alpha= R/2L. That formula applies when its a series RLC. What elements of the circuit are we talking about when we're saying series/parallel. I used to think it was the inductance and the capacitor. Does opening the switch making the inductance capacitor resistor a separate circuit has anything to do with it ?


    solution1.PNG

    Here's how I think if i'm asked to find i(t) -current through the inductance - for t > or = 0

    I get i(0) so I get a relationship like this A1+A2 =0

    then I derive the general equation of the case.
    => di/dt = ...
    then I think that di/dt isn't "anything" so I remember that the voltage of a inductance is V= L di/dt. so I multiply both sides of the newly derived equation to get V= ...
    In this circuit V(0) would be 0 because the inductance became a wire.

    Is this a wrong way to think ? It's never working for me.

    Back to the solution I posted. How did they find di(0)/dt ?
     
    Last edited: May 29, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes - this is a series circuit. There is only one path through which current may flow around the remaining RH loop when the switch opens - hence no parallel paths exist.

    If you consider the initial circuit conditions - starting with the general DE for vo(t) in terms of the LH side of the loop

    v_{o}(t)=Ri_{o}(t)+L\frac{di_{o}(t)}{dt}

    Hence

    \frac{di_{o}(t)}{dt}=\frac{1}{L}(v_{o}(t)-Ri_{o}(t))

    At t=0

    v_{o}(0)=75V

    i_{o}(0)=15mA

    \frac{di_{o}(0)}{dt}=\frac{1}{L}(v_{o}(0)-Ri_{o}(0))

    \frac{di_{o}(0)}{dt}=\frac{1}{1}(75-5000*15e^{-3})=0
     
  3. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    15e^{-3}

    is meant to be scientific format notation - for the 15mA value. I'm surprised you didn't realise that.

    perhaps I should have either written more correctly [by convention]

    15E^{-3}

    or

    15*10^{-3}

    I tend to use notations interchangeably & somewhat ambiguously perhaps.

    In any case the solution I gave is correct if you allow my intended meaning.
     
    Hitman6267 likes this.
  5. Hitman6267

    Thread Starter Member

    Apr 6, 2010
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    ok That makes sense. Thank you.
     
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