Linear Systems question...

Discussion in 'Homework Help' started by Vim0314, Mar 27, 2010.

  1. Vim0314

    Thread Starter New Member

    Sep 1, 2009
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    Hey guys. First time posting, but been reading stuff on here for a while.

    I have attached a scan from my old test and I'm having trouble figuring out part b. Could someone help me figure it out?

    I understand the cap is replaced by a voltage source and the inductor is replaced by a current source. But I am having trouble figuring out the current in the voltage source and the voltage across the current source.

    Thanks in advance!
     
  2. Ryoshima

    New Member

    Mar 19, 2009
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    0
    I believe that:

    Ic = -20 A
    Vl = 120 V

    Don't think those are right, but they are the results that I obtained. I'll keep an eye on this thread to see if anyone can confirm or deny my calculations.
     
  3. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Ryoshima, no one will deny your calculation or Vim0314 for that matter.
    Some one will just confirm the calculations and will let you know if you are wrong or right.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I have VL(0+)=-54V and IC(0+)=-9A
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You need to write a loop voltage equation that satisfies the loop conditions at t=0+. Keep in mind that VC(0+)=16V and IL(0+)=4A

    Assume a current I is flowing in the 6Ω resistor - from this and the above information you can write a suitable loop voltage equation with I as the unknown.
     
    Last edited: Mar 27, 2010
  6. Vim0314

    Thread Starter New Member

    Sep 1, 2009
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    I found the answer in the solutions manual and here it is attached...

    In blue I boxed in the correct answers I am looking for. But in yellow I have circled where I have a problem. Can someone explain to me how they figured out the Ir1? and how a 24 magically appears?

    thanks
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One could write ..... somewhat laboriously ......

    VR1+VR2+VR3+48=0

    IR1*2+16+IR3*6+48=0

    IR1*2+16+(IR1-4)*6+48=0

    IR1*2+16+IR1*6-4*6+48=0

    IR1(2+6)+16-24+48=0

    IR1(2+6)=-48-16+24

    IR1=(-48-16+24)/(2+6)=-40/8=-5
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Rather than hijack a thread it's better for all concerned if you start a new thread.

    Keep in mind also the expectation that homework forum posts will show an attempt at solving the problem.
     
    Last edited by a moderator: May 28, 2010
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