Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

alma1969

  • 3 years ago

What is an expression for the transform current through resistor R ?(see attachment)

  • This Question is Closed
  1. alma1969
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. abtrehearn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Is the source voltage an arbitrary function of time?

  3. abtrehearn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The current flows from the source through the first inductor, and then it branches off; part of the current flows through a purely capacitive branch. and the rest of it flows into the other inductor, which is in series with a resistor. At any instant, the voltages across the branches are the same. We can use this information and Kirchoff's laws to set up a system of equations to determine the currents throughout the circuit, including the current through the resistor.

  4. abtrehearn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I will be doing some pencil-and-paper calculations on this, so it may be some time before my next post here.

  5. alma1969
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Tank you for your time !!!!!!!

  6. abtrehearn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I get\[L^{2}Ci'''_R + RLCi''_R + Li'_R = e(t)\]for the time-domain equation. Since we are taking Laplace transforms of derivatives up to third-order, \[i''_R(0), i'_R(0),\]and\[i_R(0)\]enter into my calculations. Since nothing is stated about initial values for the current across the resistor, I keep those three numbers arbitrary, leaving them in symbolic form. When I take the Laplace transform of both sides of the equation I came up with, I get \[I_R(s) = E(s)/(L^{2}Cs^{3}+ RLCs^{2} + Ls)\] \[+ L^{2}Ci_R(0)s^{2}/(L^{2}Cs^{3} + RLCs^{2}+ Ls)\] \[+ (L^{2}Ci'_R(0) + RLCi_R(0))s/(L^{2}Cs^{3} + RLCs^{2} + Ls)\] \[+ (L^{2}Ci''_R(0) + RLCi'_R(0) + LI_R(0))/(L^{2}Cs^{3} + RLCs^{2} + Ls).\]I get\[L^{2}Ci'''_R + RLCi''_R + Li'_R = e(t)\]for the time-domain equation. Since we are taking Laplace transforms of derivatives up to third-order, \[i''_R(0), i'_R(0),\]and\[i_R(0)\]enter into my calculations. Since nothing is stated about initial values for the current across the resistor, I keep those three numbers arbitrary, leaving them in symbolic form. When I take the Laplace transform of both sides of the equation I came up with, I get \[I_R(s) = E(s)/(L^{2}Cs^{3}+ RLCs^{2} + Ls)\] \[+ L^{2}Ci_R(0)s^{2}/(L^{2}Cs^{3} + RLCs^{2}+ Ls)\] \[+ (L^{2}Ci'_R(0) + RLCi_R(0))s/(L^{2}Cs^{3} + RLCs^{2} + Ls)\] \[+ (L^{2}Ci''_R(0) + RLCi'_R(0) + LI_R(0))/(L^{2}Cs^{3} + RLCs^{2} + Ls).\]

  7. alma1969
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you !!!!!!!!

  8. abtrehearn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You're welcome :^)

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.