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RCL Circuit: 2 Resistors, 1 Capacitor, 2 Inductors

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    NOTE: Image uploaded with thread shows problem in clearest possible form.

    V = (45v)sin(80[pi] t)
    Resistors 1 and 2 = 50 Ohms
    Inductor 1 = 20 mH
    Inductor 2 = 2.5 mH
    Capacitor = 50 uF

    a) Find the power dissipated by each resistor.
    b) At what frequency will both resistors dissipate the same power?
    c) What would be the answer to part (b) if you swapped the inductors?
    d) Justify your answers to parts (b) and (c) with a graph of the impedance of each branch as function of frequency.
    d) What is the current delivered by the power supply at very low frequencies?

    2. Relevant equations
    They are in the second uploaded document. They are all the equations I believe for RCL Circuits.

    3. The attempt at a solution
    I'm having trouble knowing where exactly to start.
    I believe I have to try and condense the Resistors and Inductors into 1 loop but I'm not sure how to.
    Plus all the other parts of the question are intimidating to say the least for me.
    I can do simpler problems but we never covered an example like this in the course and I've searched high and low online to no avail. Please help, it will be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2015 #2

    collinsmark

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    Homework Helper
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    Hello B. Vane,

    Welcome to Physics Forums! :smile:

    I'd start with part a).

    Seriously though, start with finding the current through each resistor. Your list of impedances for different types of components should be useful. Prepare for a considerable bit of algebra involving complex numbers.

    Once you have the current through a given resistor, you should then find the expression for the power dissipated by that resistor.

    I advise coming up with an expressions for the power (or at least the current through each resistor) in terms of variables, and plug the numbers into the expression as a final step. That way, you can "reuse" much of your work between steps (part a and c for example). Otherwise you might find yourself having to start from the beginning each time something changes.

    I'm not sure what you mean.

    There are two loops. If you set the loops up smartly, you should be able to work with each loop individually for parts a), c), and d).

    (Well, the first d that is. The second d should be pretty easy though after some thought [it's one of the easier steps].)

    This problem is long, containing many parts. Give yourself time. This isn't the type of problem that you can complete in a couple minutes. Just give yourself a good chunk of time to work on it, and work through each part one step at a time. Think of it as a project.

    I have a feeling that this problem combines many of the simpler problems that you already know into one bigger problem (with many steps). Just be patient, have perseverance, and work through it step by step. :smile:
     
    Last edited: Apr 10, 2015
  4. Apr 10, 2015 #3

    Hesch

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    Gold Member

    You don't have to set up equations for loops ( Kirchhoffs Voltage Law). For example:

    Upper path: Voltage over resistor+capacitor+inductor = 45V.
    Middle path: Voltage over resistor+inductor = 45V.

    It's as simple as that.
     
    Last edited: Apr 10, 2015
  5. Apr 12, 2015 #4
    I've been working on it and here Is what I've got so far..... not really sure any of it is correct though. (Sorry its so big, can't/ don't know how to resize it)
    dWKCs6D.jpg
    I don't really have an answer for much. I guess the power disapated in R1 and R2 but I'm pretty sure my current calculated running though R2 is wrong.

    Any more guidance will be greatly appreciated!!!!
     
    Last edited by a moderator: May 1, 2017
  6. Apr 13, 2015 #5

    collinsmark

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    Hello B. Vane,

    You have the right idea up to a point.

    In the first branch (the one that has the capacitor, inductor and resistor) things look pretty good up until the point where you calculate the current at time t = 0. It is not necessary to calculate the instantaneous current in the circuit. I think the problem as asking for the "average" power dissipated by each resistor, not the instantaneous.

    Just use the rms current, when calculating the power.

    In the second branch (the one that contains an inductor and resistor), again, use the rms current, rather than the peak current, when calculating average power.
     
  7. Apr 13, 2015 #6

    NascentOxygen

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    As collinsmark said, when you use ##Power=I^2R## that formula needs the RMS voltage.

    For (b) you have to repeat the calculations, this time retaining ##\omega## in the reactances because you'll end up solving for that value of ##\omega## that makes the resistors' powers equal.
     
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