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Equivalent Circuit

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    For t<0, the circuit shown below with the voltage source at 11 volts and three resistors combined into an equivalent, Req. We need to find the Thevenin equivalent circuit for the part of the circuit connected to the inductor.

    First find Req, by combining series and parallel resistors.

    Req = ? Ω. (I already solved it)

    Find the Thevenin resistance across the inductor and the short circuit current.

    Rt = ? Ω. (I already solved for it)

    Isc = ? A. (This is what I don't have)

    2. Relevant equations

    Isc = Voc / Rt (Thevenin/Norton Theorem)

    3. The attempt at a solution
    The solved values I got (which are correct) for the resistances are:
    Req= 8.89 Ω
    Rt= 7.82 Ω
    I got Rt from 1/Rt= (1/Req) + (1/65)

    I've tried multiple times to solve for Isc. I think I mainly have a problem getting Voc or just combining voltage and current sources. I am attaching a picture of the circuit.

    How do I get Isc?
     

    Attached Files:

    • HW.png
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  2. jcsd
  3. Apr 15, 2013 #2

    gneill

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    Staff: Mentor

    When the inductor is replaced by a short circuit, what effectively happens to the 65 Ω resistor? What's the potential across Req?
     
  4. Apr 15, 2013 #3
    Does it short out the 65 Ω resistor? If it does, I had no idea, because I thought for it to short out a parallel resistor, the short circuit has to be parallel ONLY to that resistor.

    As for the potential across Req.. I guess I can use voltage division?

    Veq= [8.889Ω/(8.889Ω+65Ω)]11v, which gives me 1.323 volts.
     
  5. Apr 15, 2013 #4

    gneill

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    Staff: Mentor

    A short shorts-out anything and everything it parallels. In this case it is definitely in parallel with the 65 Ω resistor. Here's a way to look at it:
    attachment.php?attachmentid=57886&stc=1&d=1366036246.gif
    Take a close look at the above diagram. What potential is the blue node at (with respect to the bottom node)?
     

    Attached Files:

  6. Apr 15, 2013 #5
    Ahh. Thank you for taking the time to draw that out.

    Sorry, I'm getting a little confused from the different terminology. It sounds like you're asking what the voltage across the short circuit is. I'm not sure I know how to answer that. I'll take guesses though.

    I'm supposing that the short circuit practically removes the 65Ω resistor. Is this correct?

    My guess is doing Ohm's law on Req with the 65 Ω resistor removed. 11v/8.889Ω = current across I (1.237A), but I'm not sure that helps.

    I'm not sure I can apply a KCL at the bottom or top node because of the voltage source. Or am
    I supposed to do a source transformation?
     
  7. Apr 15, 2013 #6

    gneill

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    Staff: Mentor

    Correct.
    It does!
    You can do KCL on the top (blue) node. The current you want is the current through the short (same as the current from the voltage source). You now know the other currents leaving the node...
     
  8. Apr 15, 2013 #7
    KCL on the top node:

    0 = (Leaving currents) - (Entering Currents)
    0 = (Current across Req) + (Current Source) - (Current through voltage source)

    Current through voltage source = Isc

    0= 1.237A + 9A - Isc

    Isc = 10.237 A

    The website won't tell me if this is correct (I used all my attempts), but I'll assume it is. I just wanted to know where I was going wrong. I guess my pains were mainly coming from that 65Ω resistor.

    Thank you very much, gneill.
     
  9. Apr 15, 2013 #8

    gneill

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    Staff: Mentor

    Your result looks good!

    Yes, the circuit was made a bit tricky with the 65Ω resistor. But you got there in the end!
     
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