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Solving unkown impedance in circuit

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Selection_030.png

    2. Relevant equations
    V=IZ

    3. The attempt at a solution
    So my plan is as follows:
    Find the current through the 1 ohm resistor, multiplying this by the combined impedance of the capacitor & 1 ohm resistor. Now I know the voltage across the unkown impedance Z. Subtract this voltage from the 12 V source, now I know the voltage through the 9 ohm resistor. Use KCL to find the current through Z, and use V=IZ to find the unknown impedance.

    Current through 1 ohm resistor: 4(45d)
    Voltage through 1 ohm resistor & capacitor: 4(45d)(1-j1)=5.66(0d)
    Voltage through 9 ohm resistor: 12(0d)-5.66(0d) = 6.34(0d)
    Current through 9 ohm resistor: 6.34(0d)/9 = 0.7044(0d)

    By KCL:
    -Ic + Iz + Io = 0
    Iz = Ic - Io = 0.7044(0d) - 4(45d) = 3.54(53d)

    V/Iz = 5.66(0d)/3.54(53d) = 1.6(-53d) = 0.963-1.278j

    But this is wrong, and I've been trying to figure this out for a good hour now, but keep getting the same answer. So I conclude that there's something wrong with my approach.
     
  2. jcsd
  3. Apr 9, 2015 #2

    gneill

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    Staff: Mentor

    I think you want to check your KCL. There is one current flowing into the node and two flowing out. Which is which?
     
  4. Apr 9, 2015 #3
    Yes. My sign convention is (-) for in, (+) for out. Iz is the current across Z, Io is the current through the resistor/capacitor, Ic is the current through the 9 ohm resistor.
     
  5. Apr 9, 2015 #4

    gneill

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    Staff: Mentor

    The angle of your Iz looks suspicious. What was your rectangular version?
     
  6. Apr 9, 2015 #5
    0.7044 - 2sqrt(2) - 2sqrt(2)j
    -2.124-2.823j
    3.537(53d)
     
  7. Apr 9, 2015 #6

    gneill

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    Staff: Mentor

    Yeah, both terms are negative. In what quadrant should the resulting angle lie?
     
  8. Apr 9, 2015 #7
    The quadrant where x<0 and y<0, so this means that the answer is 3.537(-53d)

    Even so, this doesn't change the real answer, it is still 0.963

    But the complex part is correct. Perhaps I'll redo this problem tomorrow when I'm more awake. I really need to focus on the little things more!

    Thanks
     
  9. Apr 9, 2015 #8

    gneill

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    Staff: Mentor

    -53° is in the 4th quadrant where x > 0...
     
  10. Apr 9, 2015 #9
    Yes you are right. Sorry about that, I was up very late last night and havent' slept well lately.
     
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