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KVL rlc circuit

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Obtain the voltage Vx in the network of Fig.Q11, using the mesh current method.
    (Ans: Vx=4.35∠-194.5)

    impedance.jpg
    2. Relevant equations


    3. The attempt at a solution

    I am not getting the correct answer, are my loop equations correct ?

    10 - 2*i1 + 2j*i1 -5j(i1+i2) -5(i1+i3) = 0
    4.33 + 2.5j - 10*i2 -5j(i1+i2) +2(i3-i2) -2j(i3-i2) = 0
    -10*i3 + 2j(i3-i2) -2(i3-i2) -5(i1+i3) = 0

    Therefore,

    i1(-7-3j) + i2(-5j) -5*i3 = -10........................(1)
    i1(-5j) + i2(-12-3j) + i3(2-2j) = -4.33-2.5j.....................(2)
    -5*i1 + i2(2-2j) + i3(-17+2j) = 0....................................(3)
     
  2. jcsd
  3. Apr 7, 2015 #2

    berkeman

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    Staff: Mentor

    Some of the signs in the 2nd equation look wrong to me (but it's been a while since I used KVL -- I usually use KCL).

    Can you talk through why you chose the signs that you did for the 2nd equation?
     
  4. Apr 7, 2015 #3
    Okay, I have labeled the currents in my diagram

    4.33+2.5j, That's the voltage of the source in the second loop in complex form (Voltage rise so it's positive)
    -10*i2, Voltage drop across the resistor, hence negative sign
    -5j(i1+i2), Voltage drop across the inductor, negative sign
    +2(i3-i2), Positive sign because the loop is traveling opposite to the current direction as labeled
    -2j(i3-i2), Because impedance is negative for a capacitor and loop is travelling opposite in direction to current, so +*- = - sign
     
  5. Apr 8, 2015 #4

    NascentOxygen

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    Staff: Mentor

    One of the difficulties is that we are not told which terminal of each source is (+) and which is (-), i.e., which is the reference terminal.

    Perhaps assume the marked current is indicative of the source polarity, though I don't know whether we can generalize that to other problems like this. :oldconfused:

    I do get the same equations as you, though haven't checked them.

    Perhaps reverse the voltage source in the second loop, and see whether you get the book's answer that way?
     
  6. Apr 8, 2015 #5
    I tried reversing it but still get the same answer, maybe im solving it wrong so here are my steps. I used crammers rule

    Determinant,
    7+3j ...........5j................5
    5j................12+3j...........-2+2j
    5.................-2+2j............17-2j

    Solving that I get, -73+405j

    Then for i3,
    7+3j............5j...............10
    5j...............12+3j..........4.33+2.5j
    5................-2+2j...........0

    Solving that I get, -744.52-100.77j

    Therefore i3 = (-744.52-100.77j)/(-73+405j)
    = 0.08 + 1.82j

    So V = (0.08+1.82j)*10
    = 0.8 + 18.24j

    Or, 18.26∠87.49
     
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