Finding R and C for RC circuit

Discussion in 'Homework Help' started by TripleDeuce, Dec 3, 2010.

  1. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    0
    http://img812.imageshack.us/img812/6845/90087352.jpg

    I know omega is 6283.2 rad/sec because I am given the frequency which is 1000Hz so you just multiply that by 2∏

    I also converted the Z value given into rectangular which gave me 2000 - j1591

    but I am not sure how to do the algebra.

    Z = R - j/ωC seems like the easiest to use but R and C would still be unknown and I'm not sure what to do with the j. I know that indicates imaginary number though or more specifically the square root of -1.
     
    Last edited: Dec 3, 2010
  2. tyblu

    Member

    Nov 29, 2010
    199
    16
    You can think of the two elements, the resistor and capacitor, as general 'blocks', each with an impedance. They are arranged in series, so their impedance adds up. The impedance of the resistor doesn't change with frequency: it is R at 1kHz; that of the capacitor is Zc=-j/wC, or Zc=-j/(2∏*1000Hz*C). Together, in series, their impedance is Z=R-j/(2∏*1000Hz*C), as you mentioned.

    The real part is taken up entirely by the resistor, R; the imaginary part is taken up entirely by the capacitor, -1/(2∏*1000Hz*C). This means the Cartesian format you found, Z=2000-j1591, really gives you solutions to two different formula, each with one unknown.
    Real part: 2000 = R
    Imag part: -1591 = -1/(2∏*1000*C)
     
  3. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    Oh, I see, but I do not understand why

    -1591 = -1/(2∏*1000*C) for you are saying you can substitute -1 for j, but it is j^2 that gives you the value of -1 doesn't it?
     
  4. tyblu

    Member

    Nov 29, 2010
    199
    16
    I'm not saying you can substitute -1 for j:

    -j1591 = j*(-1591)
    and
    -j/(2∏*1000*C) = j*(-1/(2∏*1000*C))
    gives
    j*(-1591) = j*(-1/(2∏*1000*C))
    remove common factor j
    -1591 = -1/(2∏*1000*C)
     
  5. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
    26
    0
    oh, i see. Thanks. Algebra can still elude me at times.
     
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