(Vout/Vin) for RC/RL/RLC Circuits

Discussion in 'Homework Help' started by mzsyed85, Jan 10, 2009.

  1. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    Hey guys,
    I think I finally understand how to tackle RC, RL, or RLC circuits. However, I am not entirely clear on some of the derivations I have been looking at. It is my understanding that j is an operator that signifies a shift of 90 degrees - and that in an equation j = sqrt [-1]

    I have found that using the voltage divider rule instead of using complex conjugates for these circuits is much easier. But I am having a slight problem with j. I have attached the derivation below.

    When finding the magnitude of Vc/Vin, and the whole expression is squared (and then sqrt) - why doesn't the j leave a -1?
    • RC.jpg
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  2. mik3

    Senior Member

    Feb 4, 2008
    Because you don't square the j.

    The magnitude of a complex number a+jb=sqrt[(a^2)+(b^2)]
  3. vvkannan

    Active Member

    Aug 9, 2008
    hello mzsyed85,

    By meaning magnitude we are finding te distance of the point from the origin and here the point is (1,wrc) and we should not include j because it tells us that the point lies in a complex plane and then we are finding the magnitude similar to pythagoras theorem (sqrt(a^2 + b^2) = c where c is the magnitude and a and b are the points
  4. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    I don't deserve such help. Thanks for clearing that up for me