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

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mzsyed85

Joined Jan 1, 2009
10
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?
 

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vvkannan

Joined Aug 9, 2008
138
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
 
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