# (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
10
0
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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2. ### mik3 Senior Member

Feb 4, 2008
4,846
70
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
138
11
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
10
0
I don't deserve such help. Thanks for clearing that up for me