Re: R2 = ?

In article <2Py%f.4775$L.135313@xxxxxxxxxxxxxxxxxxxxx>,
gneillREM@xxxxxxxxxxxxxxxxxx says...

"jclause" <jc@xxxxxxx> wrote in message
news:123tbc2t9en1mc8@xxxxxxxxxxxxxxxxxxxxx
In article <YEw%f.4590$L.125668@xxxxxxxxxxxxxxxxxxxxx>,
gneillREM@xxxxxxxxxxxxxxxxxx says...


I see. So how do you motivate subtracting the square
of the reactance rather than adding under the square
root? The magnitude of a complex number (a +/- j*b)
is sqrt(a^2 + b^2).


Why would you not use the minus in your (a +/- j*b) example?

Because the magnitude of an imaginary number is given by
the square root of the sum of the squares of its components,
the same way that vector magnitudes are computed. It's
basically Pythagoras' theorem.


Now you have clipped out what is needed. Whatever
number "b" was (the reactance), I believe it was
1.759 *is* the magnitude.

jc


.

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