Re: vectors

From: Northstar (xxx_at_yyy.zzz)
Date: 01/29/05 

Date: Sat, 29 Jan 2005 20:29:53 -0000

In article <Vun7QR+WF28HN0X75BZ75MtMzEOy@4ax.com>, kurtzDELETE-THIS@asu.edu
says...
>
>On Sat, 29 Jan 2005 17:10:10 -0000, xxx@yyy.zzz (Northstar) wrote:
>>1
>>Sorry. The question is: R = 9.953, magnitude of Z = 32.99, what is X,
>>and how to state the equation for X? TIA
>>
>>
>
>So I gather you are saying that Z = R + j X where R = 9.953 so
>Z = 9.953 + j X. Then if |Z| = 32.99 that says:
>
>32.99 = sqrt(9.953^2 + X^2)
>
>You can solve that for X if you begin by squaring both sides.
>
>--Lynn

Thanks. I had assumed

|Z| = sqrt R^2 + X^2
|Z|^2 = R^2 + X^2
X^2 = |Z|^2 - R^2
X = sqrt (|Z|^2 - R^2) = sqrt (32.99^2 - 9.953) = 31.45
then cos angle = R / |Z| = 9.953 / 32.99 = 0.3017
and phase angle of impedance = arc cos 0.3017 = 72.44 degrees

Does that look OK? TIA