| Topic: |
Science > Physics |
| User: |
"jclause" |
| Date: |
16 Apr 2005 06:04:14 PM |
| Object: |
Need Help With Vectors |
Sorry, typo in the last two lines of the last post (corrected here).
I need help to solve for Xb in the equation:
Zt=(Ra+jXa)+(Rb+jXb)
The right hand terms (Ra+jXa)+(Rb+jXb) are impedance terms
where resistance Ra is added to reactance Xa, and
resistance Rb is added to reactance Xb, then the total
impedance is Zt.
Filling in values from measurements give the following values
Zt = 8.393
Ra = 5.685
Xa = 3.672
Rb = 1.862
Or in equation form as above
Zt=(Ra+jXa)+(Rb+jXb)
8.393 = (5.685 + j3.672) + (1.862 + jXb)
Xb = ? (apparently negative)
Any assistance will be appreciated.
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| User: "Old Man" |
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| Title: Re: Need Help With Vectors |
16 Apr 2005 07:24:53 PM |
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"jclause" <jc@aaa.bbb> wrote in message
news:11636feidpm3400@news.supernews.com...
Sorry, typo in the last two lines of the last post (corrected here).
I need help to solve for Xb in the equation:
Zt=(Ra+jXa)+(Rb+jXb)
The right hand terms (Ra+jXa)+(Rb+jXb) are impedance terms
where resistance Ra is added to reactance Xa, and
resistance Rb is added to reactance Xb, then the total
impedance is Zt.
Filling in values from measurements give the following values
Zt = 8.393
No. In general, Zt is complex: The current isn't in phase
with the voltage. More likely, this is the magnitude of zt:
| Zt | = 8.393 Ohms
Ra = 5.685
Xa = 3.672
Rb = 1.862
Or in equation form as above
Zt=(Ra+jXa)+(Rb+jXb)
8.393 = (5.685 + j3.672) + (1.862 + jXb)
This is probably incorrect. In general, Zt is complex.
Assuming that Zt is real, Xa = - Xb
Xb = ? (apparently negative)
Inductive reactance is negative. Capacitive reactance
is positive.
Any assistance will be appreciated.
Graph it. Solve for Xb from:
| Zt | = sqrt[ (Ra + Rb)^2 + (Xa + Xb)^2 ]
[Old Man]
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| User: "jclause" |
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| Title: Re: Need Help With Vectors |
18 Apr 2005 02:48:07 PM |
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In article <E86dnemxDfhHMfzfRVn-qw@prairiewave.com>, says...
This is probably incorrect. In general, Zt is complex.
Assuming that Zt is real, Xa = - Xb
Xb = ? (apparently negative)
Inductive reactance is negative. Capacitive reactance
is positive.
Any assistance will be appreciated.
Graph it. Solve for Xb from:
| Zt | = sqrt[ (Ra + Rb)^2 + (Xa + Xb)^2 ]
[Old Man]
Thanks. I stated the problem badly. Sorry.
This works:
|Zt| = sqrt [(Ra + Rb)^2 + (Xa + Xb)^2 ]
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| User: "Gregory L. Hansen" |
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| Title: Re: Need Help With Vectors |
16 Apr 2005 07:29:22 PM |
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In article <11636feidpm3400@news.supernews.com>, jclause <jc@aaa.bbb> wrote:
Sorry, typo in the last two lines of the last post (corrected here).
I need help to solve for Xb in the equation:
Zt=(Ra+jXa)+(Rb+jXb)
The right hand terms (Ra+jXa)+(Rb+jXb) are impedance terms
where resistance Ra is added to reactance Xa, and
resistance Rb is added to reactance Xb, then the total
impedance is Zt.
Filling in values from measurements give the following values
Zt = 8.393
Ra = 5.685
Xa = 3.672
Rb = 1.862
Or in equation form as above
Zt=(Ra+jXa)+(Rb+jXb)
8.393 = (5.685 + j3.672) + (1.862 + jXb)
Xb = ? (apparently negative)
Any assistance will be appreciated.
The left-hand side is real, that means the right-hand side is real.
8.393 + j*0 = (5.685+1.862) + j(3.672 + Xb)
8.393 = 7.547
0 = 3.672 + Xb
Either Zt is really supposed to be the magnitude of the impedence, or you
should say something about your error bars.
--
"And don't skimp on the mayonnaise!"
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| User: "jclause" |
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| Title: Re: Need Help With Vectors |
18 Apr 2005 01:56:31 PM |
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In article <d3sal2$4qa$1@rainier.uits.indiana.edu>,
glhansen@steel.ucs.indiana.edu says...
The left-hand side is real, that means the right-hand side is real.
8.393 + j*0 = (5.685+1.862) + j(3.672 + Xb)
8.393 = 7.547
0 = 3.672 + Xb
Either Zt is really supposed to be the magnitude of the impedence, or you
should say something about your error bars.
The question was steted wrong. Sorry.
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| User: "" |
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| Title: Re: Need Help With Vectors |
16 Apr 2005 09:22:23 PM |
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jclause wrote:
Sorry, typo in the last two lines of the last post (corrected here).
I need help to solve for Xb in the equation:
Zt=(Ra+jXa)+(Rb+jXb)
The right hand terms (Ra+jXa)+(Rb+jXb) are impedance terms
where resistance Ra is added to reactance Xa, and
resistance Rb is added to reactance Xb, then the total
impedance is Zt.
Filling in values from measurements give the following values
Zt = 8.393
Ra = 5.685
Xa = 3.672
Rb = 1.862
Or in equation form as above
Zt=(Ra+jXa)+(Rb+jXb)
8.393 = (5.685 + j3.672) + (1.862 + jXb)
Xb = ? (apparently negative)
Any assistance will be appreciated.
Z=R+jX R=Ra+Rb, X=Xa+Xb
|Z|=sq root{R^2 + X^2} with phase angle=tan^-1 (X/R)
Do the substitutions and solve for the unkown--also if you are going to
multiple post as you did with this problem--suggest you post to all
three with one posting on the "To" line of your message--that keeps all
responders on the same page.
Good luck
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| User: "jclause" |
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| Title: Re: Need Help With Vectors |
18 Apr 2005 01:57:57 PM |
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In article <1113704543.846491.296900@o13g2000cwo.googlegroups.com>,
cnctutwiler@wmconnect.com says...
Z=R+jX R=Ra+Rb, X=Xa+Xb
|Z|=sq root{R^2 + X^2} with phase angle=tan^-1 (X/R)
Do the substitutions and solve for the unkown--also if you are going to
multiple post as you did with this problem--suggest you post to all
three with one posting on the "To" line of your message--that keeps all
responders on the same page.
Good luck
OK. Thanks.
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