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XProfan X2
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05/09/13 ▲

- Page 1 -

Julian
Schmidt

[OFFTOPIC]green works somehow freundlicher
Padding reicht 10-15px.
[/OFFTOPIC]

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XProfan X2
WinXP, SP3, Core(TM) i7-3770 @ 3.40GHz

[OFFTOPIC]find? Have now time green tuned.[/OFFTOPIC]

E.T.

iF (12.05.13)

find? Have now time green tuned.

but withal green goes The Formel in a Kurzschluss

ever which indicated voices there well not...

Grüße aus Sachsen... Mario
WinXP, Win7 (64 Bit),Win8(.1),Win10, Win 11, Profan 6 - X4, XPSE, und 'nen schwarzes, blinkendes Dingens, wo ich das alles reinschütte...

Julian
Schmidt

It can too his not

R_{0}

, separate

R_{1}

and

R_{2}

in Reihenschaltung and Parallelschaltung whom equal worth having.

R_{R0} = R_{1} + R_{2}

R_{P0} = {{(R}_{1}^{-1} + {R_{2}}^{-1})}^{-1}

- Page 2 -

p.specht

supply voltage 230V ?

XProfan 11
Computer: Gerät, daß es in Mikrosekunden erlaubt, 50.000 Fehler zu machen, zB 'daß' statt 'das'...

Julian
Schmidt

not given. Evtl. is the one can itself these select.

Richi

(without plenty Nachkommastellen calculated)

we take a thrill of 230 Volt on

U=230V = Ireihe = 0,57 A = Reinzel = 202ohm

Ireihe = P/u means 133W/230V = 0,57A

Rreihe,total = u/I = 230V/0,57A = 403,508 ohm

there we 2 Widerstände in row having means: Rreihe,gesammt / 2 = 202 ohm (pi time eye) = Rreihe,single

Test: Preihe=U*I = 230V*0,57A = 131,1 Watt ( Tja The Nachkommastellen

)

Rparallel: we should now know the 2 same Widerstände The parallel geschaltet go The half-way Ohm's yield, means Reinzel=101 ohm [ Rp=(R1*R2)/(R1*R2) ]
Iparallel= P/I = 600W/230V = 2,6amp
Reinzel= 101ohm
u = still 230Volt

P=U*I = 230V*2,6A = 598 Watt in the parallel pursued

D.h. with the Reihenschaltung is the resistance 4x larger as with the Parallelschaltung
can Yes time with the Rparallel Formel each R's to charge (88ohm), then but not forget *2 for Rparallel,total means 176ohm,what there so 601 Watt yield ought to.

not to mix up let with the numbers and Ergebnissen, it are missing only many Nachkommastellen and so weichen The Results something of each other ex.
what but of my opinion to vernachlässigbar is,whether You you whom fingers on 598 Watt or on 600Watt verbrennst,that merkste not ;)
These releases having only whom sense something in feel To get!

plenty Fun and success
Richi

PS:

question:
which Leistungen go submitted, if eachone Widerständ particular eingeschaltet is.

response:
it go The LEISTUNGEN of apiece *300 (414)Watt submitted, if eachone Widerständ particular eingeschaltet is.

-pssst what still again parellel would !!!!???? And then 828 Watt correspond to would, what before yet 600 Watt were !?!?!?!!? LOL Lach ;)

*very:
600W / 2 = 300W (because 2 Widerstände)
I=P/u = 1,30434A
Reinzel=U/I = 176 ohm (pi time eye) = 300 Watt time 2 because parallel are 600Watt

The Formel was means 600W/2 ;) it stood Yes everything already in the question

p.specht

Formel for unterschiedlichen (!) both Widerstände (Spannungsangabe as Variable, R1 with the Minuszeichen, R2 with Pluszeichen calculate):

R_1,2 = U^2/P_serie * (1/2 [ + or - ] sqrt(1/4 - P_serie/P_parallel ) )

greeting

P.s.: if The Herleitung interested:
P_serie = (u1 + u2)*I_gesamt with u1+u2=U_netz, I_gesamt= U_netz/(R1+R2)
lead to first Gleichung in a Gleichungssstem: R1+R2= U_netz^2/P_serie

P_parallel = U_netz * (I1+I2) lists R1*R2/(R1+R2) = U_netz^2/P_parallel,

in the first Formel R2 as U_netz^2/P_serie - R1 squeeze out and so into second Gleichung bring into action, then The "Mitternachtsformel" (everybody can too to midnight yet know must: Solution of/ one Quadratischen Gleichung) bring into action, vereinfachen. The so found values into Leistungsformel bring into action, and FERDISCH. In Parallelschaltung yield the Einzelleistungen To 400 and 200 Watt (Normzahlen).

Computer: Gerät, daß es in Mikrosekunden erlaubt, 50.000 Fehler zu machen, zB 'daß' statt 'das'...

Richi

2 time 176 ohm In parallel Berieb:
thrill u = 230 Volt V
Stromstärke I = 2.6136363636363638 amp A
resistance Rgesamt = 88 ohm Ω because 176/2
performance P = 601.1363636363636 Watt W

In row:
thrill u = 230 Volt V
Stromstärke I = 0.5782608695652174amp A
resistance R = 397.74436090225566 ohm Ω
performance P = 133 Watt W

Julian
Schmidt

p.woodpecker (12.05.13)

in the first Formel R2 as U_netz^2/P_serie - R1 squeeze out and so into second Gleichung bring into action

How, where, what bring into action?

Have now.
R1+R2 = U_netz^2/P_reihe
and.
R1*R2/(R1+R2) = U_netz^2/P_parallel
now for R2 bring into action.
R1+(U_netz^2/P_reihe) - R1 = U_netz^2/P_reihe

and what now into second Gleichung bring into action/equate?

[OFFTOPIC]I belong well to the The PQ-Formel prefer. Rechne not often circa midnight.

[/OFFTOPIC]

05/13/13 ▲

p.specht

yet plainer:
I) Rs=R1+R2 ==> R2=Rs-R1
II) Rp=R1*R2 / (R1+R2)
Ges: R1 = X
I) in II) inserted:
Rp= X*(Rs-X) / (X+Rs-X)
Rp= X*(Rs-X) / Rs
Rp * Rs = X*(Rs-X)
Rp * Rs = X*Rs-X*X
Rp * Rs = X*Rs-X^2
X^2 - Rs*X + Rp * Rs = 0
p= -Rs
q= Rp*Rs
X_1,2 = - p/2 +\- sqrt( p^2/4 - q ) =
R_1,2 = Rs/2 +\- sqrt( (Rs^2 /4 - Rp*Rs),
with Rs=U_netz^2/Ps and Rp=U_netz^2/Pp yields
itself to Vereinfachungen The named Formel.
greeting

Computer: Gerät, daß es in Mikrosekunden erlaubt, 50.000 Fehler zu machen, zB 'daß' statt 'das'...

05/14/13 ▲

Julian
Schmidt

Jup, is correct so does it integrally well.
thanks all which The trouble made having To calculate.

05/14/13 ▲