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Caclulate capacitance

Posted on 2006-11-02
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Last Modified: 2006-11-18
i have capacitors  Cx and C1 (in parallel) and then i have capacitor C2 in series with the (Cx and C1). If C1=4.2 and C2=7.3 what must be the capacitance Cx if the ratio V0 / V1 is 93?
V1 is the voltage accross C1.
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Question by:c_hockland
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Author Comment

by:c_hockland
ID: 17864271
solution i have tried :

since Cx and C1 are in parallel then the V0  (the V of the circuit )  = V1 = Vx
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Expert Comment

by:ozo
ID: 17864602
I think there must be an unstated assumption.
Cx could be anything and you could sill make V0 / V1 anything you want my putting an appropriate charge between Cx/C1 and C2
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by:c_hockland
ID: 17864609
so , any ideas how i could start??
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Expert Comment

by:ozo
ID: 17864751
I think you need another assumption.  for example that you started with V0 = V1 = 0 and then applied V0 to the circuit, in which case the ratio
( 1/(Cx+C1) + 1/C2) / (1/(Cx+C1))  = V0 / V1
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Expert Comment

by:JR2003
ID: 17871471
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParSeriesCap.html

The capacitance of the parallel part of the circuit is C1 + Cx
We know the following facts:
Q = C/V
Q1 = Q2
V0 = V1 + V2

so we know that
(C1+Cx)/V1 = C2/V2

I take it you don't actually mean that V0/V1 = 0.93 as V0 is bigger than V1?
I'll asume you really mean that V1/V2 = 0.93?
so
V1/V2 = (C1 + Cx)/C2
so
Cx = C2(V1/V2) - C1
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Author Comment

by:c_hockland
ID: 17871479
well JR 2003 , "the question just states that the Ratio V0/V1 is to be 93" . I guess it means that V0 =93V1 . Does anything changes in the solution ?
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Accepted Solution

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JR2003 earned 2000 total points
ID: 17871687
I also should have said Q = CV not Q = C/V

V0/V1 = 93
so
V1 = V0/93

V1 + V2 = V0
so
V2 = V0(92/93)  

(C1+Cx)*V1 = C2*V2

Cx = (C2*V2)/V1 - C1

Cx = (C2*(V0(92/93)))/(V0/93) - C1

Cx = C2 * 92 - C1

Cx = 7.3 * 92 - 4.2
Cx = 667.4

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Author Comment

by:c_hockland
ID: 17871692
great! Thanks so much JR2003
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Author Comment

by:c_hockland
ID: 17871708
V2 = V0(92/93)

how did this 92 showed up?
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Author Comment

by:c_hockland
ID: 17871738
...ok i got it...thanks again
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