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find capacitance #5

Posted on 2006-11-08
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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.

solution that i tried :

using  C1 / C0 = 93
 
                4.2 / { C2 ( C1 + CX ) / ( C1 + CX + C2 ) }  = 93
 
                   { C2 ( C1 + CX ) / ( C1 + CX + C2 ) } = 4.2 / 93
 
                  substitute the values of C1 and C2  to get CX
is it correct ??
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Question by:c_hockland
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8 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 17904546
( 1/(Cx+C1) + 1/C2) / (1/(Cx+C1))  = 93
Didn't you alrady ask this in http:/Q_22047554.html
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LVL 18

Expert Comment

by:JR2003
ID: 17906029
The sum of the charges on the two parallel capacitors (Cx +C1) is the same as the charge on C2
Q = CV
so you can work out the capacitance of Cx
The voltages V1 + V2 = V0
0
 

Author Comment

by:c_hockland
ID: 17906200
the solution i provided is the solution from the solutions manual....i have a feeling that you disagree with the result though...why?
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LVL 84

Expert Comment

by:ozo
ID: 17906555
(Cx and C1) im parallel have capacitance Cx+C1

(Cx+C1) in series with C2 have capacitance
1/( 1/(Cx+C1) + 1/C2) ) = C2 ( C1 + CX ) / ( C1 + CX + C2 )
0
 

Author Comment

by:c_hockland
ID: 17906934
so Ozo , whould i use either solution ? Both look the same
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LVL 84

Accepted Solution

by:
ozo earned 2000 total points
ID: 17907013
If Cx and C1 are in parallel, V1 is the voltage accross C1+Cx
0
 

Author Comment

by:c_hockland
ID: 17907041
(Cx and C1) im parallel have capacitance Cx+C1

(Cx+C1) in series with C2 have capacitance
1/( 1/(Cx+C1) + 1/C2) ) = C2 ( C1 + CX ) / ( C1 + CX + C2 )

so i will go ahead and plug the values ,,,and that  should be it then..
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LVL 18

Expert Comment

by:JR2003
ID: 17934516
If you solve ozo's: ( 1/(Cx+C1) + 1/C2) / (1/(Cx+C1))  = 93
you still get Cx = 667.4 F as you did in: http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_22047554.html#17871687
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