Three capacitors connected to a battery

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SUMMARY

The discussion focuses on calculating the equivalent capacitance of three capacitors connected to a battery, specifically with values C3 = 10.5 µF and C1 = C2 = 2C3. The user initially calculated the total capacitance as 12.6 µF but received feedback indicating the answer was incorrect. The correct approach involves recognizing that C2 and C3 are in series, requiring the use of the formula for capacitors in series before adding C1 to find the total capacitance. Additionally, voltage does not directly influence the charge stored on the capacitors.

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Homework Statement



6611bd826734c6d4eca8948e9d698d3f6a8.jpg


Examine the figure shown. Note that C3 = 10.5 uF, and C1 = C2 = 2C3. Also the battery voltage is Vb = 39.0 volts.


Homework Equations



What is the equivalent capacitance for this arrangement?
How much charge is stored on C1?

The Attempt at a Solution



C2 and C3 are in parallel, so I used C'=C2+C3
C' is in series with C1, so the total capacitance, C'' is:

((C1)(C'))/(C1+C')

which is equal to:

((C1)(C2+C3))/(C1+C2+C3)

pluging in values, I get an answer of 12.6 uF. I am using an online homework program and am told that this is incorrect. What am I doing wrong?

For the second question, shouldn't I multiply my answer for the first part by the volts value given?

Thank you for you help.
 
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(c2+c3/2)+c1=C

c2 and c3 are in series, add up the values and divide by the number of caps then add that number to c1 and you have the total capacitance. second question, no, voltage has nothing to do with it.
 

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