Parallel Capacitors: Homework Solution for 9 μF Charge

AI Thread Summary
A 9 μF capacitor charged to 200 V is connected in parallel with an uncharged 3 μF capacitor, leading to a redistribution of charge. The initial charge on the 9 μF capacitor is calculated using Q=CV, resulting in a charge of 1.8 mC. When the 3 μF capacitor is fully charged, the voltage across both capacitors equalizes, and the remaining charge on the 9 μF capacitor is found to be approximately 1.5588 mC. The process of connecting the capacitors can be repeated multiple times to reduce the charge on the 9 μF capacitor below 50% of its initial value, but energy is not conserved during this operation. Ultimately, the voltage across the 9 μF capacitor after the process is completed will be lower than its initial voltage.
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Homework Statement



A capacitor of capacitance 9 μF is fully charged from a 200 V dc supply. The
capacitor is now disconnected from the supply and connected in parallel with an
uncharged 3 μF capacitor. Once the 3 μF capacitor is fully charged, it is removed
and discharged. a) What charge remains on the 9 μF capacitor? b) How many
times would this process have to be repeated in order to reduce the charge on the 9
μF capacitor to below 50% of its initial charge? c) What would be the pd between
the plates of the 9 μF capacitor now be?


Homework Equations



Q=CV


The Attempt at a Solution



a) I said that the 3μF capacitor will be fully charged when the voltage across both capacitors is the same,let's say Vx.

Since energy has to be conserved then Ei=Ef therefore

C1V^2 = C1Vx^2 + C2Vx^2 <=> Q1' = 1.5588*10^-3 C . Am i doing this correct ?
 
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Unfortunately, much as in elastic collisions of masses, connection of capacitors in parallel is not an operation that conserves energy; some energy is lost when the charges get shuffled into their new configuration.

What is conserved through the connection operation, though, is charge.
 
Thank you very much !
 

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