Halliday Chapter 25: Final Charge of Uncharged Capacitors Problem

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The problem involves a circuit with a 12 V battery and three uncharged capacitors (C1=4μF, C2=6μF, C3=3μF). Initially, C1 is charged to 48μC when connected to the battery. After switching to the right, the charge is redistributed among C1, C2, and C3, leading to confusion about the final charges. The calculations indicate that the total charge should remain at 48μC, but the resulting values suggest a total of 64μC, which is inconsistent. The key is to correctly apply the principles of charge conservation and equivalent capacitance to find the final charges on each capacitor.
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Hi! I need help with the following problem:
1. Figure displays a 12 V battery and three uncharged capacitors of capacitances C1=4μF , C2=6μF and C3=3μF. The switch is thrown to the left until C1 is fully charged then it is thrown to the right.What is the final charge of each capacitor?


3. initial charge of 1 is q=C1V=48μC
Switch thrown to the right:
q2=q3, V1=V2+V3, V=q/c and q1+q2+q3=48C
From these equations i found q1=24,q2=q3=12
The answer in the back of the book says that q1=32,q2=q3=16
Where is the mistake?
 

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Welcome to PF.

First of all figure the equivalent capacitance of the 2 capacitors in series.

Second figure the charge Q that will be imparted to the first capacitor C1.

You did this and got 48 from Q = V*C.

But when the switch is broken from the source, that then is your total charge that can be shared in the system.

When you connect the switch to the right you are now sharing that charge between C1 and the C2 and C3 equivalent.

Using Q = V*C , figure then the new voltage across the three.

With the resulting voltage you are in a position to determine the Q on C1 and the Q on the equivalent of C2 and C3.
 
Now i find that: q1=32μC, q2=16μC, q3=16 μC.The sum of the 3 charges is 64 μC but there are only 48μC to share.Why does this happen?

P.S. Thank you for your help!
 
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