Halliday Chapter 25

  • Thread starter george2
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  • #1
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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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Answers and Replies

  • #2
LowlyPion
Homework Helper
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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.
 
  • #3
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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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