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Capacitor Circuit: Is My Method Correct?
Four capacitors, a battery and a switch are assembled in the circuit below. Initially, the switch is set to position A and C4 is uncharged.
At t = 0, the switch is moved to B.
Find Q4, the charge on C4 when the switch is on B.
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam1/fa09/fig20.gif
Q= CV
The initial charge on the C3 capacitor = C3 * V1
Now as the switch to moves to B.
The total charges remains the same, but the voltage changes.
Then
Q = ( C3 + C4 ) V2
V2 = Q / ( C3 + C4 )
next with V2 found
Q4 = C4 V2
*****************
Ok, so that is the only way I've been able to figure this out, however, arent the capacitors in series? In that case I would have to change the equivalent capacitance equation to 1/(1/C3+1/C4) which doesn't work out. Can someone please explain what's going on?
Help would be greatly appreciated. Thanks in advance!
Homework Statement
Four capacitors, a battery and a switch are assembled in the circuit below. Initially, the switch is set to position A and C4 is uncharged.
At t = 0, the switch is moved to B.
Find Q4, the charge on C4 when the switch is on B.
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam1/fa09/fig20.gif
Homework Equations
Q= CV
The Attempt at a Solution
The initial charge on the C3 capacitor = C3 * V1
Now as the switch to moves to B.
The total charges remains the same, but the voltage changes.
Then
Q = ( C3 + C4 ) V2
V2 = Q / ( C3 + C4 )
next with V2 found
Q4 = C4 V2
*****************
Ok, so that is the only way I've been able to figure this out, however, arent the capacitors in series? In that case I would have to change the equivalent capacitance equation to 1/(1/C3+1/C4) which doesn't work out. Can someone please explain what's going on?
Help would be greatly appreciated. Thanks in advance!
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