Closing a Circuit 24.89 Giancoli

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In the discussion about closing a circuit with capacitors, a configuration is described where one capacitor (2.8 uF) is in parallel with two others (4.4 uF and 3.7 uF) that are in series. Initially, the 2.8 uF capacitor is charged while the others remain uncharged. When the switch is thrown, disconnecting the voltage source, the charge from the 2.8 uF capacitor is redistributed between it and the series combination of the other two capacitors. This results in a decrease in the charge on the 2.8 uF capacitor as it shares its charge with the series capacitors. Understanding the circuit's behavior hinges on recognizing the role of the switch and the redistribution of charge after disconnection from the voltage source.
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24.89 Giancoli

There are three capacitors. One is in parallel with two others. The two others are in series. The alone one is 2.8 uF, the other two in series are 4.4 and 3.7 uF. a 12 volt voltage is applied. Initially there is a bridge that connects to the 2.8. After it is charged, the switch is thrown. What are the final charges on each?

This one throws me because I'm not sure what happens after the switch is thrown. Does the voltage cease to be applied? or is it still applied? if it is still applied, then I'm not sure how the built-up charge in the 2.8uF capacitor matters.
 
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I can imagine a single capacitor in parallel with a series combination, but I cannot imagine what the "bridge" does or where the switch is. Can you post a circuit schematic?
 
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OK, I get it. With the switch to the left, C1 is fully charged and the other two are uncharged. When the switch is thrown to the right, the voltage source is disconnected and the charge that was initially on C1 is now shared by C1 and the series combination of C2 and C3. Obviously, C1 has less charge than before because some of it went to the series combination.
 
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