Closing a Circuit 24.89 Giancoli

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Homework Help Overview

The problem involves a circuit with three capacitors: one capacitor in parallel with two others that are in series. The values of the capacitors are given, along with a voltage source. The scenario describes the charging of the capacitors and the effect of switching the circuit configuration after charging.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of switching the circuit after charging, questioning whether the voltage remains applied and how this affects the charge distribution among the capacitors.

Discussion Status

Some participants have clarified their understanding of the circuit's behavior when the switch is thrown, indicating that the voltage source is disconnected and charge redistribution occurs. However, there are still questions regarding the initial conditions and the role of the switch.

Contextual Notes

There is uncertainty about the configuration of the circuit, particularly the placement of the switch and the "bridge" mentioned by the original poster. This may affect how the capacitors interact during the switching process.

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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?
 
GIANCOLI.ch24.p89.jpg
 
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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