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In the circuit shown, all capacitors are identical. Initially, the switch is open and the capacitor marked C_1 is the only one charged to a value Q_0. After the switch is closed and the equilibrium is reestablished, the charge on the capacitor marked C_1 is Q. Find the ratio initial change to final charge in capacitor C_1. |
Answer» SOLUTION :We BEGIN with the circuit GIVEN, assigining names to each identical capacitor. `C_(1)` begins with an unknown initial charge `Q_(0^(3))` and achieves a charge `Q` after the circuit is connected and has re-established equilibrium.  As `C_(3)` is clearly shorted, the potential difference across its plates is `0`, and as such the capacitor will contain no charge and will not afect the circuit We can therefore remove it from the schematic. This simplified circuit can be broken down into even simpler equivalne tcircuits. Leaving `C_(1)` where it is, and recognizing `C_(3)` and `C_(4)` in series with each other *yielding an equivalent capacitance of `1//2C`), and together in PARALLEL with `C_(4)`, Acknowledging the combined capacitors are equal, `C_(456)` must have an equivalent capacitance of `3//2C`, As `C_(456)` is simply in series with `C_(2)`, the resultant equivalent circuit has simply one capacitor `(C_(2456))` in series with `C_(1)`, where `C_(456)` must have an equivalent capacitance of `3//5C`, Since `Q=CV, V_(1)=Q//C` where `V_(1)` is the potential difference across `C_(1),V_(2456)` is the potential difference across `C_(2456)`, and `Q` is the charge on `C_(2456)` For equilibrium conditions, `V_(1)=V_(2)`. Therefore, `Q//C=5Q//C, 3CQ=5CQ` `Q=(3//5)Q` by the conservation of charge, `Q_(0)=Q+Q=Q+(3//5)Q=(8//5)Q`. The initial charge on the capacitor was `8//5` of the final charge on the capacitor after the switch was closed. 
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