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**Unformatted text preview: **24.20: a) and b) The equivalent resistance of the combination is 6.0 µF, therefore the −4 total charge on the network is: Q = CeqVeq (6.0 μF)(36 V) = 2.16 × 10 C. This is also the charge on the 9.0 μF capacitor because it is connected in series with the point b. So: Q9 2.16 × 10 −4 C V9 = = = 24 V. C9 9.0 × 10 −6 F Then V3 = V11 = V12 + V6 = V − V9 = 36 V − 24 V = 12 V. ⇒ Q3 = C 3V3 = (3.0 µF)(12 V) = 3.6 × 10 −5 C. ⇒ Q11 = C11V11 = (11 μF)(12 V) = 1.32 × 10 −4 C. ⇒ Q6 = Q12 = Q − Q3 − Q11 = 2.16 × 10 −4 C − 3.6 × 10 −5 C − 1.32 × 10 −4 C. = 4.8 × 10 −5 C. So now the final voltages can be calculated: Q 4.8 × 10 −5 C V6 = 6 = = 8 V. C6 6.0 × 10− 6 F Q12 4.8 × 10 − 5 C V12 = = = 4 V. C12 12 × 10 − 6 F c) Since the 3 μF, 11 μF and 6 μF capacitors are connected in parallel and are in series with the 9 μF capacitor, their charges must add up to that of the 9 μF capacitor. Similarly, the charge on the 3 μF, 11 μF and 12 μF capacitors must add up to the same as that of the 9 μF capacitor, which is the same as the whole network. In short, charge is conserved for the whole system. It gets redistributed for capacitors in parallel and it is equal for capacitors in series. ...

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