Capacitors in Series and Parallel homework

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SUMMARY

The discussion focuses on calculating the equivalent capacitance of a network of capacitors, specifically with capacitances C1 at 7.2 µF and C2 at 4.8 µF. The equivalent capacitance is determined using the formula for capacitors in series, 1/C = 1/C1 + 1/C2 + 1/C3, and then applying the parallel capacitance rules for the remaining capacitors. The charge on each capacitor is computed using the equation Q=CV, with a voltage of Vab = 450 V applied across the network. The voltage Vcd is also calculated under the same conditions.

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Homework Statement



In Fig. 25-22, each capacitance C1 is 7.2 µF, and each capacitance C2 is 4.8 µF

25-22.gif


(a) Compute the equivalent capacitance of the network between points a and b.

(b) Compute the charge on each of the three capacitors nearest a and b when Vab = 450 V.

(c) With 450 V across a and b, compute Vcd.


Homework Equations



Q=CV

The Attempt at a Solution



I tried figuring the total capacitance, but I can't tell what is series and what is parallel! In series I know that the potential differences add up in series and they are the same in parallel, and the opposite is true for charge. Please help!
 

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The three capacitors on the right hand end are in series.
Capacitances in series "add" according to the formula
1/C = 1/C1 + 1/C2 + 1/C3
Then you can replace the original circuit with a simpler equivilent one with the three capacitors replaced by the one.

Do it again - this time the two capacitors on the end are in parallel.
 
Okay, thank you so much!
 

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