# Capacitors in Series and Parallel

1. Feb 28, 2010

### SuperCass

1. The problem statement, all variables and given/known data

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

(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.

2. Relevant equations

Q=CV

3. 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!

#### Attached Files:

• ###### 25-22.gif
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2. Feb 28, 2010

### Delphi51

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.

3. Feb 28, 2010

### SuperCass

Okay, thank you so much!