Capacitors in a Network

vendredi 31 janvier 2014

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

For the capacitor network shown in Fig. P24.55, the potential difference across ab is 12.0V. Find (a) the total energy stored in this network and (b) the energy stored in the 4.80-μF capacitor.



2. Relevant equations

C= Q/V

U= Q2/2C = .5CV2 = .5QV

Capacitors in Series:

1/Ceq = 1/C1 + 1/C2 + ...

Capacitors in Parallel:

Ceq = C1 + C2 + ...



3. The attempt at a solution

I found part A:

C1 = 8.6E-6

C2 = 4.8E-6

Capacitors in Series:

1/Ceq = 1/C1 + 1/C2 + ...

1/Ceq = 1/(8.6E-6) + 1/(4.8E-6)

Ceq = 3.08E-6



C1 = 6.2E-6

C2 = 11.8E-6

Capacitors in Series:

1/Ceq = 1/C1 + 1/C2 + ...

1/Ceq = 1/(6.2E-6) + 1/(11.8E-6)

Ceq = 4.06E-6



C = 3.5E-6

Capacitors in Parallel:

Ceq = C1 + C2 + ...

Ceq = 3.5E-6 + 4.06E-6

Ceq = 7.56E-6



Capacitors in Series:

1/Ceq = 1/C1 + 1/C2 + ...

1/Ceq = 1/(7.56E-6) + 1/(3.08E-6)

Ceq = 2.19E-6



U = .5CV2

U = .5(2.19E-6)(122)

U = 1.58E-4 J



But then for part B I'm not really sure what to do. The voltage for the entire network is 12V, but I have no idea how I can use this info to find the potential energy for just one of the capacitors. Please help.




Attached Images





File Type: jpg P24.55.jpg (13.2 KB)







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