How Do You Calculate Energy Stored in Specific Capacitors Within a Network?

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SUMMARY

The discussion focuses on calculating the energy stored in a capacitor network with a potential difference of 12.0V. The total energy stored in the network was calculated using the formula U = 0.5CV², resulting in 1.58E-4 J for the equivalent capacitance of 2.19E-6 F. To find the energy stored in the 4.80-μF capacitor, the charge stored in the equivalent capacitor must be determined, which is directly related to the charge of the 4.80-μF capacitor. The user seeks clarification on how to apply the voltage across the network to find the energy specific to the 4.80-μF capacitor.

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Homework Statement


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.

Homework Equations


C= Q/V
U= Q2/2C = .5CV2 = .5QV
Capacitors in Series:
1/Ceq = 1/C1 + 1/C2 + ...
Capacitors in Parallel:
Ceq = C1 + C2 + ...

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.
 

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Determine the charge stored in the equivalent capacitor. How is it related to the charge of the 4.8μF capacitor?


ehild
 

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