- #1
Gwozdzilla
- 81
- 0
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.