TP 21.55

Given: C₁ = 15.0 mF = 15.0 x 10^-6 F

C₂ = 8.20 mF = 8.20 x 10^-6 F

C₃ = 22.0 mF = 22.0 x 10^-6 F

DV = 9.00 V

Find: E_stored - ?

First step is to find the equivalent capacitance of the network. Notice that the series combination of capacitors 2 and 3 is in parallel with the capacitor 1.

With this in mind:

1/C₂₃ = 1/C₂ + 1/C₃ = 1/(8.20 mF) + 1/(22.0 mF)

C₂₃ = 5.97 mF

C₁₂₃ = C₁ + C₂₃ = 15.0 mF + 5.97 mF = 21.0 mF

Total charge stored in the network:

Q = C₁₂₃ DV= (21.0 mF) (9.00 V) = 189 mC

Charge stored on capacitor 1:

Q₁ = C₁ DV = (15.0 mF) (9.00 V) = 135 mC

Charge stored on capacitor 2 and capacitor 3:

Q₂ = Q₃ = Q - Q₁ = 54.0 mC

Energy stored in capacitor 1:

E_stored = Q₁² /(2C₁) = (135 mC)² /(2 (15.0 mF)) = 608 x 10^-6 J = 6.08 x 10^-4 J

Energy stored in capacitor 2:

E_stored = Q₂² /(2C₂) = (54.0 mC)² /(2 (8.20 mF)) = 178 x 10^-6 J = 1.78 x 10^-4 J

Energy stored in capacitor 1:

E_stored = Q₁² /(2C₁) = (54.0 mC)² /(2 (22.0 mF)) = 66.3 x 10^-6 J = 6.63 x 10^-5 J