Question

For the system of four capacitors shown in the figure below, find the following. (Use C1 = 1.0

Answer 1

C1 + C2 = ((1 / 1 muF) + ( 1 / 7muF))^-1

C1 + C2 = 0.875 muF

C3 + C4 = ((1 / 5 muF) + (1 / 6 muF))^-1

C3 + C4 = 2.73 muF

Ceq = 0.875 muF + 2.73 muF = 3.61 muF

Energy = (1/2) (3.61 muF) (90V)²

= 14.6 mJ

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v = q / C

Each branch of the parallel combination (C1 + C2 and C3 + C4) willhave a certain charge associated with them and each branch willhave the same voltage applied, 90V, since they are inparallel.

For the C1 + C2 branch we have v = q / C₁₊₂

90 = q / 0.875 muF

q = 7.875x10^-5 C

C1 => V = q / C = (7.875x10^-5 C) /(1 muF) = 78.75V

C2 => V = q / C = (7.875x10^-5 C) /(7 muF) = 11.25 V

v = q / C₃₊₄

90 = q / 2.73 muF

q = 2.457 x 10^-4 C

C3 => V = q / C = (2.457 x 10^-4) /(5 muF) = 49.14 V

C4 => V = q / C = (2.457 x 10^-4) /(6 muF) = 40.95V

C1 = (1/2) (1 muF) (78.75 V)² = 3.1 mJ

C2 = (1/2) (7 muF) (11.25 V)² = 0.44 mJ

C3 = (1/2) (5 muF) (49.14V)² = 6 mJ

C4 = (1/2) (6 muF) (40.95V)² = 5 mJ

the sum of the individual energies = 3.1 + 0.443 + 6+ 5

= 14.54 mJ.