Six capacitors are arranged in a circuit with a DC power supply as shown in the

Question

Six capacitors are arranged in a circuit with a DC power supply as shown in the diagram below. The power supply provides 25 volts, and the capacitors have the following values; C_1 = 75 mu F, C_2 = 50 mu F, C_3 = 40 mu F, C_4 = 30 mu F, C_5 = 60 mu F, and C_4 = 90 mu F. Each of the capacitor has a parallel plate shape (2 flat square plates separated by a dielectric filled gap). What is the total (equivalent) capacitance (mu F) of this arrangement of capacitors? What is the potential energy stored in capacitor mu F? Describe three things that you could do with the given parallel plate design to increase the capacitance of C_1, and explain why each change would increase its capacitance.

Explanation / Answer

(A) c4 ans C6 are in parallel.

C = C4 + C6 = 120 uF

now C, C3 and C5 are in series.

1/C' = 1/120 + 1/40 + 1/60

C' = 20 uF

C1 and C2 are in series.

1/C" = 1/75 + 1/50

C" = 30 uF

now C' and C" are in parallel.

Ceq = C' + C" = 50 uF

(B) charge on Ceq = 50uF x 25 V = 1250 uC

charge on C' = 1250 x 20 / (20 + 30) = 500 uC

charge on C4 = 500 x 30 / (30 + 90) = 125 uC

U = Q^2 / 2 C

= (125 x 10^-6)^2 / (2 x 30 x 10^-6)

= 2.604 x 10^-4 J

(C) C = e0 er A / d

- use metrial having greater dielectric constant

- using plate that have more surface area

- decreasing distance between plates

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