Consider the system of capacitors (C1 = 6.00 µF, C2 = 6.00 µF)First row and (C3=

Question

Consider the system of capacitors (C1 = 6.00 µF, C2 = 6.00 µF)First row and (C3=2F, C4=8F)Second row. The battery has 90V +-. I have a and b parts answered.

(a) Find the equivalent capacitance of the system.
4.6 µF

(b) Find the charge on each capacitor.
270 µC (on C1)
144 µC (on C2)
270 µC (on the 6.00 µF capacitor)
144 µC (on the 2.00 µF capacitor)

(c) Find the potential difference across each capacitor.
? V (across C1)
? V (across C2)
? V (across the 6.00 µF capacitor)
? V (across the 2.00 µF capacitor)

(d) Find the total energy stored by the group.
? mJ

Explanation / Answer

(c) The potential difference across each capacitor           V1 = Q1/C1 = 270*10^-6 C/ 6.0*10^-6 F                                = 45V          V2 = Q2/C2 = 270*10^-6 C/ 6.0*10^-6 F                                = 45V          V3 = Q3 / C3 = 144*10^-6 C/2.0*10^-6 F                                = 72V           V4 = Q4/C4 = 144*10^-6 / 8*10^-6                                = 18 V (d) The energy strored in the system              U = 1/2 CV^2                  = (0.5) (4.6*10^-6F) (90V)^2 = 18.6 *10^-3 J   or 18.6 mJ                                = 45V          V3 = Q3 / C3 = 144*10^-6 C/2.0*10^-6 F                                = 72V           V4 = Q4/C4 = 144*10^-6 / 8*10^-6                                = 18 V (d) The energy strored in the system              U = 1/2 CV^2                  = (0.5) (4.6*10^-6F) (90V)^2 = 18.6 *10^-3 J   or 18.6 mJ                                = 72V           V4 = Q4/C4 = 144*10^-6 / 8*10^-6                                = 18 V (d) The energy strored in the system              U = 1/2 CV^2                  = (0.5) (4.6*10^-6F) (90V)^2 = 18.6 *10^-3 J   or 18.6 mJ

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