Two capacitors, C1 = 22.0 F and C2 = 41.0 F, are connected in series, and a 24.0

Question

Two capacitors, C1 = 22.0 F and C2 = 41.0 F, are connected in series, and a 24.0-V battery is connected across them. Find the equivalent capacitance, and the energy contained in this equivalent capacitor. equivalent capacitance F total energy stored J Find the energy stored in each individual capacitor. energy stored in C1 J energy stored in C2 J Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances?

Explanation / Answer

A. Ceq = 22(41) / [22 + 41] = 14.317 uF

B. Ueq = 1/2 (14.317 uF) (24*24) = 4.123 x 10^-9 J

C. Qeq = Ceq / V = 14.317 / 24 = 0.596 uC = Q1 = Q2

so for C1 : U1 = 1/2 (Q1^2 / C1) = 1/2 (0.356 x 10^-12) / 22x10^-6 = 0.016 x 10^-6 J

U2 = 1/2 (Q2^2 / C2) = 1/2 (0.356 x 10^-12) / 41x10^-6 = 0.0087 x 10^-6 J

D) U2 = 0.5 (41x10^-6)(24*24) = 11.81 x 10^-3 J

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