Two capacitors, C 1 = 19.0 µF and C 2 = 42.0 µF, are connected in series, and a

Question

Two capacitors, C1 = 19.0 µF and C2 = 42.0 µF, are connected in series, and a 24.0-V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.

(b) Find the energy stored in each individual capacitor.

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?

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(c) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)?
V

Which capacitor stores more energy in this situation?

Explanation / Answer

1.

Cequ = [19.0*42.0/ (19.0+42.0)] µF =13.08 µF

Energy stored U= 0.5 Cequ V^2 = 0.5 * 13.08*10^-6 * 24.0^2

U=3767.04 *10^-6 J

b.

Q=Cequ V=13.08*10^-6* 24.0 = 313.92 µC

U1= Q^2/ 2C1 = (313.93*10^-6) ^2 / (2*19.0*10^-6) = 2593.47*10^-6 J

U1= Q^2/ 2C2 = (313.93*10^-6) ^2 / (2*42.0*10^-6) =1173.23*10^-6 J

U=U1+U2 = 2593.47*10^-6 J+1173.23*10^-6 J

U=3766.69*10^-6 J

It must be true always as per energy is conserved

C.

IN parallel C= c1+c2 = (19.0+42.0)*10^-6 =61.0 *10^-6 F

U=0.5 C V^2

3766.69*10^-6 = 0.5 * 61.0*10^-6 * V^2

V=123.49 V

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