Two capacitors, C1 = 25.0 µF and C2 = 44.0 µF, are connected in series, and a 15

Question

Two capacitors, C1 = 25.0 µF and C2 = 44.0 µF, are connected in series, and a 15.0-V battery is connected across the two capacitors. (a) Find the equivalent capacitance. µF (b) Find the energy stored in this equivalent capacitance. J (c) Find the energy stored in each individual capacitor. capacitor 1 J capacitor 2 J (d) Show that the sum of these two energies is the same as the energy found in part (b). This answer has not been graded yet. (e) Will this equality always be true, or does it depend on the number of capacitors and their capacitances? This equality will always be true. This equality depends on the number of capacitors. (f) 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 (g) Which capacitor stores more energy in this situation, C1 or C2? C1 C2 The capacitors store the same amount of energy.

Explanation / Answer

A)
Ceq = (C1*C2) / (C1+C2) = 15.94 µF
B)
Ueq = 0.5*Ceq*V^2 = 0.5*1.594e-5 *15*15 =1.79e-3 J
C)
Q is the same for each capacitance
Q = Ceq*V = 1.594e-5*15 = 23.91e-5C

Energy stored in C1= 0.5Q^2/C1 = 0.5*23.91e-5* 23.91e-5/ 25e-6 =11.43e-4 J
Energy stored in C2= 0.5Q^2/C2 = 0.5*23.91e-5* 23.91e-5/ 44e-6 =6.49e-4 J

D) U1+U2 = 1.143e-3 + 0.649e-3 = 1.79e-3 = Ueq
E) This equality will always be true
F) If they were connected in parallel
V is the same for each capacitance
Ceq = C1+C2 = 69e-6 F
Ueq = 0.5*Ceq*V^2
=> V = 2Ueq/Ceq = 2*1.79e-3/6.9e-5
=> V = 7.2 V
G) The capacitor will store the same amount of energy.

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