Two capacitors are connected to a battery. The battery voltage is V=60.0 V, and

ID: 1540319 • Letter: T

Question

Two capacitors are connected to a battery. The battery voltage is V=60.0 V, and the capacitances are C1=2.00 F and C2=4.00 F.

a) Determine the energy stored in each of them when they are wired in parallel.

b)Determine the energy stored in each of them when they are wired in series.

c)For each case (a) and (b) above, show that the sum of energies stored in the capacitors is equal to the energy stored in the equivalent capacitor.

** I got 1.08 * 10-2 J for a) and 2.4 *10-3 J for b), but I don't understand c)**

Please help! Thanks in advance :)

V = Voltage across each = 60 volts

E1 = energy stored in C1 = (0.5) C1 V2 = (0.5) (2 x 10-6) (60)2 = 0.0036 J

E2 = energy stored in C2 = (0.5) C2 V2 = (0.5) (4 x 10-6) (60)2 = 0.0072 J

Ep = total energy stored in parallel = 0.0108 J

Cs = series combination of C1 and C2 = C1 C2/(C1 + C2) = 2 x 4 /(2 + 4) = 1.33 uF

V = Voltage across the combination = 60 volts

Q = charge through each capacitor as charge remains same in series = Cs V = (1.33 x 10-6) (60) = 79.8 x 10-6

E'1 = energy stored in C1 = (0.5) Q2/C1 = (0.5) (79.8 x 10-6)2 /(2 x 10-6) = 0.001592 J

E'2 = energy stored in C2 = (0.5) Q2/C2 = (0.5) (79.8 x 10-6)2 /(4 x 10-6) = 0.000796 J

Es = total energy stored in series = 0.001592 + 0.000796 J = 0.0024 J

part c)

Cp = parallel combination = 2 + 4 = 6 uF

Ep = energy stored in parallel combination = (0.5) Cp V2 = (0.5) (6 x 10-6) (60)2 = 0.0108 J

Cs = parallel combination = 1.33 uF

Es = energy stored in series combination = (0.5) Cs V2 = (0.5) (1.33 x 10-6) (60)2 = 0.024 J

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