Two capacitors, C 1 = 16.0 µF and C 2 = 45.0 µF, are connected in series, and a

Question

Two capacitors, C1 = 16.0 µF and C2 = 45.0 µF, are connected in series, and a 3.0-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 (J).

(d) Show that the sum of these two energies is the same as the energy found in part (b).

(e) Will this equality always be true, or does it depend on the number of capacitors and their capacitances?

(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)?

(g) Which capacitor stores more energy in this situation, C1 or C2?

Explanation / Answer

given C1 = 16.0 µF and C2 = 45.0 µF

Ceq = (C1*C2) / (C1+C2) = (11.803) µF
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B
Ueq = 0.5*Ceq*V^2 = 0.5*11.803 *3*3 =53 µJ
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C
The capacitors are in series connection. So, voltage across them would be different, but the charge on each would be same. So, C1/C2=V2/V1=16/45 (From C=Q*V)
So,
=>V2/(V-V1)=16/45
=>V2=0.786885
=>V1=V- V2 = 2.21311

So, energy in C1 is (1/2)*(C1)*(V1*V1)= 39.182 µJ
energy in C2 = (1/2)*(C2)*(V2*v2) = 13.931 µJ  
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D) sum of energies = 53.1137 which is same as B)

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E) The equality is independant of number of capacitances as it follows law of conservation of energy.

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F) If connected in parallel the net capacitance is 16 + 45 µF = 61 µF

Ueq = 0.5*Ceq*V^2 = 53 µJ

V= 1.31822 volts

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G) In this situation as V is constant C2 stores more energy as energy is proportional to C.

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