Two capacitors, C 1 = 23.0 µF and C 2 = 37.0 µF, are connected in series, and a

Question

Two capacitors, C1 = 23.0 µF and C2 = 37.0 µF, are connected in series, and a 6.0-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.

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

(e) 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

Explanation / Answer

(a)
Equivalent capacitance,
Ceq = (23.0*37.0)/(23.0+37.0) uF
Ceq = 14.18 uF

(b)
Energy Stored, U = 1/2*CV^2
U =  1/2 * 14.18 * 10^-6 * 6.0^2 J
U = 2.55*10^-4 J

(c)
Energy stored in each individual capacitor,
Q = Ceq * V
Q = 14.18 * 6.0 = 85.08 uC

Charge remains same on each capacitor in the series circut.

V1 = Q/C1 = 85.08/23.0 = 3.70 V
V2 = Q/C2 = 95.08/37.0 = 2.30 V

U1 = 1/2*C1V1^2 = 1/2 * 23.0 * 10^-6 * 3.70^2 J
U1 = 1.57 * 10^-4 J

U2 =  1/2*C2V2^2 = 1/2 * 37.0 * 10^-6 * 2.30^2 J
U2 = 0.978 * 10^-4 J

(d)
U1 + U2 = (1.57 + 0.978) * 10^-4 J
U1 + U2 = 2.55 * 10^-4 J

Hence proved !!

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