Question

Two capacitors, C₁ = 15.0 µF and C₂ = 44.0 µF, are connected in series, and a 18.0-V battery is connected across them.

(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.

equivalent capacitance ________uF

total energy stored ________J

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

energy stored in C1 ______J

energy stored in C2 ______J

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?

(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

Answer 1

remember these about Combination of capacitors

if C1 and C2 are the individual capacitor that re connected in series,

then their resultant capaciatnce 1/Cnet = 1/C1+1/C2 or Cnet = C1C2/(C1+C2)

if c1 and C2 are connected in parallel, then their effective capacitancen is Cnet = C1+C2

also in sereis combination same amount of charge passes through each of the capacitiors

and in parallel combination, same PD (voltage) exists across each of the capacitor

so here

a. 1/Cs = (1/15) + (1/44)   

Cs = 11.18 uF

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Total energy Stored U = 0.5 CV^2

U = 0.5 * 11.18 e -6 * 18*18

U = 1.811 mJ ----------------------1

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in series combination, Charge reamins same AND VOLATGE DEVIDES

as Q = CV

C1v1 = C2V2

V2/V1 = C1/C2 = 15/44 = 0.349

V2 = 0.341 V1

but V1 + V2 = 18

1.341 V1 = 18

V1 = 18/1.341

V1 = 13.42 V

V2 = 4.57 V

energy stored U1 = 0.5* 15e -6 *13.41 *13.41 = 1.348 mJ

energy stored U2 = 0.5* 44e -6 * 4.57*4.57 = 0.459 mJ

from 1
total enenry = 1.811 mJ = 1.348 + 0.459 = 1.811 mJ

no this is not always true

as in paralell it changes

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c, Cnet in parallel = C1 + C2 = 15+44 = 59 uF

V in parallel remains same = V1 = V2 = V = 18 V