Two capacitors, C_1 = 25.0 mu F and C_2 = 44.0 mu F, are connected in series, an

Question

Two capacitors, C_1 = 25.0 mu F and C_2 = 44.0 mu F, are connected in series, and a 6.00-V battery is connected across the two capacitors. Find the equivalent capacitance. mu F Find the energy stored in this equivalent capacitance. Find the energy stored in each individual capacitor. capacitor 1 J capacitor 2 J 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. 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. 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) Ceq in series = 25*44/(25+44) = 15.94

b) Energy stored = 1/2 * Ceq *V^2 = (15.94*10^-6*6^2 )/2 = 2.87*10^-4 J

c) Vdrop in first capacitor = (C2/(C1+C2))*V = (44/69)*6 = 3.83

Energy stored in first capacitor= 1/2 C1 Vdrop^2 = 1.83*10^-4 J

Energy stored in second capacitor= 1/2 C2 Vdrop^2 = 1/2 *44*10^-6 *(6-3.83)^2 = 1.04*10^-4 J

d) Total energy stored in 2 resistor =( 1.83+1.04)*10^-4 = 2.87*10^-4 J

e) This equation always holds true if connected in series and do not depend on number of capacitor

f) Equivalent capacitance in parallel = 25+44 =69

1/2 Ceq V^2 = 2.87*10^-4

V= 2.88 V

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