Two capacitors, C 1 = 16.0 ? F and C 2 = 38.0 ? F, are connected in series, and

Question

Two capacitors,

C1 = 16.0 ?F

and

C2 = 38.0 ?F,

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.

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

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?

This answer has not been graded yet.

(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

Explanation / Answer

C1 = 16 ?F

C2 = 38 ?F

Connected in series

Voltage , V = 18 V

a) Equivalent capacitance = C1C2/(C1+C2) = 16*38/(16+38) = 11.259 ?F

Energy stored in equivalnet capacitor = 0.5 C V^2 = 0.5 * (11.259 x 10^-6 ) *18^2 = 0.001824 J

b) Voltage across C1 = V1 = (C2/(C1+C2))*V = (38/(16+38))*18 = 12.67 V

Energy stored in C1 = 0.5 C1 V1^2 = 0.5 *(16*10^-6)*12.67^2 = 0.001284 J

Voltage across C2 = V2 = V - V1 = 18 - 12.67 = 5.33 V

Energy stored in C2 = 0.5 C2 V2^2 = 0.5 * (38*10^-6)*5.33^2 = 0.00054 J

Energy stored in C1 + Energy stored in C2 = 0.001284 +  0.00054 = 0.001824 J , which is equal to the value calculated in Part A . , It doesn't depend on the number of capacitors and their capacitances. This equality is always true

c) Capacitors in parallel, Equivalent capacitance = 16 + 38 = 54 ?F

Energy stored = E = 0.5 CV^2

V = sqrt(2*E/C) = sqrt(2* 0.001824 / (54*10^-6)) = 8.22 V

Potential difference required = 8.22 V

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