Two capacitors, C1 = 19.0 µF and C2 = 39.0 µF, are connected in series, and a 18

Question

Two capacitors, C1 = 19.0 µF and C2 = 39.0 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 ( µF)
total energy stored (J)

(b) Find the energy stored in each individual capacitor. energy stored in C1 3 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)?
6 V

Explanation / Answer

HELLO!! Q = CV energy = (1/2) C V^2 a) In series, charge on each capacitor must be the same. Q = C1 V1 = C2 V2 V1 + V2 = V Two equations, two unknowns (V1,V2) Equivalence capacitance : Ceq = Q/V = C1 V1/V = C2 V2 / V b) energy_j = (1/2) Cj (Vj)^2; j = {1,2} Total energy = (1/2) [C1 V1^2 + C2 V2^2] = (1/2) Ceq V^2 Always true because charge must be conserved. c) Ceq = Sum of individual capacitances. energy = (1/2) Ceq V^2 : use energy computed in (b) and Ceq computed in (c) to solve for V. Larger capacitance stores more energy. Hope this helps! Have a nice day!!

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