The capacitors, C_1 = 19.0 mu F and C_2 = 31.0 mu F, are connected in series, an

Question

The capacitors, C_1 = 19.0 mu F and C_2 = 31.0 mu F, are connected in series, and a 15.0 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)? Which capacitor stores more energy in this situation, C_1 or C_2? C_1 C_2 Both C_1 and C_2 store the same amount of energy.

Explanation / Answer

a)

1/c = (1/19) + (1/31)

c = 11.78 uF

Energy stored = (1/2)*c*v^2

= 0.5 x 11.78 x 10^-6 x (18)^2

= 1.9 x 10^-3 J

b)

q = cv

q = 11.78 x 10^-6 x 18

q = 2.12 x 10^-4 C

voltage across 19 microf cap

v = q/c

v = (2.12 x 10^-4) /(19 x 10^-6)

v = 11.16 v

Energy stored in this cap = 0.5*19*10^-6*11.16^2 = 1.11 x 10^-3 joules

voltage across 31 v

v = (2.12 x 10^-4) / (31 x 10^-6)

v = 6.84v

Energy stored in this cap = 0.5*31*10^-6*6.84^2 = 7.24 x 10^-4 J

c)

Ceq = Sum of individual capacitances.

energy = (1/2) Ceq V^2 : use energy computed in (b) and Ceq computed to solve for V.

Larger capacitance stores more energy.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.