Two capacitors, C_1 = 16.0 mu F and C_2 = 35.0 mu F, are connected in series, an

Question

Two capacitors, C_1 = 16.0 mu F and C_2 = 35.0 mu F, are connected in series, and a 21.0-V battery is connected across them. (a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor. equivalent capacitance mu F total energy stored J (b) Find the energy stored in each individual capacitor. energy stored in C_1 J energy stored in C_2 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 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)
Ceq = (C1*C2) / (C1+C2) = (800/60) µF = 13.3 F

Ueq = 0.5*Ceq*V^2 = 0.5*13.3 F *21*21 =2.93e-3 J

B)

Q is the same for each capacitance
Q = Ceq*V = 13.3 F *21 = 2.79e-4C

Energy stored in C1= 0.5Q^2/C1 = 0.5*2.79e-4* 2.79e-4/ 16e-6 =0.002432 J
Energy stored in C2 = 0.5Q^2/C2

= 0.5*2.79e-4* 2.79e-4/ 35e-6

                             = 0.00111 J

C)

The potential difference of the capacitor is = 21.0V.

Capacitor C1 is store 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.