Two capacitors, one that has a capacitance of 8 muF and one that has a capacitan

Question

Two capacitors, one that has a capacitance of 8 muF and one that has a capacitance of 24 muF are first discharged and then are connected in series. The series combination is then connected across the terminals of a 10-V battery. Next, they are carefully disconnected so that they are not discharged and they are then reconnected to each other--positive plate to positive plate and negative plate to negative plate. (a) Find the potential difference across each capacitor after they are reconnected. (b) Find the energy stored in the capacitor before they are disconnected from the battery, and find the energy stored after they are reconnected

Explanation / Answer

when they are in series

Net capacitnace is Cnet = C1*C2/(C1+C2) = (8*24)/(8+24) = 6 uF

net charge on ech capacitor is same Qnet = Cnet*V = 6*10^-6*10 = 60 uC

a) potential difference across 8uF is V1 = Q/C

Q = Q1+Q2 = 60+60 = 120 uC

and C = C1+C2 = 8+24 = 32 uF

V1= 120uC/32uF = 3.75 V

across 24 uF capacitor is also 3.75 V

b) energy stored is U1 = 0.5*C*V^2 = 0.5*6*10^-6*10^2 = 3*10^-4 J

reconnected

U2 = Q^2/(2*C) = 120^2*10^-12/(2*32*10^-6) = 225*10^-6 J = 2.25*10^-4 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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