QuestionDetails: A 170-pF capacitor and a 680-pF capacitor are both charged to 1

Question

QuestionDetails: A 170-pF capacitor and a 680-pF capacitor are both charged to 1.90 kV. They are then disconnected from the voltagesource and are connected together, positive plate to negative plateand negative plate to positive plate. (a) Find the resulting potential differenceacross each capacitor. V170 pF = 1 kV V680 pF = 2 kV
(b) Find the energy lost when the connections are made.
3 µJ QuestionDetails: (a) Find the resulting potential differenceacross each capacitor. V170 pF = 1 kV V680 pF = 2 kV
(b) Find the energy lost when the connections are made.
3 µJ V170 pF = 1 kV V680 pF = 2 kV

Explanation / Answer

potential across eachcapacitor=v=[C1V1-C2V2]/C1+C2=[680*1900-170*1900]/680+170 =                                                    = 1140V=1.140KV ENERGY LOST=   1/2[C1C2/C1+C2](V1+V2)2=981920000J

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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