A 80-pF capacitor and a 320-pF capacitor are both charged to 1.70 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. kV V80 pF = V320 pF= kV (b) Find the energy lost when the connections are made.

A 80-pF capacitor and a 320-pF capacitor are both charged to 1.70 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. kV V80 pF = V320 pF= kV (b) Find the energy lost when the connections are made.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter26: Capacitance And Dielectrics

Section: Chapter Questions

Problem 26.14P: What If? The two capacitors of Problem 13 (C1 = 5.00 F and C2 = 12.0 F) are now connected in series...

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