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

The answers to 1 and 2 are the same, since they are connected(they will have the same potential). . The idea is that charge is conserved, so first find theinitial charge on each cap: .    q = CV =   170 * 1.9 = 323 . and            680 * 1.9   = 1292 . These are connected neg to pos, so the net charge on the twocaps after they are connected is .        1292 - 323 =   969 . This charge is split proportionally between the two caps, sothe charge on the 170 cap is .         969 * 170 / (170 +680) =    193.8 . And finally, the potential across the 170 cap is .    V = Q / C = 193.8 / 170 =     1.14  kV     (this is also the potential across the680 cap) . So the answer to 1 and 2   is    1.14 kV . #3...

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