A 170-pF capacitor and a 680-pF capacitor are both charged to 1.90 kV. They are

Question

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 (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

a) the charge after applied 1.90kV is: Q170_before = 170*10-12 F*1.90*103 V = 323 nC Q680_before = 680*10-12 F*1.90*103 V = 1292 nC after their disconnected and connected in series, wehave: total charge Q = Q170_before+Q680_before = 323nC+1292nC = 1615nC since their connected, now their charge is equal tothe total change.   Q170_after =Q680_after = Q = 1615nC V170pf =  Q170_after/170pF =1615nC/170pF = 9.5*103V = 9.5kV V680pf =  Q680_after/680pF =1615nC/680pF = 2.375*103 V = 2.375kV

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